# NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3

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**NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles (Ex 9.3) Exercise 9.3**

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**Access NCERT solutions for Class 9 Maths Chapter 9 – Areas of Parallelograms and Triangles**

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**NCERT Solutions for Class 9 Maths Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3**

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**Access Other Exercise Solutions of Class 9 Maths Chapter 9 – Area of Parallelograms and Triangles**

Students learn about the concept of area and its associated formulae in Chapter 9 “Areas of Parallelograms and Triangles,” as well as a wide range of topics, such as figures on the same base and between the same parallels, parallelograms on the same base and also parallelograms between the same parallels, and triangles on the same base and between the same parallels. The uses of the concept Areas of Parallelograms and Triangles are countless. If students choose to continue studying Mathematics, they will use concepts learned in this chapter to solve problems in a variety of themes like Probability and Trigonometry. Students will get an understanding of these formulae by using the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.3 and applying them when answering problems in subsequent exercises. When answering related questions and understanding topics like Areas of Parallelograms and Triangles, they would have an advantage if they had previously studied the implicit formulae. Students must use the NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.3 for any assistance they need with Chapter 9 Areas of Parallelograms and Triangles Exercise 9.3 of Class 9. The NCERT Solutions For Class 9 Maths Chapter 9 Exercise 9.3 can always be used by students to help them complete their assignments.

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**Q.1 **In the given figure, E is any point on median AD of a ∆ABC. Show that ar(ABE) = ar(ACE).

**Ans**

$\begin{array}{l}\mathrm{Given}:\mathrm{In}\text{}\mathrm{\Delta}\text{ABC, AD is median and E is any point on AD.}\\ \text{To prove :}\mathrm{ar}\left(\mathrm{ABE}\right)=\mathrm{ar}\left(\mathrm{ACE}\right)\\ \text{Proof: In}\mathrm{\Delta ABC},\text{AD is median.}\\ \text{So, ar}\left(\mathrm{\Delta ABD}\right)=\text{ar}\left(\mathrm{\Delta ACD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta EBC},\text{ED is median.}\\ \text{So, ar}\left(\mathrm{\Delta EBD}\right)=\text{ar}\left(\mathrm{\Delta ECD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{Subtracting equation}\left(\mathrm{ii}\right)\text{from equation}\left(\mathrm{i}\right),\text{we get}\\ \text{ar}\left(\mathrm{\Delta ABD}\right)-\text{ar}\left(\mathrm{\Delta EBD}\right)=\text{ar}\left(\mathrm{\Delta ACD}\right)-\text{ar}\left(\mathrm{\Delta ECD}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Area}\left(\mathrm{\Delta}\text{ABE}\right)=\text{Area}\left(\mathrm{\Delta}\text{ACE}\right)\end{array}$

**Q.2 **

$\begin{array}{l}\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{ABC},\mathrm{E}\mathrm{is}\mathrm{the}\mathrm{mid}\u2013\mathrm{point}\mathrm{of}\mathrm{median}\mathrm{AD}.\mathrm{Show}\\ \mathrm{that}\mathrm{ar}\left(\mathrm{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right).\end{array}$

**Ans**

$\begin{array}{l}\text{Given:In}\mathrm{\Delta ABC},\text{AD is the median and E is the mid-point of AD.}\\ \mathrm{To}\text{prove :}\mathrm{ar}\left(\mathrm{BED}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Proof : In}\mathrm{\Delta ABC},\text{AD is median.}\\ \text{So, ar}\left(\mathrm{\Delta ABD}\right)=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}..\left(\mathrm{i}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta BDA},\text{BE is median.}\\ \text{So, ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\text{Median divides a triangle into two triangles of equal areas.}\right]\\ \text{From equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{2}\left\{\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\right\}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BED}\right)=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \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}\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}Hence proved.}\end{array}$

**Q.3 **Show that the diagonals of a parallelogram divide it into four triangles of equal area.

**Ans**

Given: ABCD is a parallelogram; in which diagonals AC and BD bisect each other at O. To prove: area(DAOB)= area(DBOC)= area(DCOD)= area(DDOA)

**Q.4 **In the figure shown below, ABC and ABD are two triangles on the same base AB.

If line- segment CD is bisected by AB at O, show that ar(ABC) = ar (ABD).

