NCERT Solutions Class 10 maths Chapter 6

Q.1 Fill in the blanks using the correct word given in brackets:
(i) All circles are ­­­­­­­­_______. (congruent, similar)
(ii) All squares are______. (similar, congruent)
(iii) All _______triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are _______and
(b) their corresponding sides are_______. (equal, proportional)

Ans.

(i) All circles are­­­­­­ similar.
(ii) All squares are ­­­­­­similar.
(iii) All equilateral triangles are similar.
(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

Q.2 Give two different examples of pair of
(i) similar figures. (ii) non-similar figures.

Ans.

(i)
(a) Any two circles are similar.
(b) Any two squares are similar.

(ii)
(a) A trapezium and a parallelogram are not similar.
(b) An acute angle triangle and an obtuse angle triangle are not similar.

Q.3 State whether the following quadrilaterals are similar or not:

Ans.

Quadrilaterals PQRS and ABCD are not similar as their corresponding angles are not equal.

Q.4 In the following figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Ans.

(i)

$\begin{array}{l}\text{It is given that}\mathrm{DE}\parallel \mathrm{BC}.\\ \text{By using Basic Proportionality Theorem, we get}\\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\\ \Rightarrow \frac{1.5}{3}=\frac{1}{\mathrm{EC}}\\ \Rightarrow \mathrm{EC}=\frac{3\times 1}{1.5}=2\\ \therefore \mathrm{EC}=2\text{cm}\end{array}$

(ii)

$\begin{array}{l}\text{It is given that}\mathrm{DE}\left|\right|\mathrm{BC}.\\ \text{By using Basic Proportionality Theorem, we get}\\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\\ \Rightarrow \frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}=\frac{1}{3}\\ \Rightarrow \mathrm{AD}=\frac{7.2}{3}=2.4\\ \therefore \mathrm{AD}=2.4\text{cm}\end{array}$

Q.5

$\begin{array}{l}\text{E and F are points on the sides PQ and PR respectively of a}\mathrm{\Delta }\text{PQR. For each of the following}\\ \text{cases, state whether EF}\parallel \text{QR:}\\ \text{(i)}\mathrm{PE}=3.9\text{cm,}\mathrm{EQ}=3\text{cm,}\mathrm{PF}=3.6\text{cm and}\mathrm{FR}=2.4\text{cm}\\ \text{(ii)}\mathrm{PE}=4\text{cm,}\mathrm{QE}=4.5\text{cm,}\mathrm{PF}=8\text{cm and}\mathrm{RF}=9\text{cm}\\ \text{(iii)}\mathrm{PQ}=1.28\text{cm,}\mathrm{PR}=2.56\text{cm,}\mathrm{PE}=0.18\text{cm and}\mathrm{PF}=0.36\text{cm}\end{array}$

Ans.

(i)
As per given information, we have the following triangle.

$\begin{array}{l}\text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{3.9}{3}=1.3\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{3.6}{2.4}=\frac{3}{2}=1.5\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}\ne \frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{does not divide}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is not parallel to}\mathrm{QR}.\end{array}$

(ii)
As per given information, we have the following triangle.

$\begin{array}{l}\text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{4}{4.5}=\frac{40}{45}=\frac{8}{9}\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{8}{9}\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{divides}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is parallel to}\mathrm{QR}.\end{array}$

(iii)
As per given information, we have the following triangle.

$\begin{array}{l}\left(\mathrm{iv}\right)\\ \text{In the above triangle},\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{0.18}{1.1}=\frac{18}{110}=\frac{9}{55}\\ \text{and}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{0.36}{2.2}=\frac{36}{220}=\frac{9}{55}\\ \text{Hence,}\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{i.e., the line}\mathrm{EF}\text{divides}\mathrm{PQ}\text{and}\mathrm{PR}\text{in}\\ \text{the same ratio and so}\mathrm{EF}\text{is parallel to}\mathrm{QR}.\end{array}$

Q.6

$\begin{array}{l}\text{In the following figure, if}\mathrm{LM}\parallel \mathrm{CB}\text{and}\mathrm{LN}\parallel \mathrm{CD},\text{prove that}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}.\end{array}$

Ans.

$\begin{array}{l}\text{In the given figure,}\mathrm{LM}\parallel \mathrm{CB}\\ \text{So, using basic proportionality theorem in}\mathrm{\Delta }\text{ABC, we get}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AL}}{\mathrm{AC}}...\text{(1)}\\ \text{Also, in the given figure,}\mathrm{LN}\parallel \mathrm{CD}\\ \text{So, using basic proportionality theorem in}\mathrm{\Delta }\text{ACD, we get}\\ \frac{\mathrm{AN}}{\mathrm{AD}}=\frac{\mathrm{AL}}{\mathrm{AC}}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}.\end{array}$

Q.7

$\begin{array}{l}\text{In the following figure,}\mathrm{DE}\parallel \mathrm{AC}\text{and}\mathrm{DF}\parallel \mathrm{AE}.\\ \text{Prove that}\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}.\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABC,}\mathrm{DE}\parallel \mathrm{AC}\\ \text{So, using basic proportionality theorem, we get}\\ \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BE}}{\mathrm{EC}}...\text{(1)}\\ \text{In}\mathrm{\Delta }\text{BAE,}\mathrm{DF}\parallel \mathrm{AE}\\ \text{So, using basic proportionality theorem, we get}\\ \frac{\mathrm{BD}}{\mathrm{DA}}=\frac{\mathrm{BF}}{\mathrm{FE}}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\end{array}$

Q.8

$\text{In the following figure,}\mathrm{DE}\parallel \mathrm{OQ}\text{and}\mathrm{DF}\parallel \mathrm{OR}.\text{\hspace{0.17em}Show that}\mathrm{EF}\parallel \mathrm{QR}.$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{POQ,}\mathrm{DE}\parallel \mathrm{OQ}\\ \text{So, using basic proportionality theorem, we get}\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}}...\text{(1)}\\ \text{In}\mathrm{\Delta }\text{POR,}\mathrm{DF}\parallel \mathrm{OR}\\ \text{So, using basic proportionality theorem, we get}\\ \frac{\mathrm{PF}}{\mathrm{FR}}=\frac{\mathrm{PD}}{\mathrm{DO}}...\text{(2)}\\ \text{From (1) and (2), we get}\\ \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\\ \text{Hence, by converse of basic proportionality theorem,}\mathrm{EF}\parallel \mathrm{QR}.\end{array}$

Q.9 In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB ∥ PQ and AC ∥ PR.Show that BC ∥ QR.

Ans.

$\begin{array}{l}\text{In ΔPOQ, AB\hspace{0.33em}ll\hspace{0.33em}PQ (Given)}\\ \text{Therefore, by Basic Proportionality Theorem,}\\ \frac{\text{AP}}{\text{AO}}\text{=}\frac{\text{OB}}{\text{BQ}}...\text{(i)}\\ \text{In ΔPOR, AC\hspace{0.33em}ll\hspace{0.33em}PR (Given)}\\ \text{Therefore, by Basic Proportionality Theorem,}\\ \frac{\text{AP}}{\text{AO}}\text{=}\frac{\text{OC}}{\text{CR}}...\text{(ii)}\\ \text{From (i) and (ii),}\\ \frac{\text{OB}}{\text{BQ}}\text{=}\frac{\text{OC}}{\text{CR}}\\ \text{Hence, by using Converse of Basic Proportionality Theorem}\\ \text{in ΔOQR, we find that BC\hspace{0.33em}ll\hspace{0.33em}QR.}\end{array}$

Q.10 Using Basic Proportionality Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Ans.

We consider ΔABC drawn below in which D is the mid-point of side AB and DE is the line segment drawn parallel to BC.

$\begin{array}{l}\text{By using Basic Proportionality Theorem, we get}\\ \frac{\text{AD}}{\text{DB}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}...\text{(1)}\\ \text{D is the mid-point of AB. Therefore, AD=DB}\\ \text{From (1), we have}\\ \frac{\text{AD}}{\text{DB}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\\ \Rightarrow \frac{\text{AD}}{\text{AD}}\text{\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\\ \Rightarrow \text{1\hspace{0.33em}=\hspace{0.33em}}\frac{\text{AE}}{\text{EC}}\\ \mathrm{\Rightarrow }\text{EC\hspace{0.33em}=\hspace{0.33em}AE}\\ \text{Therefore, E is the mid-point of AC.}\end{array}$

Q.11 Using converse of Basic Proportionality Theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
(Recall that you have done it in Class IX)

Ans.

We consider ΔABC drawn below in which DE is the line segment joining the mid-points D and E of sides AB and AC respectively.

$\begin{array}{l}\text{We have}\\ \mathrm{AD}=\mathrm{DB}\Rightarrow \frac{\mathrm{AD}}{\mathrm{DB}}=1\\ \text{and}\\ \mathrm{AE}=\mathrm{EC}\Rightarrow \frac{\mathrm{AE}}{\mathrm{EC}}=1\\ \text{Therefore,}\\ \frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\\ \text{Hence, by converse of Basic Proportionality Theorem,}\mathrm{DE}\parallel \mathrm{BC}.\end{array}$

Q.12

$\begin{array}{l}\mathrm{ABCD}\text{is a trapezium in which}\mathrm{AB}\parallel \mathrm{DC}\text{and its diagonals intersect each other at the point}\mathrm{O}.\\ \text{Show that}\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}.\end{array}$

Ans.

$\begin{array}{l}\text{Given}:\\ \text{ABCD is a trapezium in which AB is parallel to CD and diagonals intersect each other at O}.\end{array}$

$\begin{array}{l}\text{To prove}:\\ \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\\ \text{Construction}:\\ \text{Draw}\mathrm{OF}\parallel \mathrm{AB}\text{which meets AD in F.}\\ \text{In}\mathrm{\Delta ABD}\text{,}\mathrm{OF}\parallel \mathrm{AB}\\ \text{So, by converse of Basic\hspace{0.17em}\hspace{0.17em}Proportionality Theorem,}\\ \frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{DF}}{\mathrm{FA}}...\text{(i)}\\ \text{Now in}\mathrm{\Delta ADC},\mathrm{OF}\parallel \mathrm{CD}[\mathrm{S}\text{ince}\mathrm{AB}\parallel \mathrm{CD}]\\ \therefore \frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{OA}}...\text{(ii)}\\ \text{Comparing equation (i) and (ii), we get}\\ \frac{\mathrm{DO}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{OA}}\\ \Rightarrow \frac{\mathrm{OA}}{\mathrm{OB}}=\frac{\mathrm{CO}}{\mathrm{DO}}\end{array}$

Q.13 The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO = CODO. Show that ABCD is a trapezium.

Ans.

