NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.5

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.5 are designed to help students understand and solve proportionality and similarity problems based on triangles. These solutions provide step-by-step guidance for applying theorems like the Basic Proportionality Theorem (BPT), AA (Angle-Angle) similarity criterion, and SSS (Side-Side-Side) similarity criterion to solve real-life problems and proofs.

Exercise 6.5 focuses on applications of the properties of similar triangles and using theorems in geometric problems. This exercise helps students practice problems involving:

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles Exercise 6.5

Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

$\begin{array}{l}\text{In the following figure, ABD is a triangle right angled\hspace{0.17em}at A and AC\hspace{0.33em}}\perp \mathrm{B}\text{D. }\\ \text{Show that}\\ \left(\text{i}\right){\text{ AB}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}}.\text{\hspace{0.33em}BD}\\ \left(\text{ii}\right){\text{ AC}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BC\hspace{0.33em}.\hspace{0.33em}DC}\\ \left(\text{iii}\right){\text{ AD}}^{\text{2}}\text{\hspace{0.33em}=\hspace{0.33em}BD\hspace{0.33em}.\hspace{0.33em}CD}\end{array}$

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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.

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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Triangles

Medium

Q.

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

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• Similar triangles

• Theorems on proportionality

• Using proportional sides and angles

By practicing this exercise, students gain strong reasoning and problem-solving skills, which are essential for both board exams and future mathematical understanding.

Q1. 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

Solution:

As per the Pythagorean theorem, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

(i) 7 cm, 24 cm, 25 cm

$$25^2 = 625, \quad 7^2 + 24^2 = 49 + 576 = 625$$

252=625,72+242=49+576=625

$\therefore$

∴ This is a right triangle.
Length of hypotenuse = 25 cm.

(ii) 3 cm, 8 cm, 6 cm

$$8^2 = 64, \quad 3^2 + 6^2 = 9 + 36 = 45$$

82=64,32+62=9+36=45

Since

$64 \neq 45$

64=45, this is not a right triangle.

(iii) 50 cm, 80 cm, 100 cm

$$100^2 = 10000, \quad 50^2 + 80^2 = 2500 + 6400 = 8900$$

1002=10000,502+802=2500+6400=8900

Since

$10000 \neq 8900$

10000=8900, this is not a right triangle.

(iv) 13 cm, 12 cm, 5 cm

$$13^2 = 169, \quad 12^2 + 5^2 = 144 + 25 = 169$$

132=169,122+52=144+25=169

$\therefore$

∴ This is a right triangle.
Length of hypotenuse = 13 cm.

Conclusion:
(i) and (iv) are right triangles.

Q2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that $PM^2 = QM \times MR$

PM2=QM×MR.

Solution:

As we know, if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

Let us apply similarity of triangles to show that

$PM^2 = QM \times MR$

PM2=QM×MR.
Using AA similarity, we have:

$$\triangle PQR \sim \triangle PQM, \quad \triangle PQR \sim \triangle PRM$$

△PQR∼△PQM,△PQR∼△PRM

By the property of similar triangles, we know that the corresponding sides are proportional:

$$\frac{PM}{QM} = \frac{PQ}{PR} \quad \text{and} \quad \frac{PM}{MR} = \frac{PR}{PQ}$$

QMPM​=PRPQ​andMRPM​=PQPR​

Thus, multiplying the corresponding sides of the two proportionalities, we get:

$$PM^2 = QM \times MR$$

PM2=QM×MR

Q3. In Fig. 6.53, ABD is a triangle right angled at A and AC ⊥ BD. Show that:

(i)

$AB^2 = BC \times BD$

AB2=BC×BD
(ii)

$AC^2 = BC \times DC$

AC2=BC×DC
(iii)

$AD^2 = BD \times CD$

AD2=BD×CD

Solution:

Using the property that if a perpendicular is drawn from the right angle of a triangle to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other, we can solve each part:

(i)

$$\frac{AB}{BD} = \frac{BC}{AB} \quad \text{(Using similarity of triangles)}$$

BDAB​=ABBC​(Using similarity of triangles)

By the property of similar triangles:

$$AB^2 = BC \times BD$$

AB2=BC×BD

(ii)

$$\frac{AC}{BC} = \frac{DC}{AC} \quad \text{(Using similarity of triangles)}$$

BCAC​=ACDC​(Using similarity of triangles)

By the property of similar triangles:

$$AC^2 = BC \times DC$$

AC2=BC×DC

(iii)

$$\frac{AD}{BD} = \frac{CD}{AD} \quad \text{(Using similarity of triangles)}$$

BDAD​=ADCD​(Using similarity of triangles)

By the property of similar triangles:

$$AD^2 = BD \times CD$$

AD2=BD×CD

Q4. ABC is an isosceles triangle right angled at C. Prove that $AB^2 = AC^2 + 2AC^2$

AB2=AC2+2AC2.