**Ans**

$\begin{array}{l}\text{Given:\Delta ABC and}\mathrm{\Delta ADB}\text{have same base AB and OC}=\text{OD.}\\ \text{To prove:}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta ABD}\right)\\ \text{Proof:In}\mathrm{\Delta ADC},\text{AO is median, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area}\left(\mathrm{\Delta AOC}\right)=\text{area}\left(\mathrm{\Delta AOD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\text{Median divides a triangle into two triangles of equal areas.}\right]\\ \text{In}\mathrm{\Delta BDC},\text{BO is median, so}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}area}\left(\mathrm{\Delta BOC}\right)=\text{area}\left(\mathrm{\Delta BOD}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Median}\text{divides a triangle into two triangles of equal areas.}\right]\\ \text{Adding equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{area}\left(\mathrm{\Delta AOC}\right)+\text{area}\left(\mathrm{\Delta BOC}\right)=\text{area}\left(\mathrm{\Delta AOD}\right)+\text{area}\left(\mathrm{\Delta BOD}\right)\\ \text{area}\left(\mathrm{\Delta ABC}\right)=\text{area}\left(\mathrm{\Delta DAB}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

**Q.5 **

$\begin{array}{l}\mathrm{D},\mathrm{E}\mathrm{and}\mathrm{F}\mathrm{are}\mathrm{respectively}\mathrm{the}\mathrm{mid}\u2013\mathrm{points}\mathrm{of}\mathrm{the}\mathrm{sides}\\ \mathrm{BC},\mathrm{CA}\mathrm{and}\mathrm{AB}\mathrm{of}\mathrm{a}\mathrm{\Delta ABC}.\mathrm{Show}\mathrm{that}\\ \left(\mathrm{i}\right)\mathrm{BDEF}\mathrm{is}\mathrm{a}\mathrm{parallelogram}.\\ \left(\mathrm{ii}\right)\mathrm{ar}\left(\mathrm{DEF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{BDEF}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\end{array}$

**Ans**

$\begin{array}{l}\text{Given}:\text{In}\mathrm{\Delta}\text{ABC, D, E and F are the mid-points of BC, CA and AB}\\ \text{respectively.}\\ \text{To prove:}\left(\mathrm{i}\right)\mathrm{BDEF}\text{is a parallelogram.}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{ii}\right)\text{\hspace{0.17em}}\mathrm{ar}\left(\mathrm{DEF}\right)=\frac{1}{4}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\mathrm{ar}\left(\mathrm{BDEF}\right)=\frac{1}{2}\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Proof}:\left(\mathrm{i}\right)\text{In}\mathrm{\Delta}\text{ABC, F and E are the mid-points of AB and AC}\\ \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}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}respectively, so by mid-point theorem, we have}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} FE}=\frac{1}{2}\mathrm{BC}\text{and FE}\parallel \text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow \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}}\mathrm{FE}=\mathrm{BD}\text{and FE}\parallel \mathrm{BD}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\because \mathrm{D}\text{is the mid-point}\\ \text{of BC.}\end{array}\right]\\ \text{BDEF is a parallelogram because one pair of opposite sides}\\ \text{is parallel and equal.}\\ \text{(ii) Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}FE}=\frac{1}{2}\mathrm{BC}\text{and FE}\parallel \text{BC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{So},\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}\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}}\mathrm{FE}=\mathrm{DC}\text{and FE}\parallel \text{DC}\\ \text{Thus, EFDC is a parallelogram.\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Since},\text{one pair of opposite sides}\\ \text{is equal and parallel.}\end{array}\right]\\ \text{Similarly},\text{\hspace{0.17em}DEAF is a parallelogram.}\\ \text{Since},\text{diagonal of a parallelogram divides it into two congruent}\\ \text{triangles. So,}\\ \text{ar}\left(\mathrm{\Delta AFE}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\text{, ar}\left(\mathrm{\Delta BFD}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\text{and}\\ \text{ar}\left(\mathrm{\Delta DFE}\right)=\text{ar}\left(\mathrm{\Delta DCE}\right)\\ \Rightarrow \text{\hspace{0.17em}ar}\left(\mathrm{\Delta AFE}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)=\text{ar}\left(\mathrm{\Delta BFD}\right)=\text{ar}\left(\mathrm{\Delta DFE}\right)\\ \Rightarrow \text{ar}\left(\mathrm{\Delta DFE}\right)=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ [\because \mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\text{ar}\left(\mathrm{\Delta AFE}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)+\text{ar}\left(\mathrm{\Delta BFD}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)]\\ \left(\mathrm{iii}\right)\mathrm{a}\mathrm{r}\left(\mathrm{BDEF}\right)=\text{ar}\left(\mathrm{\Delta BFD}\right)+\text{ar}\left(\mathrm{\Delta DFE}\right)\\ \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}\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}}=\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)+\frac{1}{4}\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \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}\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}}=\frac{1}{2}\text{ar}\left(\mathrm{\Delta ABC}\right)\end{array}$