$\begin{array}{l}\text{Given}:\\ \text{Diagonals of a quadrilateral ABCD intersect each other at point O such that}\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}.\end{array}$

$\begin{array}{l}\text{To prove}:\\ \mathrm{AB}\parallel \text{CD}\\ \text{Construction}:\\ \text{Draw}\mathrm{OF}\parallel \mathrm{AB}\text{which meets AD in F.}\\ \text{In}\mathrm{\Delta ABD}\text{,}\mathrm{OF}\parallel \mathrm{AB}\\ \text{So, by converse of Basic\hspace{0.17em}\hspace{0.17em}Proportionality Theorem,}\\ \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{DF}}{\mathrm{FA}}...\text{(i)}\\ \mathrm{Given}\mathrm{that}\\ \frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\\ \Rightarrow \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{AO}}\\ \Rightarrow \frac{\mathrm{DF}}{\mathrm{FA}}=\frac{\mathrm{CO}}{\mathrm{AO}}\text{[From (i)]}\\ \Rightarrow \text{OF}\parallel \text{DC [By converse of Basic Proportionality Theorem]}\\ \text{Also, OF}\parallel \mathrm{AB}\\ \mathrm{Therefore},\mathrm{AB}\parallel \text{DC}\\ \mathrm{Hence},\text{ABCD is a trapezium.}\end{array}$

Q.14 State which pairs of triangles in the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form.

Ans.

$\begin{array}{l}\text{(i)}\\ \text{Corresponding angles are equal in the given triangles. i.e.,}\\ \angle \mathrm{A}=\angle \mathrm{P}=60\mathrm{°},\\ \angle \mathrm{B}=\angle \mathrm{Q}=80\mathrm{°}\\ \text{and}\\ \angle \mathrm{C}=\angle \mathrm{R}=40\mathrm{°}.\\ \text{Therefore, by AAA similarity criterion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}.\\ \text{(ii)}\\ \text{Ratios of the corresponding sides of the given pair of}\\ \text{triangles are equal.}\\ \text{i.e.,}\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{CA}}{\mathrm{PQ}}=\frac{1}{2}\\ \text{Therefore, by SSS similarity criterion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta QRP}.\\ \text{(iii)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \\ \text{(iv)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \text{(v)}\\ \text{The given pair of triangles are not similar as their}\\ \text{corresponding sides are not proportional.}\\ \text{(vi)}\\ \text{In}\mathrm{\Delta DEF},\\ \angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=180\mathrm{°}\\ \Rightarrow 70\mathrm{°}+80\mathrm{°}+\angle \mathrm{F}=180\mathrm{°}\\ \Rightarrow \angle \mathrm{F}=180\mathrm{°}-150\mathrm{°}=30\mathrm{°}\\ \text{In}\mathrm{\Delta PQR},\\ \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\mathrm{°}\\ \Rightarrow \angle \mathrm{P}+80\mathrm{°}+30\mathrm{°}=180\mathrm{°}\\ \Rightarrow \angle \mathrm{P}=180\mathrm{°}-110\mathrm{°}=70\mathrm{°}\\ \text{We find out that corresponding angles are equal in}\mathrm{\Delta DEF}\\ \text{and}\mathrm{\Delta PQR}.\text{i.e.,}\\ \angle \mathrm{D}=\angle \mathrm{P}=70\mathrm{°},\\ \angle \mathrm{E}=\angle \mathrm{Q}=80\mathrm{°}\\ \text{and}\\ \angle \mathrm{F}=\angle \mathrm{R}=30\mathrm{°}.\\ \text{Therefore, by AAA similarity criterion,}\mathrm{\Delta DEF}~\mathrm{\Delta PQR}.\end{array}$

Q.15

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Ans.

$\begin{array}{l}\mathrm{DOB}\text{is a straight line.}\\ \therefore \angle \text{BOC}+\angle \mathrm{D}\text{OC}=180\mathrm{°}\\ \Rightarrow \angle \text{DOC}=180\mathrm{°}-\angle \mathrm{B}\text{OC}=180\mathrm{°}-125\mathrm{°}=55\mathrm{°}\\ \text{In}\mathrm{\Delta }\text{DOC,}\\ \angle \mathrm{D}\text{OC}+\angle \text{DC}\mathrm{O}+\angle \text{O}\mathrm{D}\text{C}=180\mathrm{°}\\ \Rightarrow \angle \mathrm{D}\text{CO}=180\mathrm{°}-\angle \mathrm{D}\text{OC}-\angle \text{O}\mathrm{D}\text{C}\\ \Rightarrow \angle \mathrm{D}\text{CO}=180\mathrm{°}-55\mathrm{°}-70\mathrm{°}=55\mathrm{°}\\ \text{It is given that}\mathrm{\Delta }\text{ODC}~\mathrm{\Delta }\text{OBA and so corresponding angles}\\ \text{in these triangles are equal.}\\ \therefore \angle \text{OAB}=\angle \text{DCO}=55\mathrm{°}\end{array}$

Q.16

$\begin{array}{l}\text{Diagonals AC and BD of a trapezium ABCD with AB}\parallel \text{DC intersect each other at the point O. Using}\\ \text{a similarity criterion for two triangles, show that}\frac{\text{OA}}{\text{OC}}=\frac{\text{OB}}{\text{OD}}.\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{DOC and}\mathrm{\Delta }\text{BOA,}\\ \angle \text{CDO}=\angle \text{ABO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DCO}=\angle \text{BAO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DOC}=\angle \text{BOA [Vertically opposite angles]}\\ \therefore \mathrm{\Delta }\text{DOC}~\mathrm{\Delta }\text{BOA [AAA similarity critarion]}\\ \Rightarrow \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}}\text{[Corresponding sides are proportional]}\\ \Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\end{array}$

Q.17

\text{In the following figure,}\frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\text{and}\angle \text{1}=\angle 2.\text{Show that}\Delta \text{PQS}\sim \Delta \text{TQR}\text{.}

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{PQR,}\\ \angle \text{1}=\angle 2\text{(Given)}\\ \Rightarrow \text{PQ}=\text{PR}...\text{(1)}\\ \text{It is given that}\\ \frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{PR}}\\ \Rightarrow \frac{\text{QR}}{\text{QS}}=\frac{\text{QT}}{\text{QP}}\text{[From (1), PR}=\text{QP] \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{(2)}\\ \angle \text{Q is common in}\mathrm{\Delta }\text{PQS and}\mathrm{\Delta }\text{TQR and the sides including}\\ \text{this angle in both triangles are proportional as shown in}\\ \text{equation (2).}\\ \text{Therefore, by SAS critarion,}\mathrm{\Delta }\text{PQS}~\mathrm{\Delta }\text{TQR}.\end{array}$

Q.18

$\text{S and T are points on sides PR and QR of}\mathrm{\Delta }\text{PQR such that}\angle \text{P}=\angle \text{RTS. Show that}\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS.}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{RPQ and}\mathrm{\Delta }\text{RTS,}\\ \angle \mathrm{P}=\angle \mathrm{T}\text{(Given)}\\ \text{and}\angle \mathrm{R}=\angle \mathrm{R}\\ \text{Therefore, by AA criterion,}\mathrm{\Delta }\text{RPQ}~\mathrm{\Delta }\text{RTS}.\end{array}$

Q.19

$\text{In the following figure, if}\mathrm{\Delta }\text{ABE}\cong \mathrm{\Delta }\text{ACD, show that}\mathrm{\Delta }\text{ADE}~\mathrm{\Delta }\text{ABC.}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABE}\cong \mathrm{\Delta }\text{ACD.}\\ \text{Therefore, by CPCT,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}=\text{AC}\\ \text{and AD}=\text{AE}\\ \text{So,}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\\ \text{In}\mathrm{\Delta }\text{ADE and}\mathrm{\Delta }\text{ABC,}\\ \angle \mathrm{A}=\angle \mathrm{A}\\ \text{and}\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\\ \text{Therefore, by SAS criterion,}\mathrm{\Delta }\text{ADE}~\mathrm{\Delta }\text{ABC}.\end{array}$

Q.20

$\begin{array}{l}\text{In the following figure, altitudes AD and CE of}\mathrm{\Delta }\text{ABC intersect each other at the point P. Show that:}\\ \text{(i)}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{CDP}\\ \text{(ii)}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{CBE}\\ \text{(iii)}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{ADB}\\ \text{(iv)}\mathrm{\Delta }\text{PDC}~\mathrm{\Delta }\text{BEC}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{AEP and}\mathrm{\Delta }\text{CDP,}\\ \angle \mathrm{APE}=\angle \mathrm{CPD}\text{(Vertically opposite angles)}\\ \text{and}\angle \text{AEP}=\angle \text{CDP}=90\mathrm{°}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{CDP}.\\ \text{(ii)}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{CBE,}\\ \angle \text{ADB}=\angle \text{CEB}=90\mathrm{°}\\ \text{and}\angle \text{ABD}=\angle \text{CBE}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{CBE}.\\ \text{(iii)}\\ \text{In}\mathrm{\Delta }\text{AEP and}\mathrm{\Delta }\text{ADB,}\\ \angle \text{AEP}=\angle \text{ADB}=90\mathrm{°}\\ \text{and}\angle \text{PAE}=\angle \text{BAD}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{AEP}~\mathrm{\Delta }\text{ADB}.\\ \text{(iv)}\\ \text{In}\mathrm{\Delta }\text{PDC and}\mathrm{\Delta }\text{BEC,}\\ \angle \text{PDC}=\angle \text{BEC}=90\mathrm{°}\\ \text{and}\angle \text{DCP}=\angle \text{ECB}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{PDC}~\mathrm{\Delta }\text{BEC}.\end{array}$

Q.21

$\begin{array}{l}\text{E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD}\\ \text{at F. Show that}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB.}\end{array}$

Ans.