**Q.6** In figure shown below, diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. If AB = CD, then show that:

(i) ar (DOC)= ar (AOB)

(ii) ar (DCB)= ar (ACB)

(iii) DA||CB or ABCD is a parallelogram.

**Ans**

$\begin{array}{l}\text{Given:In quadrilateral ABCD, AC and BD are diagonals which}\\ \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}\hspace{0.17em} \hspace{0.17em}intersect at O,}\mathrm{OB}\text{}=\text{}\mathrm{OD}\text{and}\mathrm{AB}\text{}=\text{}\mathrm{CD}.\\ \text{To prove:\hspace{0.17em}\hspace{0.17em}}\left(\mathrm{i}\right)\text{}\mathrm{ar}\text{}\left(\mathrm{DOC}\right)=\text{}\mathrm{ar}\text{}\left(\mathrm{AOB}\right)\mathrm{}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{ar}\text{}\left(\mathrm{DCB}\right)=\text{}\mathrm{ar}\text{}\left(\mathrm{ACB}\right)\\ \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}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\left(\mathrm{iii}\right)\text{}\mathrm{DA}\left|\right|\mathrm{CB}\text{}\mathrm{or}\text{}\mathrm{ABCD}\text{}\mathrm{is}\text{}\mathrm{a}\text{}\mathrm{parallelogram}.\\ \text{Construction}:\text{Draw BM}\perp \text{AC and DN}\perp \text{AC.}\\ \text{Proof:}\left(\mathrm{i}\right)\text{In}\mathrm{\Delta BOM}\text{and}\mathrm{\Delta DON}\\ \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}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BOM}=\angle \mathrm{DON}\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}\hspace{0.17em}}\left[\mathrm{Vertical}\text{opposite angles}\right]\\ \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}\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{BMO}=\angle \mathrm{DNO}\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}\hspace{0.17em}}\left[\mathrm{each}\text{\hspace{0.17em}90\xb0}\right]\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{OB}=\mathrm{OD}\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}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\therefore \mathrm{\Delta BOM}\cong \mathrm{\Delta DON}\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}}\left[\mathrm{By}\text{ A.A.S}\right]\\ \Rightarrow \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{BM}=\mathrm{DN}\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}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{C.P.C.T.}\right]\\ \text{and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BOM}\right)=\mathrm{ar}\left(\mathrm{\Delta DON}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \mathrm{In}\text{}\mathrm{\Delta BMA}\text{and}\mathrm{\Delta DNC}\\ \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{BM}=\mathrm{DN}\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}\hspace{0.17em}}\left[\text{Proved above}\right]\\ \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}\hspace{0.17em}}\angle \mathrm{BMA}=\angle \mathrm{DNC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{each}\text{\hspace{0.17em}90\xb0}\right]\\ \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}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}}\mathrm{AB}=\mathrm{CD}\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}\hspace{0.17em}}\left[\mathrm{Given}\right]\\ \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}}\mathrm{\Delta BMA}\cong \mathrm{\Delta DNC}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{By}\text{R.H.S.}\right]\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BMA}\right)=\mathrm{ar}\left(\mathrm{\Delta DNC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \mathrm{ar}\left(\mathrm{\Delta BOM}\right)+\mathrm{ar}\left(\mathrm{\Delta BMA}\right)=\mathrm{ar}\left(\mathrm{\Delta DON}\right)+\mathrm{ar}\left(\mathrm{\Delta DNC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOB}\right)=\mathrm{ar}\left(\mathrm{\Delta DOC}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)\\ \left(\mathrm{ii}\right)\text{Since},\text{}\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)\\ \mathrm{Adding}\text{ar}\left(\mathrm{\Delta BOC}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta DOC}\right)+\text{ar}\left(\mathrm{\Delta BOC}\right)=\mathrm{ar}\left(\mathrm{\Delta AOB}\right)+\text{ar}\left(\mathrm{\Delta BOC}\right)\\ \Rightarrow \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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta DCB}\right)=\mathrm{ar}\left(\mathrm{\Delta ABC}\right)\\ \left(\mathrm{iii}\right)\text{ar}\left(\mathrm{\Delta DCB}\right)=\text{ar}\left(\mathrm{\Delta ABC}\right)\\ \text{If two triangles having same base and equal area, then}\\ \text{they lie between same parallels.}\\ \text{So, BC}\parallel \text{AD}\\ \text{or DA}\parallel \mathrm{BC}\\ \mathrm{As}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta AOB}\cong \mathrm{\Delta COD}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}}\angle \mathrm{OAB}=\angle \mathrm{OCD}\\ \Rightarrow \angle \mathrm{CAB}=\angle \mathrm{ACD}\\ \Rightarrow \mathrm{AB}\parallel \mathrm{CD}\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}\hspace{0.17em}}\left[\mathrm{Alternate}\text{angles are equal.}\right]\\ \text{So},\text{ABCD is a parallelogram because opposite side}\\ \text{are parallel.}\end{array}$