$\begin{array}{l}\text{In the above figure, ABCD is a parallelogram in which AB}\parallel \text{DC.}\\ \text{Opposite angles in parallelogram are equal. Therefore,}\\ \angle \text{EAB}=\angle \text{BCF}\\ \text{Also,}\\ \angle \text{ABE}=\angle \text{CFB [Alternate interior angles]}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{CFB}.\end{array}$

Q.22

$\begin{array}{l}\text{In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively.}\\ \text{Prove that:}\\ \text{(i)}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}\\ \text{(ii)}\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{AMP,}\\ \angle \text{ABC}=\angle \text{AMP}=90\mathrm{°}\\ \text{and,}\\ \angle \text{CAB}=\angle \text{PAM}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}.\\ \text{(ii)}\\ \text{Corresponding sides are proportional in similar triangles.}\\ \text{Therefore,}\\ \mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{AMP}\\ \Rightarrow \frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}\end{array}$

Q.23

$\begin{array}{l}\text{CD and GH are respectively the bisectors of}\mathrm{\Delta }\text{ACB and}\mathrm{\Delta }\text{EGF such that D and H lie on sides AB and FE}\\ \text{of}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{EFG respectively. If}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG, show that:}\\ \text{(i) \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\\ \text{(ii)\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ \text{(iii)\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{FEG.}\\ \angle \text{A}=\angle \text{F,}\angle \text{B}=\angle \text{E,}\angle \text{BCA}=\angle \text{EGF}\\ \text{Also,}\\ \frac{1}{2}\angle \text{BCA}=\frac{1}{2}\angle \text{EGF}\\ \Rightarrow \angle \text{ACD}=\angle \text{FGH and}\angle \text{DCB}=\angle \text{HGE}\\ \therefore \text{By AA similarity critarion,}\\ \mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{HGF and}\mathrm{\Delta }\text{DCB}~\mathrm{\Delta }\text{HGE}\\ \Rightarrow \frac{\text{CD}}{\text{GH}}=\frac{\text{AC}}{\text{FG}}\end{array}$

Q.24

\begin{array}{l}\text{In the following figure, E is a point on side CB produced of an isosceles triangle ABC}\\ \text{with AB}=\text{AC}\text{. If AD}\perp \text{BC and EF}\perp \text{AC, prove that}\Delta \text{ABD}\sim \Delta \text{ECF}\text{.}\end{array}

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{ECF,}\\ \angle \text{BDA}=\angle \text{CFE}=90\mathrm{°}\\ \text{and}\angle \text{ABD}=\angle \text{ECF [AB}=\text{AC}\Rightarrow \angle \text{B}=\angle \text{C]}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{ECF}.\end{array}$

Q.25

$\begin{array}{l}\text{Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and}\\ \text{median PM of}\mathrm{\Delta }\text{PQR (see the following figure). Show that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC}\\ \text{and}\\ \frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{It is given that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AD}}{\text{PM}}\\ \Rightarrow \frac{\text{AB}}{\text{PQ}}=\frac{\frac{1}{2}\text{BC}}{\frac{1}{2}\text{QR}}=\frac{\text{AD}}{\text{PM}}\\ \Rightarrow \frac{\text{AB}}{\text{PQ}}=\frac{\text{BD}}{\text{QM}}=\frac{\text{AD}}{\text{PM}}\\ \text{Therefore, by SSS similarity critarion,}\mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{PQM and so}\\ \text{in}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{PQR,}\angle \text{B}=\angle \text{Q and}\frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}.\\ \text{Therefore, by SAS similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Q.26

$\text{D is a point on the side BC of a triangle ABC such that}\angle \text{ADC}=\angle {\text{BAC. Show that CA}}^{\text{2}}=\text{CB}·\text{CD.}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{BAC and}\mathrm{\Delta }\text{ADC,}\\ \angle \text{ADC}=\angle \text{BAC (Given)}\\ \angle \text{ACD}=\angle \text{BCA (Common angle)}\\ \therefore \mathrm{\Delta }\text{ADC}~\mathrm{\Delta }\text{BAC (By AA similarity critarion)}\\ \text{We know that corresponding sides of similar triangles}\\ \text{are proportional.}\\ \therefore \frac{\text{CA}}{\text{CD}}=\frac{\text{CB}}{\text{CA}}\\ \Rightarrow {\text{\hspace{0.17em}\hspace{0.17em}CA}}^2=\text{CB}\cdot \text{CD}\end{array}$

Q.27

$\begin{array}{l}\text{Sides AB and AC and median AD of a triangle ABC}\\ \text{are respectively proportional to sides PQ and PR and}\\ \text{median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that,}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\\ \text{We extend AD and PM up to point E and L respectively,}\\ \text{such that AD}=\text{DE and PM}=\text{ML. We join B to E, C to E,}\\ \text{Q to L and R to L.}\end{array}$

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{In quadrilateral ABEC, diagonals AE and BC bisect each}\\ \text{other at point D. Therefore, quadrilateral ABEC is a}\\ \text{parallelogram.}\\ \therefore \text{AC}=\text{BE and AB}=\text{EC}\\ \text{Similarly, we can prove that quadrilateral PQLR is}\\ \text{a parallelogram and PR}=\text{QL, PQ}=\text{LR.}\\ \text{It is given that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}=\frac{\text{AD}}{\text{PM}}\\ \Rightarrow \frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QL}}=\frac{\text{2AD}}{\text{2PM}}\\ \Rightarrow \frac{\text{AB}}{\text{PQ}}=\frac{\text{BE}}{\text{QL}}=\frac{\text{AE}}{\text{PL}}\\ \text{Therefore, by SSS similarity critarion,}\mathrm{\Delta }\text{ABE}~\mathrm{\Delta }\text{PQL and so}\\ \text{in}\mathrm{\Delta }\text{ABE and}\mathrm{\Delta }\text{PQL,}\\ \angle \text{BAE}=\angle \text{QPL}...\text{(1)}\\ \text{Similarly, we can show that}\mathrm{\Delta }\text{AEC}~\mathrm{\Delta }\text{PLR and}\\ \angle \text{CAE}=\angle \text{RPL}...\text{(2)}\\ \text{Adding (1) and (2), we get}\\ \angle \text{BAE}+\angle \text{CAE}=\angle \text{QPL}+\angle \text{RPL}\\ \angle \text{CAB}=\angle \text{RPQ}\\ \text{Also, we have}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AC}}{\text{PR}}\\ \text{Therefore, by SAS similarity critarion,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR.}\end{array}$

Q.28

$\begin{array}{l}\text{A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower}\\ \text{casts a shadow 28 m long. Find the height of the tower.}\end{array}$

Ans.

$\begin{array}{l}\text{Let CD is a pole and DF is its shadow}\text{.}\\ \text{Let AB is a tower and BE is its shadow}\text{.}\\ \text{At the same time in a day, the sun rays will fall on pole}\\ \text{and tower at the same angle}\text{.}\\ \text{Therefore,}\\ \text{}\text{}\text{}\angle \text{DCF}=\angle \text{BAE}\\ \text{Also,}\text{}\angle \text{CDF}=\angle \text{ABE}=90°\\ \therefore \Delta \text{ABE}~\Delta \text{CDF [By AA similarity critarion]}\\ \text{Corresponding sides are proportional in similar triangles}\text{.}\\ \text{Therefore,}\\ \frac{\text{AB}}{\text{CD}}=\frac{\text{BE}}{\text{DF}}\\ \Rightarrow \frac{\text{AB}}{\text{6}}=\frac{\text{28}}{\text{4}}=7\\ \Rightarrow \text{AB}=42\\ \text{Therefore, the height of the tower is 42 m.}\end{array}$

Q.29

$\begin{array}{l}\text{If AD and PM are medians of triangles ABC and PQR, respectively where}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR, prove that}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}.\end{array}$

Ans.

$\begin{array}{l}\text{It is given that AD and PM are medians}.\text{Therefore,}\\ \frac{1}{2}\text{BC}=\text{BD}=\text{DC and}\frac{1}{2}\text{QR}=\text{QM}=\text{MR}\\ \text{Also,}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{PQR}\\ \text{Therefore,}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\text{BC}}{\text{QR}}=\frac{\text{AC}}{\text{PR}}\text{and}\angle \text{A}=\angle \text{P,}\angle \text{B}=\angle \text{Q,\hspace{0.17em}\hspace{0.17em}}\angle \text{C}=\angle \text{R}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta }\text{PQM,}\\ \frac{\text{AB}}{\text{PQ}}=\frac{\frac{1}{2}\text{BC}}{\frac{1}{2}\text{QR}}=\frac{\text{BD}}{\text{QM}}\text{and}\angle \text{B}=\angle \text{Q}\\ \therefore \mathrm{\Delta }\text{ABD}~\mathrm{\Delta }\text{PQM [By SAS similarity critarion]}\\ \Rightarrow \frac{\text{AB}}{\text{PQ}}=\frac{\text{AD}}{\text{PM}}\end{array}$

Q.30

$\text{Let}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }{\text{DEF and their areas be, respectively, 64 cm}}^{\text{2}}{\text{and 121 cm}}^{\text{2}}\text{. If EF}=\text{15.4 cm, find BC.}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta }\text{DEF.}\\ \therefore \frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{DEF}\right)}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}\\ \mathrm{Given}\mathrm{that}\\ \mathrm{EF}=15.4\text{\hspace{0.17em}\hspace{0.17em}cm}\\ \text{ar}\left(\mathrm{\Delta }\text{ABC}\right)=64{\text{cm}}^2\\ \text{ar}\left(\mathrm{\Delta DEF}\right)=121{\text{cm}}^2\\ \therefore \frac{\text{ar}\left(\mathrm{\Delta }\text{ABC}\right)}{\text{ar}\left(\mathrm{\Delta }\text{DEF}\right)}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}\\ \Rightarrow \frac{64}{121}=\frac{{\mathrm{BC}}^2}{{(15.4)}^2}\\ \Rightarrow \frac{8}{11}=\frac{\mathrm{BC}}{15.4}\\ \Rightarrow \mathrm{BC}=\frac{8\times 15.4}{11}\\ \Rightarrow \mathrm{BC}=11.2\end{array}$

Q.31 Diagonals of a trapezium ABCD with AB ║ DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta C}\text{OD and}\mathrm{\Delta A}\text{OB,}\\ \angle \text{ODC}=\angle \text{OBA [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \text{DCO}=\angle \text{BAO [Alternate interior angles as AB}\parallel \text{CD]}\\ \angle \mathrm{C}\text{OD}=\angle \mathrm{A}\text{OB [Vertically opposite angles]}\\ \therefore \mathrm{\Delta C}\text{OD}~\mathrm{\Delta A}\text{OB [AAA similarity critarion]}\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)}{\mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right)}=\frac{{\mathrm{CD}}^2}{{\mathrm{AB}}^2}\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)}{\mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right)}=\frac{{\mathrm{CD}}^2}{{\left(2\mathrm{CD}\right)}^2}=\frac{1}{4}\text{[Given AB}=2\mathrm{CD}]\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta A}\text{OB}\right):\mathrm{ar}\left(\mathrm{\Delta C}\text{OD}\right)=4:1\end{array}$

Q.32 In the following figure, ABC and DBC are two triangles on the same base BC.
If AD intersects BC at O, show that ar(ΔABC) (ΔDBC) = AO DO.

Ans.

$\text{We draw perpendiculars AP and DM on line BC.}$

$\begin{array}{l}\text{We know that area of a triangle}=\frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}\\ \therefore \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBC}\right)}=\frac{\frac{1}{2}\mathrm{BC}\times \mathrm{AP}}{\frac{1}{2}\mathrm{BC}\times \mathrm{DM}}=\frac{\mathrm{AP}}{\mathrm{DM}}\\ \mathrm{In}\mathrm{\Delta }\text{APO and}\mathrm{\Delta DM}\text{O,}\\ \angle \mathrm{APO}=\angle \mathrm{DMO}=90\mathrm{°}\\ \angle \mathrm{A}\text{OP}=\angle \mathrm{D}\text{OM [Vertically opposite angles]}\\ \therefore \mathrm{\Delta }\text{APO}~\mathrm{\Delta DM}\text{O [By AA similarity critarion]}\\ \Rightarrow \frac{\mathrm{AP}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}\end{array}$

Q.33 If the areas of two similar triangles are equal, prove that they are congruent.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{\Delta ABC}~\mathrm{\Delta PQR},\text{then we have}\\ \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}{\mathrm{ar}\left(\mathrm{\Delta ABC}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2\text{[Given}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta PQR}\right)]\\ \Rightarrow 1={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{QR}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{PR}}\right)}^2\\ \Rightarrow 1=\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\\ \Rightarrow \mathrm{AB}=\mathrm{PQ},\mathrm{BC}=\mathrm{QR}\text{and}\mathrm{AC}=\mathrm{PR}\\ \therefore \mathrm{\Delta ABC}\cong \mathrm{\Delta PQR}\text{[By SSS congruence criterion]}\end{array}$

Q.34

$\begin{array}{l}\text{D, E and F are respectively the mid-points of sides AB, BC and CA of}\mathrm{\Delta }\text{ABC. Find the ratio of the areas}\\ \text{of}\mathrm{\Delta }\text{DEF and}\mathrm{\Delta }\text{ABC.}\end{array}$

Ans.