**Q.7 **D and E are points on sides AB and AC respectively of ΔABC such that ar(DBC)=ar(EBC). Prove that DE || BC.

**Ans**

$\begin{array}{l}\text{Given : In\hspace{0.17em}\hspace{0.17em}}\mathrm{\Delta ABC},\text{D and E are the points on AB and AC}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}respectively and}\mathrm{ar}\left(\mathrm{DBC}\right)=\mathrm{ar}\left(\mathrm{EBC}\right).\\ \text{To prove : DE}\parallel \text{BC}\\ \text{Proof : Since,}\mathrm{ar}\left(\mathrm{DBC}\right)=\mathrm{ar}\left(\mathrm{EBC}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and both triangles have same base BC.}\\ \text{So, DE}\parallel \text{BC\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\because \mathrm{If}\text{two triangles have equal area and same}\\ \text{base, then they lie between same parallels.}\end{array}\right]\end{array}$

**Q.8 **XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ABE)= ar(ACF).

**Ans**

$\begin{array}{l}\text{Given:In}\mathrm{\Delta ABC},\text{XY}\parallel \mathrm{BC},\text{BE}\parallel \mathrm{AC}\text{and CF}\parallel \mathrm{AB}.\\ \text{To prove:\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABE}\right)=\text{}\mathrm{ar}\left(\mathrm{ACF}\right)\\ \text{Proof:Since, parallelograms BCYE and BCFX have same base}\\ \text{BC and lie between same parallels.So,}\\ \text{ar}\left(\text{BCYE}\right)=\mathrm{ar}\left(\text{BCFX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \text{\Delta AEB and parallelogram BCYE have same base and lie between}\\ \text{same parallels. So,}\\ \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}ar}\left(\mathrm{\Delta ABE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCYE}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}..\left(\mathrm{ii}\right)\\ \text{Similarly,}\\ \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}ar}\left(\mathrm{\Delta ACF}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCFX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right),\text{we have}\\ \text{}\frac{1}{2}\text{ar}\left(\text{BCYE}\right)=\frac{1}{2}\mathrm{ar}\left(\text{BCFX}\right)\\ \text{From equation}\left(\mathrm{ii}\right)\text{and equation}\left(\mathrm{iii}\right),\text{we have}\\ \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}ar}\left(\mathrm{\Delta ABE}\right)=\text{ar}\left(\mathrm{\Delta ACF}\right)\\ \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}\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}Hence Proved.}\end{array}$

**Q.9 **The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q and then parallelogram PBQR is completed (see the figure below). Show that ar (ABCD) = ar (PBQR).

**Ans**

$\begin{array}{l}\mathrm{Given}:\mathrm{ABCD}\text{and PBQR are parallelogram. AQ}\parallel \mathrm{CP}.\\ \mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)\\ \mathrm{Construction}:\text{Join AC and PQ.}\\ \text{Proof: Since, AQ}\parallel \text{CP}\\ \text{}\mathrm{\Delta ACP}\text{and}\mathrm{\Delta QPC}\text{have same base CP and both triangles are}\\ \text{between same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ACP}\right)=\mathrm{ar}\left(\mathrm{\Delta QPC}\right)\\ \mathrm{Subtracting}\text{ar}\left(\mathrm{\Delta BCP}\right)\text{from both sides, we get}\\ \mathrm{ar}\left(\mathrm{\Delta ACP}\right)-\text{ar}\left(\mathrm{\Delta BCP}\right)=\mathrm{ar}\left(\mathrm{\Delta QPC}\right)-\text{ar}\left(\mathrm{\Delta BCP}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ABC}\right)=\mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\times \mathrm{ar}\left(\mathrm{\Delta ABC}\right)=2\times \mathrm{ar}\left(\mathrm{\Delta QBP}\right)\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{PBQR}\right)\\ [\because \mathrm{Diagonal}\text{divides a parallelogram in two triangles of equal area.}]\end{array}$