$\begin{array}{l}\text{D and E are mid-points of}\mathrm{\Delta }\text{ABC.}\\ \therefore \mathrm{DE}\parallel \mathrm{AC}\text{and DE}=\frac{1}{2}\mathrm{AC}\\ \mathrm{In}\mathrm{\Delta }\text{BED and}\mathrm{\Delta }\text{BCA,}\\ \angle \text{BED}=\angle \text{BCA [Corresponding angles]}\\ \angle \text{BDE}=\angle \text{BAC [Corresponding angles]}\\ \angle \text{EBD}=\angle \text{CBA [Common angles]}\\ \therefore \mathrm{\Delta }\text{BED}~\mathrm{\Delta }\text{BCA [AAA similarity criterion]}\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)}{\mathrm{ar}\left(\mathrm{\Delta }\text{BCA}\right)}={\left(\frac{\mathrm{DE}}{\mathrm{AC}}\right)}^2={\left(\frac{\frac{1}{2}\mathrm{AC}}{\mathrm{AC}}\right)}^2=\frac{1}{4}\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{BCA}\right)\\ \mathrm{Similarly},\\ \mathrm{ar}\left(\mathrm{\Delta CF}\text{E}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{CBA}\right)\text{and}\mathrm{ar}\left(\mathrm{\Delta ADF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)\\ \text{Also,}\\ \mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)=\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)-\left[\mathrm{ar}\left(\mathrm{\Delta }\text{BED}\right)+\mathrm{ar}\left(\mathrm{\Delta CF}\text{E}\right)+\mathrm{ar}\left(\mathrm{\Delta ADF}\right)\right]\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)=\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)-\frac{3}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta D}\text{EF}\right)}{\mathrm{ar}\left(\mathrm{\Delta }\text{ABC}\right)}=\frac{1}{4}\end{array}$

Q.35 Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{us}\mathrm{take}\text{two triangles ABC and PQR such that}\mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}.\\ \mathrm{Let}\text{AD and PS be the medians of}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta PQR}\text{respectively.}\\ \text{Now,}\\ \mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}\\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\text{and}\angle \text{A}=\angle \mathrm{P}\text{,}\angle \mathrm{B}=\angle \mathrm{Q},\angle \mathrm{C}=\angle \mathrm{R}.\\ \mathrm{Since}\text{AD and PS are medians, so we have}\\ \text{BD}=\mathrm{DC}=\frac{\mathrm{BC}}{2}\text{and QS}=\mathrm{SR}=\frac{\mathrm{QR}}{2}\\ \text{In}\mathrm{\Delta }\text{ABD and}\mathrm{\Delta PQS},\\ \angle \mathrm{B}=\angle \mathrm{Q}\text{and}\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QS}}\\ \therefore \mathrm{\Delta }\text{ABD}~\mathrm{\Delta PQS}\text{[SAS similarity criterion]}\\ \Rightarrow \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QS}}=\frac{\mathrm{AD}}{\mathrm{PS}}\\ \text{Now,}\\ \mathrm{\Delta }\text{ABC}~\mathrm{\Delta PQR}\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{PQ}}\right)}^2\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta PQR}\right)}={\left(\frac{\mathrm{AD}}{\mathrm{PS}}\right)}^2\end{array}$

Q.36

$\begin{array}{l}\text{Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the}\\ \text{equilateral triangle described on one of its diagonals.}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\text{ABCD is a square and length of its one side is}\mathrm{a}\text{unit.}\\ \text{Therefore, length of its diagonal}=\sqrt{2}\mathrm{a}\\ \text{Equilateral triangles as per question are formed here as}\\ \mathrm{\Delta }\text{ABE and}\mathrm{\Delta DBF}.\\ \text{Length of a s}\mathrm{ide}\mathrm{of}\text{equilateral}\mathrm{\Delta }\text{ABE}=\mathrm{a}\\ \text{and}\\ \text{length of a s}\mathrm{ide}\mathrm{of}\text{equilateral}\mathrm{\Delta DBF}=\sqrt{2}\mathrm{a}\\ \mathrm{We}\text{know that equilateral triangles are similar to each other.}\\ \text{Therefore, ratio of their areas is equal to ratio of squares of}\\ \text{their sides.}\\ \mathrm{i}.\mathrm{e}.,\frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BE}\right)}{\mathrm{ar}\left(\mathrm{\Delta DBF}\right)}={\left(\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}\right)}^2=\frac{1}{2}\end{array}$

Q.37

$\begin{array}{l}\text{ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles}\\ \text{ABC and BDE is}\\ \text{(A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that}\mathrm{\Delta }\text{ABC and}\mathrm{\Delta }\text{BDE are equilateral and D is the}\\ \text{mid point of BC. Therefore, BD}=\frac{\mathrm{BC}}{2}\\ \mathrm{We}\text{know that equilateral triangles are similar to each other}\\ \text{and so ratio of their areas is equal to square of the ratio of}\\ \text{their sides.}\\ \therefore \frac{\mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right)}{\mathrm{ar}\left(\mathrm{\Delta BDE}\right)}={\left(\frac{\mathrm{AB}}{\mathrm{BD}}\right)}^2={\left(\frac{\mathrm{BC}}{\mathrm{BD}}\right)}^2\text{[}\mathrm{\Delta A}\text{BC is equilateral]}\\ ={\left(\frac{\mathrm{BC}}{\frac{\mathrm{BC}}{2}}\right)}^2=\frac{4}{1}\\ \Rightarrow \mathrm{ar}\left(\mathrm{\Delta A}\text{BC}\right):\mathrm{ar}\left(\mathrm{\Delta BDE}\right)=4:1\\ \text{Hence, the correct answer is (C).}\end{array}$

Q.38

$\begin{array}{l}\text{Sides of two similar triangles are in the ratio 4}:\text{9}.\text{\hspace{0.17em}Areas of these triangles are in the ratio}\\ \left(\text{A}\right)\text{2}:\text{3}\left(\text{B}\right)\text{4}:\text{9}\left(\text{C}\right)\text{81}:\text{16}\left(\text{D}\right)\text{16}:\text{81}\end{array}$

Ans.

$\begin{array}{l}\mathrm{It}\text{is given that sides of two similar triangles are in the}\\ \text{ratio 4}:\text{9.}\\ \text{Therefore,}\\ \text{Ratio of areas of these triangles}=\frac{4^2}{9^2}=\frac{16}{81}\\ \text{Hence, the correct answer is (D).}\end{array}$

Q.39 Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm

Ans.

(i) Given sides of the triangle are 7 cm, 24 cm and 25 cm.
Squares of the given sides of the triangle are 49 cm², 576 cm² and 625 cm².
Now,
49 cm² + 576 cm² = 625 cm²
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 25 cm.

(ii)

$\begin{array}{l}\text{Given sides of the triangle are 3 cm},\text{8 cm and 6 cm}.\\ {\text{Squares of the given sides of the triangle are 9 cm}}^{\text{2}},{\text{64 cm}}^2\\ {\text{and 36 cm}}^2\text{.}\\ \text{Now,}\\ 9+36\ne 64\\ \text{We find that sum of the squares of two sides is not equal to}\\ \text{the square of third side.}\\ \text{Therefore, triangle of the given sides is not a right triangle.}\\ \\ \text{(iii)}\\ \text{Given sides of the triangle are 50 cm},\text{80 cm and 100 cm}.\\ {\text{Squares of the given sides of the triangle are 2500 cm}}^{\text{2}},\\ {\text{6400 cm}}^2{\text{and 10000 cm}}^2\text{.}\\ \text{Now,}\\ 2500+6400\ne 10000\\ \text{We find that sum of the squares of two sides is not equal to}\\ \text{the square of third side.}\\ \text{Therefore, triangle of the given sides is not a right triangle.}\end{array}$

(iv) Given sides of the triangle are 13 cm, 12 cm and 5 cm.
Squares of the given sides of the triangle are 169 cm², 144 cm² and 25 cm².
Now, 144 c m² + 25 cm² = 169 cm²
Therefore, by converse of Pythagoras theorem, the given triangle is a right triangle.
Also, we know that hypotenuse is the longest side in a right triangle. Thus, length of the hypotenuse is 13 cm.

Q.40 PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that ( PM )² = QM . MR.

Ans.

$\begin{array}{l}\text{Let}\angle \text{MPR}=\mathrm{\theta }\\ \text{In}\mathrm{\Delta }\text{MPR,}\\ \angle \text{MRP}=180\mathrm{°}-90\mathrm{°}-\mathrm{\theta }=90\mathrm{°}-\mathrm{\theta }\\ \text{Similarly, in}\mathrm{\Delta }\text{MPQ,}\\ \angle \text{MPQ}=90\mathrm{°}-\angle \text{MPR}=90\mathrm{°}-\mathrm{\theta }\\ \angle \text{MQP}=180\mathrm{°}-90\mathrm{°}-(90\mathrm{°}-\mathrm{\theta })=\mathrm{\theta }\\ \text{Now, in}\mathrm{\Delta }\text{MPR and}\mathrm{\Delta }\text{MQP, we have}\\ \angle \text{MQP}=\angle \text{MPR,}\angle \text{MPQ}=\angle \text{MRP and}\angle \text{PMQ}=\angle \text{PMR.}\\ \text{Therefore, by AAA similarity critarion,}\mathrm{\Delta }\text{MPR}~\mathrm{\Delta }\text{MQP}.\\ \text{Therefore, we have}\\ \frac{\text{QM}}{\text{PM}}=\frac{\text{MP}}{\text{MR}}\\ \Rightarrow {\text{PM}}^2=\text{QM}\times \text{MR}\end{array}$

Q.41 In the following figure, ABD is a triangle right angled at A and AC ⊥ BD. Show thati AB2 = BC . BDii AC2 = BC . DCiii AD2 = BD . CD