**Q.10 **Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at O. Prove that ar(AOD) = ar(BOC).

**Ans**

$\begin{array}{l}\text{Given:In trapezium ABCD, AB}\parallel \text{CD.}\\ \text{To prove: ar (AOD) = ar (BOC)}\\ \text{Proof: Area of two triangles is equal if they have same base}\\ \text{and lie between same parallels.}\\ \text{Here, AB}\parallel \text{CD,}\mathrm{\Delta ADC}\text{and}\mathrm{\Delta BDC}\text{have same base.}\\ \therefore \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta ADC}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)\\ \mathrm{\Rightarrow}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOD}\right)+\mathrm{ar}\left(\mathrm{\Delta DOC}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)+\mathrm{ar}\left(\mathrm{\Delta DOC}\right)\\ \mathrm{\Rightarrow}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta AOD}\right)=\mathrm{ar}\left(\mathrm{\Delta BDC}\right)\end{array}$

**Q.11 **In figure shown below, ABCDE is a pentagon.

A line through B parallel to AC meets DC produced at F. Show that

(i) ar(ACB) = ar (ACF)

(ii) ar(AEDF) = ar (ABCDE)

**Ans**

$\begin{array}{l}\text{Given: ABCDE is a pentagon and BF}\parallel \mathrm{AC}.\\ \text{To prove :}\left(\mathrm{i}\right)\text{}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\mathrm{}\\ \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}\hspace{0.17em} \hspace{0.17em}}\left(\mathrm{ii}\right)\text{}\mathrm{a}\mathrm{r}\left(\mathrm{AEDF}\right)=\mathrm{ar}\left(\mathrm{ABCDE}\right)\\ \text{Proof:}\left(\mathrm{i}\right)\text{}\mathrm{Since},\text{\hspace{0.17em}}\mathrm{AC}\parallel \mathrm{BF},\text{}\mathrm{\Delta ACB}\text{and}\mathrm{\Delta ACF}\text{have same base AC.}\\ \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}So, \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\\ \left[\begin{array}{l}\text{Areas of two triangles are equal if they have same base}\\ \text{and lie between same parallels.}\end{array}\right]\\ \left(\mathrm{ii}\right)\mathrm{W}\mathrm{e}\text{proved above:}\\ \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}\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}}\mathrm{ar}\left(\mathrm{ACB}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)\\ \text{Adding ar}\left(\mathrm{AEDC}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{ACB}\right)+\text{ar}\left(\mathrm{AEDC}\right)=\mathrm{ar}\left(\mathrm{ACF}\right)+\text{ar}\left(\mathrm{AEDC}\right)\\ \Rightarrow \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}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{ABCDE}\right)=\text{ar}\left(\mathrm{AEDF}\right)\\ \mathrm{or}\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}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{AEDF}\right)=\text{ar}\left(\mathrm{ABCDE}\right)\\ \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}\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}Hence proved.}\end{array}$

**Q.12** A villager Itwaari has a plot of land of the shape of a quadrilateral. The Gram Panchayat of the village decided to take over some portion of his plot from one of the corners to construct a Health Centre. Itwaari agrees to the above proposal with the condition that he should be given equal amount of land in lieu of his land adjoining his plot so as to form a triangular plot. Explain how this proposal will be implemented.