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{ADB and}\mathrm{\Delta }\text{CAB, we have}\\ \angle \text{DAB}=\angle \text{ACB}=90\mathrm{°}\\ \text{and \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{ABD}=\angle \text{CBA [Common angle]}\\ \\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{ADB}~\mathrm{\Delta }\text{CAB}.\\ \text{Therefore, we have}\\ \frac{\text{AB}}{\text{CB}}=\frac{\text{BD}}{\text{AB}}\\ \Rightarrow {\text{AB}}^2=\text{BC}\cdot \text{BD}\\ \text{(ii)}\\ \text{Let}\angle \text{CAB}=\mathrm{\theta }\\ \text{In}\mathrm{\Delta }\text{CBA, we have}\\ \angle \text{CBA}=180\mathrm{°}-90\mathrm{°}-\mathrm{\theta }=90\mathrm{°}-\mathrm{\theta }\\ \text{In}\mathrm{\Delta }\text{CAD, we have}\\ \angle \text{CAD}=90\mathrm{°}-\angle \text{CAB}=90\mathrm{°}-\mathrm{\theta }\\ \angle \text{CDA}=180\mathrm{°}-90\mathrm{°}-(90\mathrm{°}-\mathrm{\theta })=\mathrm{\theta }\\ \\ \text{In}\mathrm{\Delta }\text{CBA and}\mathrm{\Delta }\text{CAD, we have}\\ \angle \text{CBA}=\angle \text{CAD},\angle \text{CAB}=\angle \text{CDA and}\angle \text{ACB}=\angle \text{DCA}=90\mathrm{°}\\ \\ \text{Therefore, by AAA similarity critarion,}\mathrm{\Delta }\text{CBA}~\mathrm{\Delta }\text{CAD}.\\ \text{Therefore, we have}\\ \frac{\text{AC}}{\text{DC}}=\frac{\text{BC}}{\text{AC}}\\ \Rightarrow {\text{AC}}^2=\text{BC}\cdot \text{DC}\\ \text{(iii)}\\ \text{In}\mathrm{\Delta }\text{DCA and}\mathrm{\Delta }\text{DAB,}\\ \angle \text{DCA}=\angle \text{DAB}=90\mathrm{°},\angle \text{CDA}=\angle \text{ADB}\\ \text{Therefore, by AA similarity critarion,}\mathrm{\Delta }\text{DCA}~\mathrm{\Delta }\text{DAB}.\\ \text{Therefore, we have}\\ \frac{\text{DC}}{\text{DA}}=\frac{\text{DA}}{\text{DB}}\\ \Rightarrow {\text{AD}}^2=\text{BD}\cdot \text{CD}\end{array}$

Q.42 ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

Ans. Given that ABC is an isosceles triangle right angled at C.
Therefore, AC = BC

Using Pythagoras theorem in the given triangle,
we have
AB² = AC² + BC² = AC² + AC² = 2AC²

Q.43 ABC is an isosceles triangle with AC = BC. If AB² = 2AC², prove that ABC is a right triangle.

Ans.

Given that ABC is an isosceles triangle with AC = BC and AB² = 2AC².
Therefore,
AB² = 2AC² = AC² + BC²
Therefore, by converse of Pythagoras theorem, ABC is a right triangle.

Q.44 ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Ans.

$\begin{array}{l}\text{It is given that ABC is an equilateral triangle of side 2a}.\\ \text{Let AD is an altitude}.\\ \text{We know that altitude bisects opposite side in an}\\ \text{equilateral triangle}.\\ \text{Therefore},\text{BD}=\text{CD}=\text{a}.\\ \text{Using Pythagoras theorem in}\mathrm{\Delta }\text{ADB, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^2={\text{BD}}^2+{\text{AD}}^2\\ \Rightarrow {\text{AD}}^2={\text{AB}}^2-{\text{BD}}^2={\left(2\mathrm{a}\right)}^2-{\mathrm{a}}^2=4{\mathrm{a}}^2-{\mathrm{a}}^2=3{\mathrm{a}}^2\\ \Rightarrow \text{AD}=\mathrm{a}\sqrt{3}\\ \text{We know that all the altitudes in an equilateral triangle}\\ \text{are of same length. Therefore, length of each altitude is a}\sqrt{3}\text{.}\end{array}$

Q.45 Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Ans.

$\begin{array}{l}\text{Let ABCD is a rhombus.}\\ \text{We know that diagonals of a rhombus bisect each other}\\ \text{at right angles.}\\ \text{Using Pythagoras Theorem in}\mathrm{\Delta }\text{AOB,}\mathrm{\Delta }\text{BOC,}\mathrm{\Delta }\text{COD and}\mathrm{\Delta }\text{AOD,}\\ \text{we get}\\ {\text{AB}}^2={\text{OA}}^2+{\text{OB}}^2\text{,}\\ {\text{BC}}^2={\text{OB}}^2+{\text{OC}}^2,\\ {\text{CD}}^2={\text{OC}}^2+{\text{OD}}^2,\\ {\text{DA}}^2={\text{OA}}^2+{\text{OD}}^2\\ \text{Adding all these equations, we get}\\ {\text{AB}}^2+{\text{BC}}^2+{\text{CD}}^2+{\text{DA}}^2=2({\text{OA}}^2+{\text{OB}}^2+{\text{OC}}^2+{\text{OD}}^2)\\ =2\left[2{\left(\frac{\text{AC}}{2}\right)}^2+2{\left(\frac{\text{BD}}{2}\right)}^2\right]\\ ={\text{AC}}^2+{\text{BD}}^2\end{array}$

Q.46

$\begin{array}{l}\text{In\hspace{0.33em}the\hspace{0.33em}following\hspace{0.33em}figure,\hspace{0.33em}O\hspace{0.33em}is\hspace{0.33em}a\hspace{0.33em}point\hspace{0.33em}in\hspace{0.33em}the\hspace{0.33em}interior\hspace{0.33em}of\hspace{0.33em}a\hspace{0.33em}triangle\hspace{0.33em}ABC,}\\ \text{OD\hspace{0.33em}}\perp \text{\hspace{0.33em}BC,\hspace{0.33em}OE\hspace{0.33em}}\perp \text{\hspace{0.33em}AC\hspace{0.33em}and\hspace{0.33em}OF\hspace{0.33em}}\perp \text{\hspace{0.33em}AB.\hspace{0.33em}}\\ {\text{(i) OA}}^{\text{2}}{\text{+OB}}^{\text{2}}{\text{+OC}}^{\text{2}}-{\text{\hspace{0.33em}OD}}^{\text{2}}-{\text{OE}}^{\text{2}}-{\text{OF}}^{\text{2}}{\text{\hspace{0.33em}=\hspace{0.33em}AF}}^{\text{2}}{\text{+BD}}^{\text{2}}{\text{+CE}}^{\text{2}}\text{,}\\ {\text{(ii)\hspace{0.33em} AF}}^{\text{2}}{\text{+BD}}^{\text{2}}{\text{+CE}}^{\text{2}}{\text{\hspace{0.33em}=\hspace{0.33em}AE}}^{\text{2}}{\text{+CD}}^{\text{2}}{\text{+BF}}^{\text{2}}\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{We join O to A},\text{B and C}.\\ \text{(i)}\\ \text{In}\mathrm{\Delta }\text{AOF,}\\ {\text{OA}}^2={\text{OF}}^2+{\text{AF}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{In}\mathrm{\Delta }\text{BOD,}\\ {\text{OB}}^2={\text{OD}}^2+{\text{BD}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{In}\mathrm{\Delta }\text{COE,}\\ {\text{OC}}^2={\text{OE}}^2+{\text{EC}}^{\text{2}}\text{[By Pythagoras theorem]}\\ \text{Adding these equations, we get}\\ {\text{OA}}^2+{\text{OB}}^2+{\text{OC}}^2={\text{OF}}^2+{\text{AF}}^{\text{2}}+{\text{OD}}^2+{\text{BD}}^{\text{2}}+{\text{OE}}^2+{\text{EC}}^{\text{2}}\\ \Rightarrow {\text{OA}}^2+{\text{OB}}^2+{\text{OC}}^2-{\text{OD}}^2-{\text{OE}}^2-{\text{OF}}^2={\text{AF}}^{\text{2}}+{\text{BD}}^{\text{2}}+{\text{EC}}^{\text{2}}\\ \text{(ii)}\\ \text{We have form above result,}\\ {\text{AF}}^{\text{2}}+{\text{BD}}^{\text{2}}+{\text{EC}}^{\text{2}}={\text{OA}}^2+{\text{OB}}^2+{\text{OC}}^2-{\text{OD}}^2-{\text{OE}}^2-{\text{OF}}^2\\ =({\text{OA}}^2-{\text{OE}}^2)+({\text{OB}}^2-{\text{OF}}^2)+({\text{OC}}^2-{\text{OD}}^2)\\ ={\text{AE}}^2+{\text{BF}}^2+{\text{CD}}^2\end{array}$

Q.47 A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Ans.

$\begin{array}{l}\text{Let AB is a ladder and AC is wall}.\text{A represents the window}\\ \text{and BC is the distance of the foot of the ladder from the}\\ \text{base of the wall}.\\ \text{ABC is a right angle triangle. Therefore, using Pythagoras}\\ \text{theorem, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^2={\text{BC}}^2+{\text{AC}}^2\\ \Rightarrow {10}^2={\text{BC}}^2+8^2\\ \Rightarrow {\text{BC}}^2=100-64=36\\ \Rightarrow \text{BC}=6\\ \text{Therefore, the distance of the foot of the ladder from the}\\ \text{base of the wall is 6 m.}\end{array}$

Q.48 A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Ans.

$\begin{array}{l}\text{Let AC is a pole and AB is a guy wire with stake at B}.\\ \text{Then, ABC is a right angle triangle. Therefore, using Pythagoras}\\ \text{theorem, we get}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AB}}^2={\text{BC}}^2+{\text{AC}}^2\\ \Rightarrow {24}^2={\text{BC}}^2+{18}^2\\ \Rightarrow {\text{BC}}^2={24}^2-{18}^2=576-324=252\\ \Rightarrow \text{BC}=\sqrt{252}=\sqrt{4\times 9\times 7}=6\sqrt{7}\\ \text{Therefore, the distance of the foot of the ladder from the}\\ \text{base of the wall is 6}\sqrt{7}\text{m.}\end{array}$

Q.49 An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west ata speed of 1200 km per hour. How far apart will be the two planes after 112  hours?

Ans.

$\begin{array}{l}\text{Distance travelled by the plane flying towards north in}\\ \text{1}\frac{1}{2}\text{hours}=\text{1}000\times 1\frac{1}{2}\text{km}=1500\text{km}\\ \text{Distance travelled by the plane flying towards west in}\\ \text{1}\frac{1}{2}\text{hours}=\text{12}00\times 1\frac{1}{2}\text{km}=1800\text{km}\\ \text{We represent thsese distances by OA and OB.}\\ \text{Also, AB represents the distance between the two planes.}\\ \text{Using Pythagoras theorem, we get}\\ {\text{AB}}^2={\text{OA}}^2+{\text{OB}}^2={1500}^2+{1800}^2=22,50,000+32,40,000\\ \Rightarrow {\text{AB}}^2=54,90,000\\ \Rightarrow \text{AB}=\sqrt{54,90,000}=\sqrt{9\times 6,10,000}=300\sqrt{61}\\ \text{Therefore, distance between the two planes}=300\sqrt{61}\text{km}\end{array}$

Q.50

$\begin{array}{l}\text{Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet}\\ \text{of the poles is 12 m, find the distance between their tops.}\end{array}$

Ans.

$\begin{array}{l}\text{Let AB and CD be two poles of heights 6 m and 11m}\\ \text{respectively}.\text{Let their feet are on the ground at points B}\\ \text{and D}.\\ \text{It is given that distance between feet is 12 m}.\text{Therefore},\\ \text{BD}=\text{12 m}.\\ \text{Also},\\ \text{AB}=\text{DE}=\text{6m}\\ \text{and}\\ \text{CE}=\text{CD}-\text{DE}=\text{11 m}-\text{6 m}=\text{5 m}\\ \text{Now},\text{AEC is a right triangle}.\text{Therefore},\text{by Pythagoras theorem},\\ \text{AC}=\sqrt{{\text{AE}}^2+{\text{CE}}^2}=\sqrt{{\text{12}}^2+{\text{5}}^2}=\sqrt{\text{1}44+2\text{5}}=\sqrt{\text{169}}=13\text{m}\\ \text{Therefore,}\\ \text{Distance between tops of the two poles}=13\text{m}\end{array}$

Q.51

$\begin{array}{l}\text{D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.}\\ \text{Prove that}{\mathrm{AE}}^2+{\mathrm{BD}}^2={\mathrm{AB}}^2+{\mathrm{DE}}^2\text{.}\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ACE,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AE}}^2={\text{AC}}^2+{\text{CE}}^2...\text{(1)}\\ \text{In}\mathrm{\Delta }\text{DCB,}\\ {\text{BD}}^2={\text{CD}}^2+{\text{BC}}^2...\text{(2)}\\ \text{On adding (1) and (2), we have}\\ {\text{AE}}^2+{\text{BD}}^2={\text{AC}}^2+{\text{CE}}^2+{\text{CD}}^2+{\text{BC}}^2...\left(3\right)\\ \text{In}\mathrm{\Delta }\text{ACB,}\\ {\text{AB}}^2={\text{AC}}^2+{\text{BC}}^2...\left(4\right)\\ \text{In}\mathrm{\Delta }\text{DCE,}\\ {\text{DE}}^2={\text{CD}}^2+{\text{CE}}^2...\left(5\right)\\ \text{On adding (4) and (5), we have}\\ {\text{AB}}^2+{\text{DE}}^2={\text{AC}}^2+{\text{BC}}^2+{\text{CD}}^2+{\text{CE}}^2...\left(6\right)\\ \text{From (3) and (6), we find that}\\ {\text{AE}}^2+{\text{BD}}^2={\text{AB}}^2+{\text{DE}}^2\end{array}$

Q.52

$\begin{array}{l}\text{The perpendicular from A on side BC of a}\mathrm{\Delta }\text{ABC intersects BC at D such that}\mathrm{DB}=3\mathrm{CD}\text{(see the}\\ \text{following figure). Prove that}2{\mathrm{AB}}^2=2{\mathrm{AC}}^2+{\mathrm{BC}}^2.\end{array}$

Ans.

$\begin{array}{l}\text{In}\mathrm{\Delta }\text{ADC,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^2={\text{CD}}^2+{\text{AD}}^2\\ \Rightarrow {\text{AD}}^2={\text{\hspace{0.17em}AC}}^2-{\text{CD}}^2...\text{(1)}\\ \text{In}\mathrm{\Delta }\text{ADB,}\\ {\text{AB}}^2={\text{AD}}^2+{\text{DB}}^2\\ \Rightarrow {\text{AD}}^2={\text{AB}}^2-{\text{DB}}^2...\text{(2)}\\ \text{From (1) and (2), we have}\\ {\text{AC}}^2-{\text{CD}}^2={\text{AB}}^2-{\text{DB}}^2...\left(3\right)\\ \text{Now, it is given that DB}=\text{3CD}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC}=\text{CD}+\text{DB}\\ \Rightarrow \text{BC}=\text{CD}+\text{3CD}=4\text{CD}\\ \Rightarrow \text{CD}=\frac{\text{BC}}{4}...\left(4\right)\\ \text{Again,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}BC}=\text{CD}+\text{DB}\\ \Rightarrow \text{BC}=\frac{\text{DB}}{3}+\text{DB}=\frac{4\text{DB}}{3}\\ \Rightarrow \text{DB}=\frac{3\text{BC}}{4}...\left(5\right)\\ \text{From (3), (4) and (5), we have}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AC}}^2-{\text{CD}}^2={\text{AB}}^2-{\text{DB}}^2\\ \Rightarrow {\text{AC}}^2-\frac{{\text{BC}}^2}{16}={\text{\hspace{0.17em}AB}}^2-\frac{9{\text{BC}}^2}{16}\\ \Rightarrow 16{\text{AC}}^2-{\text{BC}}^2=16{\text{AB}}^2-9{\text{BC}}^2\\ \Rightarrow 16{\text{AB}}^2=16{\text{AC}}^2-{\text{BC}}^2+9{\text{BC}}^2=16{\text{AC}}^2+8{\text{BC}}^2\\ \Rightarrow 2{\text{AB}}^2=2{\text{AC}}^2+{\text{BC}}^2\end{array}$

Q.53

$\text{In an equilateral triangle ABC, D is a point on side BC such that BD}=\frac{1}{3}{\text{BC. Prove that 9AD}}^{\text{2}}={\text{7AB}}^{\text{2}}\text{.}$

Ans.

$\begin{array}{l}\text{It is given that ABC is an equilateral triangle.}\\ \text{Let length of each side of}\mathrm{\Delta }\text{ABC be a.}\\ \text{It is given that D is a point on side BC such that}\\ \text{BD}=\frac{1}{3}\text{BC}=\frac{\mathrm{a}}{3}\text{.}\\ \text{Let AE is an altitude of equilateral triangle ABC. So,}\\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}=\frac{\mathrm{a}}{2}\text{.}\\ \text{Also, AE}=\sqrt{{\text{AC}}^2-{\text{EC}}^2}=\sqrt{{\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{4}}=\frac{\sqrt{3}}{2}\mathrm{a}\\ \text{Again, DE}=\text{BE}-\text{BD}=\frac{\mathrm{a}}{2}-\frac{\mathrm{a}}{3}=\frac{\mathrm{a}}{6}\\ \text{In}\mathrm{\Delta }\text{ADE,}\\ {\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}AD}}^2={\text{AE}}^2+{\text{DE}}^2\text{[By Pythagoras Theorem]}\\ \Rightarrow {\text{AD}}^2={\left(\frac{\sqrt{3}}{2}\mathrm{a}\right)}^2+\frac{{\mathrm{a}}^2}{36}=\frac{3}{4}{\mathrm{a}}^2+\frac{{\mathrm{a}}^2}{36}=\frac{27{\mathrm{a}}^2+{\mathrm{a}}^2}{36}=\frac{28}{36}{\mathrm{a}}^2=\frac{7}{9}{\mathrm{a}}^2\\ \Rightarrow {\text{AD}}^2=\frac{7}{9}{\mathrm{a}}^2=\frac{7}{9}{\text{AB}}^2\text{[AB}=\text{a]}\\ \Rightarrow 9{\text{AD}}^2=7{\text{AB}}^2\end{array}$

Q.54 In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Ans.

$\begin{array}{l}\text{Let ABC is an equilateral triangle and AE is its altitude.}\\ \text{Let length of each side of}\mathrm{\Delta }\text{ABC be a.}\\ \text{So,}\\ \text{BE}=\text{EC}=\frac{1}{2}\text{BC}=\frac{\mathrm{a}}{2}\text{.}\\ \text{Using Pythagoras theorem in}\mathrm{\Delta }\text{ACE, we have}\\ {\text{AE}}^2={\text{AC}}^2-{\text{EC}}^2={\mathrm{a}}^2-\frac{{\mathrm{a}}^2}{4}=\frac{3}{4}{\mathrm{a}}^2\\ \Rightarrow 3{\mathrm{a}}^2=4{\text{AE}}^2\\ \text{Therefore, three times the square of one side of an}\\ \text{equilateral triangle is equal to four times the square}\\ \text{of one of its altitude.}\end{array}$

Q.55

$\begin{array}{l}\text{Tick the correct answer and justify: In}\mathrm{\Delta }\text{ABC, AB}=\text{6}\sqrt{\text{3}}\text{cm, AC}=\text{12 cm and BC}=\text{6 cm.}\\ \text{The angle B is :}\\ \text{(A) 120° \hspace{0.17em} (B) 60°}\\ \text{(C) 90° (D) 45°}\end{array}$

Ans.

$\begin{array}{l}\text{It is given that in}\mathrm{\Delta }\text{ABC},\\ \text{AB}=\text{6}\sqrt{\text{3}}\text{cm, AC}=\text{12 cm and BC}=\text{6 cm}\\ \Rightarrow {\text{AB}}^2={\text{108 cm}}^2{\text{, AC}}^2={\text{144 cm}}^2{\text{and BC}}^2=36{\text{cm}}^2\\ \text{Now,}\\ {\text{AB}}^2+{\text{BC}}^2={\text{108 cm}}^2+36{\text{cm}}^2=144{\text{cm}}^2={\text{AC}}^2\\ \text{Therefore, by converse of Pythagoras theorem, we find that}\\ \text{in}\mathrm{\Delta }\text{ABC, angle B is 90°.}\\ \text{Therefore, correct answer is (C).}\end{array}$

Q.56

$\text{In the following figure, PS is the bisector of}\angle \text{QPR of}\mathrm{\Delta }\text{PQR. Prove that}\frac{\mathrm{QS}}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{PR}}.$

Ans.

$\begin{array}{l}\text{We draw a line segment RT parallel to SP which}\\ \text{intersects extended line segment QP at point T.}\\ \text{It is given that PS is angle bisector of}\angle \text{QPR.}\\ \text{Therefore,}\\ \angle \text{QPS}=\angle \text{SPR (1)}\\ \text{Also,}\\ \angle \text{SPR}=\angle \text{PRT (As PS}\parallel \text{TR) (2)}\\ \angle \text{QPS}=\angle \text{QTR (As PS}\parallel \text{TR) (3)}\\ \text{Using these equations we get,}\\ \angle \text{PRT}=\angle \text{QTR}\\ \text{So, PT}=\text{PR}\\ \text{Now in}\mathrm{\Delta }\text{QTR and}\mathrm{\Delta }\text{QPS,}\\ \angle \text{QSP}=\angle \text{QRT (As PS}\parallel \text{TR)}\\ \angle \text{QPS}=\angle \text{QTR (As PS}\parallel \text{TR)}\\ \angle \text{Q is common}\\ \therefore \mathrm{\Delta }\text{QTR}~\mathrm{\Delta }\text{QPS}\\ \Rightarrow \frac{\text{QS}}{\text{SR}}=\frac{\text{QP}}{\text{PT}}\\ \Rightarrow \frac{\text{QS}}{\text{SR}}=\frac{\text{QP}}{\text{PR}}\text{[PT}=\text{PR}]\end{array}$

Q.57

$\begin{array}{l}\text{In the following figure, D is a point on hypotenuse AC of}\mathrm{\Delta }\text{ABC, DM}\perp \text{BC and DN}\perp \text{AB. Prove that:}\\ \text{(i)}{\mathrm{DM}}^2=\mathrm{DN}.\mathrm{MC}\text{(ii)}{\mathrm{DN}}^2=\mathrm{DM}.\mathrm{AN}\end{array}$

Ans.