**Ans**

$\begin{array}{l}\mathrm{Given}:\mathrm{Let}\text{ABCD be the shape of the plot of land of Itwaari and}\\ \text{AED is his adjacent plot. ED be the side of his adjoining land.}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{To}\text{prove: ar}\left(\mathrm{\Delta AOB}\right)=\text{ar}\left(\mathrm{\Delta EOD}\right)\\ \mathrm{Construction}:\text{}\\ \text{The proposal may be implemented as follows}.\mathrm{}\\ \text{Join diagonal BD and draw a line parallel to BD through point A}.\text{}\\ \text{Let it meet}\mathrm{}\text{the extended side CD of ABCD at point E}.\text{Join BE and}\\ \text{AD}.\text{Let them intersect each other at O}.\text{Then},\text{portion}\mathrm{\Delta}\text{AOB can}\\ \text{be cut from the original field so that the new shape of the field}\\ \text{will be}\mathrm{\Delta}\text{BCE}.\end{array}$

$\begin{array}{l}\text{Proof: Since,}\mathrm{\Delta EDB}\text{}\mathrm{and}\text{}\mathrm{\Delta ADB}\text{have same base and lie}\\ \text{between same parallels. So,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta EDB}\right)=\text{ar}\left(\mathrm{\Delta ADB}\right)\\ \mathrm{Subtracting}\text{ar}\left(\mathrm{\Delta BOD}\right)\text{from both sides, we get}\\ \text{ar}\left(\mathrm{\Delta EDB}\right)-\text{ar}\left(\mathrm{\Delta BOD}\right)=\text{ar}\left(\mathrm{\Delta ADB}\right)-\text{ar}\left(\mathrm{\Delta BOD}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ar}\left(\mathrm{\Delta EOD}\right)=\text{ar}\left(\mathrm{\Delta BOA}\right)\\ \mathrm{Adding}\text{ar}\left(\mathrm{BCDO}\right)\text{both sides, we get}\\ \text{ar}\left(\mathrm{\Delta EOD}\right)+\text{ar}\left(\mathrm{BCDO}\right)=\text{ar}\left(\mathrm{\Delta BOA}\right)+\text{ar}\left(\mathrm{BCDO}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \hspace{0.17em}}\mathrm{ar}\left(\mathrm{\Delta BCE}\right)=\mathrm{ar}\left(\mathrm{ABCD}\right)\\ \mathrm{Thus},\text{we can replace land in the shape of}\mathrm{\Delta BOA}\text{by the land}\\ \text{in the shape of}\mathrm{\Delta EOD}\text{to implement the proposal of Itwaari.}\end{array}$

**Q.13 **ABCD is a trapezium with AB || DC. A line parallel to AC intersects AB at X and BC at Y. Prove that ar (ADX) = ar(ACY).

**Ans**

$\begin{array}{l}\text{Given}:\text{}\mathrm{ABCD}\text{is a trapezium in which AB}\parallel \text{CD and XY}\parallel \mathrm{AC}.\\ \mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{ADX}\right)=\mathrm{ar}\left(\mathrm{ACY}\right)\\ \text{Construction}:\mathrm{Join}\text{CX.}\\ \text{Proof:Since, AB}\parallel \mathrm{CD},\text{then}\\ \text{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACX}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{\hspace{0.17em}}\left(\mathrm{i}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\begin{array}{l}\mathrm{Both}\text{}\mathrm{\Delta s}\text{have same base and}\\ \text{lie between same parallels.}\end{array}\right]\\ \text{and AC}\parallel \mathrm{XY},\text{then}\\ \text{ar}\left(\mathrm{\Delta ACX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\text{\hspace{0.17em}}\left(\mathrm{ii}\right)\\ \text{From equation}\left(\mathrm{i}\right)\text{and equqation}\left(\mathrm{ii}\right),\text{we have}\\ \text{ar}\left(\mathrm{\Delta ADX}\right)=\mathrm{ar}\left(\mathrm{\Delta ACY}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Hence proved.}\end{array}$

**Q.14 **In Fig.9.28, AP||BQ||CR. Prove that ar(AQC) = ar(PBR).

**Ans**

$\mathrm{Given}:\mathrm{In}\text{given figure,}\mathrm{AP}\left|\left|\mathrm{BQ}\right|\right|\mathrm{CR}$