(i)

$\begin{array}{l}\text{We have, DN}\parallel \text{CB, DM}\parallel \text{AB and}\angle \text{B}=90°\\ \text{So},\text{DNMB is a rectangle.}\\ \therefore \text{DN}=\text{MB and DM}=\text{NB}\\ \angle \text{2}+\angle \text{3}=\text{9}0°...\left(\text{1}\right)\\ \text{In}\Delta \text{CDM,}\\ \angle \text{1}+\angle \text{2}+\angle \text{DMC}=\text{18}0°\\ \angle \text{1}+\angle \text{2}=\text{9}0°...\left(\text{2}\right)\\ \text{In}\Delta \text{DMB}\\ \angle \text{3}+\angle \text{DMB}+\angle \text{4}=\text{18}0°\\ \angle \text{3}+\angle \text{4}=\text{9}0°...\left(\text{3}\right)\\ \text{From equation}\left(\text{1}\right)\text{and}\left(\text{2}\right),\\ \angle \text{1}=\angle \text{3}\\ \text{From equation}\left(\text{1}\right)\text{and}\left(\text{3}\right),\\ \angle \text{2}=\angle \text{4}\\ \therefore \Delta \text{DCM}~\Delta \text{BDM}\\ \Rightarrow \frac{\text{DM}}{\text{BM}}=\frac{\text{MC}}{\text{DM}}\\ \Rightarrow {\text{DM}}^2=\text{BM}\times \text{MC}=\text{DN}\times \text{MC [BM}=\text{DN]}\\ \\ \text{(ii)}\\ \text{In}\Delta \text{DBN,}\\ \angle \text{5}+\angle \text{7}=\text{9}0°...\left(\text{4}\right)\\ \text{In}\Delta \text{DAN,}\\ \angle \text{6}+\angle \text{8}=\text{9}0°...\left(\text{5}\right)\\ \text{In}\Delta \text{DAB,}\\ \angle \text{5}+\angle \text{6}=\text{9}0°...\left(\text{6}\right)\\ \text{From equation}\left(\text{4}\right)\text{and}\left(\text{6}\right),\text{we get}\\ \angle \text{6}=\angle \text{7}\\ \text{From equation}\left(\text{5}\right)\text{and}\left(\text{6}\right),\text{we get}\\ \angle \text{5}=\angle \text{8}\\ \therefore \Delta \text{BND}~\Delta \text{DNA [By AA similarity critarion]}\\ \Rightarrow \frac{\text{AN}}{\text{DN}}=\frac{\text{DN}}{\text{NB}}\\ \Rightarrow {\text{DN}}^2=\text{AN}\times \text{NB}=\text{AN}\times \text{DM [As NB}=\text{DM]}\end{array}$

Q.58 

$\begin{array}{l}\text{In the following figure, ABC is a triangle in which}\angle \text{ABC > 90° and AD}\perp \text{CB produced. Prove that}\\ {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2+2\mathrm{BC}.\mathrm{BD}.\end{array}$

Ans.

Applying Pythagoras theorem in ΔADB, we get
AB² = AD² + DB² ….(1)
Applying Pythagoras theorem in ΔACD, we get
AC² = AD² + DC²
⇒AC² = AD² + ( BD + BC )²
⇒AC² = AD² + DB² + BC² + 2BD x BC
Now using equation ( 1 ), we get
AC² = AB² + BC² + 2BD × BC

Q.59

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Ans.

$\begin{array}{l}\text{A}\mathrm{pplying}\mathrm{Pythagoras}\mathrm{theorem}\text{i}\mathrm{n}\mathrm{\Delta ADB},\text{we get}\\ {\mathrm{AD}}^2+{\mathrm{BD}}^2={\mathrm{AB}}^2\\ \Rightarrow {\mathrm{AD}}^2={\mathrm{AB}}^2-{\mathrm{BD}}^2\mathrm{}...\left(1\right)\\ \text{A}\mathrm{pplying}\mathrm{Pythagoras}\mathrm{theorem}\text{i}\mathrm{n}\mathrm{\Delta ADC},\text{we get}\\ {\mathrm{AD}}^2+{\mathrm{DC}}^2={\mathrm{AC}}^2\mathrm{}\\ \mathrm{Now}\mathrm{using}\mathrm{equation}\left(1\right),\text{we get}\\ {\mathrm{AB}}^2-{\mathrm{BD}}^2+{\mathrm{DC}}^2={\mathrm{AC}}^2\\ \Rightarrow {\mathrm{AB}}^2-{\mathrm{BD}}^2+{(\mathrm{BC}-\mathrm{BD})}^2={\mathrm{AC}}^2\\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2-{\mathrm{BD}}^2+{\mathrm{BC}}^2+{\mathrm{BD}}^2-2\mathrm{BC}\times \mathrm{BD}\\ \Rightarrow {\mathrm{AC}}^2={\mathrm{AB}}^2+{\mathrm{BC}}^2-2\mathrm{BC}\times \mathrm{BD}\end{array}$

Q.60

$\begin{array}{l}\text{In the following figure, AD is a median of a triangle ABC and AM}\perp \text{BC. Prove that:}\\ {\text{(i) AC}}^{\text{2}}={\text{AD}}^{\text{2}}+\text{BC.DM}+{\left(\frac{\text{BC}}{2}\right)}^{\text{2}}\\ {\text{(ii) AB}}^{\text{2}}={\text{AD}}^{\text{2}}-\text{BC . DM}+{\left(\frac{\text{BC}}{2}\right)}^{\text{2}}\\ {\text{(iii) AC}}^{\text{2}}+{\text{AB}}^{\text{2}}={\text{2AD}}^{\text{2}}+\frac{1}{2}{\text{BC}}^{\text{2}}\end{array}$

Ans.

We have the following figure.

$\begin{array}{l}\text{(i)}\\ \mathrm{}\text{Applying Pythagoras theorem i}\mathbf{n}\mathrm{\Delta }\mathbf{AMD},\text{we get}\\ {\mathbf{AM}}^2+{\mathbf{DM}}^2={\mathbf{AD}}^2\mathrm{}...\left(\mathbf{1}\right)\\ \text{Applying Pythagoras theorem i}\mathbf{n}\mathrm{\Delta }\mathbf{AMC},\text{we get}\\ {\mathbf{AM}}^2+{\mathbf{MC}}^2={\mathbf{AC}}^2\mathrm{}\\ \Rightarrow {\mathbf{AM}}^2+{(\mathbf{DM}+\mathbf{DC})}^2={\mathbf{AC}}^2\\ \Rightarrow ({\mathbf{AM}}^2+{\mathbf{DM}}^2)+{\mathbf{DC}}^2+\mathbf{2}\mathbf{DM}.\mathbf{DC}={\mathbf{AC}}^2\\ \mathbf{Using}\mathbf{equation}\left(\mathbf{1}\right)\text{,}\mathbf{we}\mathbf{ge}\mathrm{t}\\ {\mathbf{AD}}^2+{\mathbf{DC}}^2+\mathbf{2}\mathbf{DM}.\mathbf{DC}={\mathbf{AC}}^2\\ \Rightarrow {\mathbf{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2+\mathbf{2}\mathbf{DM}.\frac{\text{BC}}{2}={\mathbf{AC}}^2\\ \Rightarrow {\mathbf{AD}}^2+{\left(\frac{\text{BC}}{2}\right)}^2+\mathbf{DM}\cdot \text{BC}={\mathbf{AC}}^2\\ \text{(ii)}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{ABM, we get}\\ {\mathrm{AB}}^2={\mathrm{AM}}^2+{\mathrm{MB}}^2\\ =({\mathrm{AD}}^2-{\mathrm{DM}}^2)+{\mathrm{MB}}^2\\ =({\mathrm{AD}}^2-{\mathrm{DM}}^2)+{(\mathrm{BD}-\mathrm{DM})}^2\\ ={\mathrm{AD}}^2-{\mathrm{DM}}^2+{\mathrm{BD}}^2+{\mathrm{DM}}^2-2\mathrm{BD}\times \mathrm{DM}\\ ={\mathrm{AD}}^2+{\mathrm{BD}}^2-2\mathrm{BD}\times \mathrm{DM}\\ ={\mathrm{AD}}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2-2\frac{\mathrm{BC}}{2}\times \mathrm{DM}\\ ={\mathrm{AD}}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2-\mathrm{BC}\times \mathrm{DM}\\ \text{(iii)}\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{AMB, we get}\\ {\mathrm{AM}}^2+{\mathrm{MB}}^2={\mathrm{AB}}^2...\left(1\right)\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta AMC}\text{, we get}\\ {\mathrm{AM}}^2+{\mathrm{MC}}^2={\mathrm{AC}}^2...\left(2\right)\\ \text{Adding equations (1) and (2), we get}\\ 2{\mathrm{AM}}^2+{\mathrm{MB}}^2+{\mathrm{MC}}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2\\ \Rightarrow 2{\mathrm{AM}}^2+{(\mathrm{BD}-\mathrm{DM})}^2+{(\mathrm{MD}+\mathrm{DC})}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2\\ \Rightarrow 2{\mathrm{AM}}^2+{\mathrm{BD}}^2+{\mathrm{DM}}^2-2\mathrm{BD}.\mathrm{DM}+{\mathrm{MD}}^2+{\mathrm{DC}}^2+2\mathrm{DM}.\mathrm{DC}={\mathrm{AB}}^2+{\mathrm{AC}}^2\\ \Rightarrow 2{\mathrm{AM}}^2+2{\mathrm{DM}}^2+{\mathrm{BD}}^2+{\mathrm{DC}}^2+2\mathrm{DM}(-\mathrm{BD}+\mathrm{DC})={\mathrm{AB}}^2+{\mathrm{AC}}^2\\ \Rightarrow 2({\mathrm{AM}}^2+{\mathrm{DM}}^2)+{\left(\frac{\mathrm{BC}}{2}\right)}^2+{\left(\frac{\mathrm{BC}}{2}\right)}^2+2\mathrm{DM}(-\frac{\mathrm{BC}}{2}+\frac{\mathrm{BC}}{2})={\mathrm{AB}}^2+{\mathrm{AC}}^2\\ \Rightarrow 2{\mathrm{AD}}^2+\frac{{\mathrm{BC}}^2}{2}={\mathrm{AB}}^2+{\mathrm{AC}}^2\end{array}$

Q.61 Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Ans.

Let ABCD be a parallelogram. We draw perpendiculars AF on CD and DE on extended side BA.