$\begin{array}{l}\mathrm{To}\text{prove:}\mathrm{ar}\left(\mathrm{AQC}\right)=\mathrm{ar}\left(\mathrm{PBR}\right)\\ \text{Proof:Since,\hspace{0.17em}\hspace{0.17em}}\mathrm{AP}\left|\left|\mathrm{BQ}\right|\right|\\ \text{So, ar}\left(\mathrm{\Delta BPQ}\right)=\text{ar}\left(\mathrm{\Delta ABQ}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{i}\right)\\ \left[\mathrm{Both}\text{triangle}\mathrm{s}\text{have same base and lie between same parallels.}\right]\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{BQ}\parallel \mathrm{CR}\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}ar}\left(\mathrm{\Delta BQR}\right)=\text{ar}\left(\mathrm{\Delta BQC}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}...\left(\mathrm{ii}\right)\\ \left[\mathrm{Both}\text{triangle}\mathrm{s}\text{have same base and lie between same parallels.}\right]\\ \mathrm{Adding}\text{equation}\left(\mathrm{i}\right)\text{and equation}\left(\mathrm{ii}\right),\text{we get}\\ \text{ar}\left(\mathrm{\Delta BPQ}\right)+\text{ar}\left(\mathrm{\Delta BQR}\right)=\text{ar}\left(\mathrm{\Delta ABQ}\right)+\text{ar}\left(\mathrm{\Delta BQC}\right)\\ \Rightarrow \text{ar}\left(\mathrm{\Delta BPR}\right)=\text{ar}\left(\mathrm{\Delta AQC}\right)\\ \Rightarrow \text{\hspace{0.17em}ar}\left(\mathrm{\Delta AQC}\right)=\text{ar}\left(\mathrm{\Delta BPR}\right)\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{Hence}\text{proved.}\end{array}$

**Q.15 **Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar(AOD)=ar (BOC). Prove that ABCD is a trapezium.

**Ans**

$\begin{array}{l}\text{Given}:\text{ABCD is a quadrilateral and its diagonals AC and BD}\\ \text{intersect at O and}\mathrm{ar}\left(\mathrm{AOD}\right)=\mathrm{ar}\left(\mathrm{BOC}\right).\\ \mathrm{To}\text{prove:}\mathrm{ABCD}\text{is a trapezium.}\\ \text{Proof:}\mathrm{It}\text{ is given that}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{AOD}\right)=\mathrm{ar}\left(\mathrm{BOC}\right)\\ \text{Adding}\mathrm{ar}\left(\mathrm{AOB}\right)\text{both sides, we get}\\ \mathrm{ar}\left(\mathrm{AOD}\right)+\mathrm{ar}\left(\mathrm{AOB}\right)=\mathrm{ar}\left(\mathrm{BOC}\right)+\mathrm{ar}\left(\mathrm{AOB}\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{ar}\left(\mathrm{ABD}\right)=\mathrm{ar}\left(\mathrm{ABC}\right)\\ \text{Since},\text{\hspace{0.17em}both triagles have same base AB and having equal area.}\\ \text{So they will lie between same parallels.}\\ \text{Thus, AB}\parallel \text{CD}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}ABCD is a trapezium.}\end{array}$

**Q.16 **In the figure shown below, ar (DRC) = ar (DPC) and ar(BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

**Ans**

$\begin{array}{l}\text{Given}:\text{In given figure,}\\ \text{ar}\left(\text{DRC}\right)=\text{ar}\left(\text{DPC}\right)\text{and}\text{ar}\left(\text{BDP}\right)=\text{ar}\left(\text{ARC}\right).\\ \text{To prove: The}\text{quadrilaterals ABCD and DCPR are trapeziums}\text{.}\\ \text{Proof: It is given that}\\ \text{ar}\left(\text{DRC}\right)=\text{ar}\left(\text{DPC}\right)\\ or\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ar}\left(\text{DPC}\right)=\text{ar}\left(\text{DRC}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(i\right)\\ \text{Since, both triangles have same base and equal area}\text{. Then,}\\ \text{DC}\parallel \text{RP}\\ \Rightarrow \text{DCPR is a trapezium}\text{.}\\ \text{Given that}\\ \text{ar}\left(\text{BDP}\right)=\text{ar}\left(\text{ARC}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{\dots}\left(ii\right)\\ \text{Subtracting equation}\left(i\right)\text{from equation}\left(ii\right),\text{we get}\\ \text{ar}\left(\text{BDP}\right)-\text{ar}\left(\text{DPC}\right)=\text{ar}\left(\text{ARC}\right)-\text{ar}\left(\text{DRC}\right)\\ \Rightarrow \text{ar}\left(\text{BDC}\right)=\text{ar}\left(\text{ACD}\right)\\ \text{Since, both triangles have same base and equal area}\text{. Then,}\\ \text{AB}\parallel \text{DC}\\ \Rightarrow \text{ABCD is a trapezium}\text{.}\\ \text{In this way, we proved that DCPR and ABCD are trapeziums}\text{.}\end{array}$

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