$\begin{array}{l}\text{Applying Pythagoras theorem in}\mathrm{\Delta }\text{DEA, we get}\\ {\mathrm{DE}}^2+{\mathrm{EA}}^2={\mathrm{DA}}^2...\left(\mathrm{i}\right)\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta DEB},\text{we get}\\ {\mathrm{DE}}^2+{\mathrm{EB}}^2={\mathrm{DB}}^2\\ {\mathrm{DE}}^2+{(\mathrm{EA}+\mathrm{AB})}^2={\mathrm{DB}}^2\\ ({\mathrm{DE}}^2+{\mathrm{EA}}^2)+{\mathrm{AB}}^2+2\mathrm{EA}\times \mathrm{AB}={\mathrm{DB}}^2\\ {\mathrm{DA}}^2+{\mathrm{AB}}^2+2\mathrm{EA}\times \mathrm{AB}={\mathrm{DB}}^2...\left(\mathrm{ii}\right)\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta ADF},\text{we get}\\ {\mathrm{AD}}^2={\mathrm{AF}}^2+{\mathrm{FD}}^2\\ \text{Applying Pythagoras theorem in}\mathrm{\Delta AFC},\text{we get}\\ {\mathrm{AC}}^2={\mathrm{AF}}^2+{\mathrm{FC}}^2\\ ={\mathrm{AF}}^2+{(\mathrm{DC}-\mathrm{FD})}^2\\ ={\mathrm{AF}}^2+{\mathrm{DC}}^2+{\mathrm{FD}}^2-2\mathrm{DC}\times \mathrm{FD}\\ =({\mathrm{AF}}^2+{\mathrm{FD}}^2)+{\mathrm{DC}}^2-2\mathrm{DC}\times \mathrm{FD}\\ {\mathrm{AC}}^2={\mathrm{AD}}^2+{\mathrm{DC}}^2-2\mathrm{DC}\times \mathrm{FD}...\left(\mathrm{iii}\right)\\ \text{Since ABCD is a parallelogram}\\ \mathrm{AB}=\mathrm{CD}...\left(\mathrm{iv}\right)\\ \mathrm{and},\mathrm{BC}=\mathrm{AD}...\left(\mathrm{v}\right)\\ \text{In}\mathrm{\Delta }\text{DEA and}\mathrm{\Delta }\text{ADF,}\\ \angle \mathrm{DEA}=\angle \mathrm{AFD}=90\mathrm{°},\angle \mathrm{EAD}=\angle \mathrm{ADF}(\mathrm{EA}\parallel \mathrm{DF})\\ \mathrm{Therefore},\angle \mathrm{FAD}=\angle \mathrm{EDA}\\ \text{AD is common in both triangles.}\\ \mathrm{Therefore},\text{by ASA congruence critarion, we get}\\ \mathrm{\Delta EAD}\cong \mathrm{\Delta FDA}\\ \Rightarrow \mathrm{EA}=\mathrm{DF}...\text{(vi)}\\ \text{Adding equation (ii) and (iii), we get}\\ {\mathrm{DA}}^2+{\mathrm{AB}}^2+2\mathrm{EA}\times \mathrm{AB}+{\mathrm{AD}}^2+{\mathrm{DC}}^2-2\mathrm{DC}\times \mathrm{FD}={\mathrm{DB}}^2+{\mathrm{AC}}^2\\ \Rightarrow {\mathrm{DA}}^2+{\mathrm{AB}}^2+{\mathrm{AD}}^2+{\mathrm{DC}}^2+2\mathrm{EA}\times \mathrm{AB}-2\mathrm{DC}\times \mathrm{FD}={\mathrm{DB}}^2+{\mathrm{AC}}^2\\ \Rightarrow {\mathrm{BC}}^2+{\mathrm{AB}}^2+{\mathrm{AD}}^2+{\mathrm{DC}}^2+2\mathrm{EA}\times \mathrm{AB}-2\mathrm{AB}\times \mathrm{EA}={\mathrm{DB}}^2+{\mathrm{AC}}^2\\ \text{(Using equations (iv), (v) and (vi))}\\ \Rightarrow {\mathrm{AB}}^2+{\mathrm{BC}}^2+{\mathrm{CD}}^2+{\mathrm{DA}}^2={\mathrm{AC}}^2+{\mathrm{BD}}^2\\ \mathrm{Hence},\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{squares}\mathbf{of}\mathbf{the}\mathbf{diagonals}\mathbf{of}\mathrm{a}\mathbf{parallelogram}\mathbf{is}\\ \mathbf{equal}\mathbf{to}\mathbf{the}\mathbf{sum}\mathbf{of}\mathbf{the}\mathbf{squares}\mathbf{of}\mathbf{its}\mathbf{sides}.\end{array}$

Q.62

$\begin{array}{l}\text{In the following figure, two chords AB and CD intersect each other at the point P. Prove that :}\\ \text{(i)}\mathrm{\Delta }\text{APC}~\mathrm{\Delta }\text{DPB (ii) AP.PB}=\text{CP.DP}\end{array}$

Ans.

$\begin{array}{l}\text{Let us join CB.}\\ \text{(i) In}\mathrm{\Delta }\text{APC and}\mathrm{\Delta }\text{DPB,}\\ \angle \text{APC}=\angle \text{DPB [Vertically opposite angles]}\\ \angle \text{CAP}=\angle \text{BDP [Angles in same segment for chord CB]}\\ \therefore \mathrm{\Delta }\text{APC}~\mathrm{\Delta }\text{DPB [By AA similarity criterion]}\\ \text{(ii)}\\ \text{We know that corresponding sides of similar triangles}\\ \text{are proportional.}\\ \text{Therefore,}\\ \mathrm{\Delta }\text{APC}~\mathrm{\Delta }\text{DPB}\\ \Rightarrow \frac{\text{AP}}{\text{DP}}=\frac{\text{PC}}{\text{PB}}\\ \Rightarrow \text{AP}\cdot \text{PB}=\text{PC}\cdot \text{DP}\end{array}$

Q.63

$\begin{array}{l}\text{In the following figure, two chords AB and CD of a circle intersect each other at the point P}\\ \text{(when produced) outside the circle. Prove that}\\ \text{(i)}\mathrm{\Delta }\text{PAC}~\mathrm{\Delta }\text{PDB (ii)}\mathrm{PA}\cdot \mathrm{PB}=\mathrm{PC}\cdot \mathrm{PD}\end{array}$

Ans.

$\begin{array}{l}\text{(i)}\\ \text{In}\mathrm{\Delta }\text{PAC and}\mathrm{\Delta }\text{PDB,}\\ \angle \text{P}=\angle \text{P (common)}\\ \angle \text{PAC}=\angle \text{PDB}\\ \text{(Exterior angle of a cyclic quadrilateral is}\\ \text{equal to opposite interior angle)}\\ \mathrm{\Delta }\text{PAC}~\mathrm{\Delta }\text{PDB}\\ \text{(ii)}\\ \text{We know that corresponding sides of similar triangles}\\ \text{are proportional.}\\ \therefore \mathrm{\Delta }\text{PAC}~\mathrm{\Delta }\text{PDB}\\ \Rightarrow \frac{\text{PA}}{\text{PD}}=\frac{\text{PC}}{\text{PB}}\\ \Rightarrow \text{PA}\cdot \text{PB}=\text{PC}\cdot \text{PD}\end{array}$

Q.64 In the following figure, D is a point on side BC of ΔABC such that BD CD = AB AC . Prove that AD is the bisector of ∠BAC.

Ans.

$\begin{array}{l}\text{Let us extend BA to P such that AP}=\text{AC. Join PC.}\\ \text{It is given that}\\ \frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AC}}\\ \Rightarrow \frac{\text{BD}}{\text{CD}}=\frac{\text{AB}}{\text{AP}}\\ \text{Therefore, by converse of basic proportinality theorem, AD}\parallel \text{PC.}\\ \text{Therefore},\\ \angle \text{BAD}=\angle \text{APC}...\text{(1)}\\ \angle \text{DAC}=\angle \text{ACP}...\text{(2)}\\ \text{Also,}\\ \text{\hspace{0.17em}\hspace{0.17em} AP}=\text{AC}\\ \Rightarrow \angle \text{APC}=\angle \text{ACP}\\ \Rightarrow \angle \text{APC}=\angle \text{ACP}=\angle \text{DAC [From (2)]}\\ \Rightarrow \angle \text{BAD}=\angle \text{DAC \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[From (1)]}\\ \text{Therefore, AD is the bisector of angle BAC.}\end{array}$

Q.65 Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see the following figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Ans.

$\begin{array}{l}\text{Let AB be the height of the tip of fishing rod from water}\\ \text{surface.}\\ \text{Let BC be the horizontal distance of fly from the tip of}\\ \text{fishing rod.}\\ \text{AC is the length of string.}\\ \text{On applying Pythagoras theorem in}\mathrm{\Delta }\text{ABC, we get}\\ {\text{AC}}^{\text{2}}={\text{AB}}^{\text{2}}+{\text{BC}}^{\text{2}}\\ \Rightarrow {\text{AB}}^{\text{2}}={\left(\text{1.8}\right)}^{\text{2}}+{\left(\text{2.4}\right)}^{\text{2}}\\ \Rightarrow {\text{AB}}^{\text{2}}=\text{3.24}+\text{5.76}\\ \Rightarrow {\text{AB}}^{\text{2}}=\text{9.00}\\ \Rightarrow \text{AB}=\sqrt{\text{9}}=3\\ \text{Therefore, length of the string is 3 m.}\\ \text{It is given that string is pulled at the rate of 5 cm per second.}\\ \text{So, string pulled in 12 seconds}=12\times 5=60\text{cm}=0.6\text{m}\end{array}$

$\begin{array}{l}\text{Let after 12 second Fly be at point D.}\\ \text{Length of string out after 12 second is AD}\\ \text{AD\hspace{0.33em}=\hspace{0.33em}AC\hspace{0.33em}–\hspace{0.33em}string pulled by Nazima in 12 seconds}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}3.00\hspace{0.33em}–\hspace{0.33em}0.6}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2.4}\\ \text{In ΔADB,}\\ {\text{AB}}^{\text{2}}{\text{\hspace{0.33em}+\hspace{0.33em}BD}}^{\text{2}}{\text{\hspace{0.33em}=\hspace{0.33em}AD}}^{\text{2}}\\ \Rightarrow {\left(\text{1.8}\right)}^{\text{2}}{\text{\hspace{0.33em}+\hspace{0.33em}BD}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}}{\left(\text{2.4}\right)}^{\text{2}}\\ \Rightarrow {\text{BD}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}5.76\hspace{0.33em}}–\text{\hspace{0.33em}3.24\hspace{0.33em}=\hspace{0.33em}2.52}\\ \text{⇒ BD\hspace{0.33em}=\hspace{0.33em}1.587}\\ \text{Horizontal distance of fly\hspace{0.33em}=\hspace{0.33em}BD\hspace{0.33em}+\hspace{0.33em}1.2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}1.587\hspace{0.33em}+\hspace{0.33em}1.2}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2.787}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=\hspace{0.33em}2.79 m}\end{array}$