Solution:

In the given triangle

$ABC$

ABC, since it is isosceles with

$AC = BC$

AC=BC and right angled at

$C$

C, applying the Pythagorean theorem:

$$AB^2 = AC^2 + BC^2$$

AB2=AC2+BC2

Since

$AC = BC$

AC=BC, we substitute:

$$AB^2 = AC^2 + AC^2 = 2AC^2$$

AB2=AC2+AC2=2AC2

Thus,

$AB^2 = 2AC^2$

AB2=2AC2.

Q5. ABC is an isosceles triangle with AC = BC. If $AB^2 = AC^2 + 2AC^2$

AB2=AC2+2AC2, prove that $ABC$

$ABC$ is a right triangle.

Solution:

We are given that

$AC = BC$

AC=BC and

$AB^2 = AC^2 + 2AC^2$

AB2=AC2+2AC2.
Simplifying the equation:

$$AB^2 = AC^2 + 2AC^2 = 3AC^2$$

AB2=AC2+2AC2=3AC2

Now, applying the Pythagorean theorem, if

$ABC$

ABC is a right triangle, the sum of the squares of the two smaller sides should equal the square of the hypotenuse:

$$AB^2 = AC^2 + BC^2$$

AB2=AC2+BC2

Since

$AC = BC$

AC=BC, we get:

$$AB^2 = AC^2 + AC^2 = 2AC^2$$

AB2=AC2+AC2=2AC2

Since

$AB^2 = 3AC^2$

AB2=3AC2, we can conclude that

$ABC$

ABC is a right triangle because the equation satisfies the Pythagorean theorem.

Q6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.

Solution:

In an equilateral triangle, the altitude bisects the side. So, the side length of the equilateral triangle is

$2a$

2a.
Let the altitude be

$AD$

AD, which bisects

$BC$

BC.
In the right triangle

$ABD$

ABD, applying the Pythagorean theorem:

$$AB^2 = AD^2 + BD^2$$

AB2=AD2+BD2

Since

$BD = a$

BD=a (half of the side length):

$$(2a)^2 = AD^2 + a^2$$

(2a)2=AD2+a2

$$4a^2 = AD^2 + a^2$$

4a2=AD2+a2

$$AD^2 = 3a^2$$

AD2=3a2

$$AD = \sqrt{3}a$$

AD=3​a

Thus, the altitude is

$AD = \sqrt{3}a$

AD=3​a.

Q7. Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

Solution:

In a rhombus, the diagonals bisect each other at right angles.
Let the diagonals of the rhombus be

$AC$

AC and

$BD$

BD.
In triangle

$ABC$

ABC, applying the Pythagorean theorem:

$$AB^2 = AC^2 + BC^2$$

AB2=AC2+BC2

Since the diagonals bisect each other, we have:

$$AB^2 + BC^2 + CD^2 + DA^2 = AC^2 + BD^2$$

AB2+BC2+CD2+DA2=AC2+BD2

Thus, the sum of the squares of the sides is equal to the sum of the squares of the diagonals.

FAQs: Class 10 Maths Chapter 6 – Triangles Exercise 6.5

Q1. What is the focus of Exercise 6.5?
Answer:
Exercise 6.5 focuses on solving problems based on the properties of similar triangles, including the application of the Basic Proportionality Theorem (BPT) and the similarity criteria such as AA, SAS, and SSS.

Q2. What are similar triangles?
Answer:
Two triangles are said to be similar if their corresponding angles are equal, and their corresponding sides are proportional. This means the ratio of corresponding sides is the same.

Q3. What is the Basic Proportionality Theorem (BPT)?
Answer:
The Basic Proportionality Theorem (BPT) states that if a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Q4. How do I apply similarity properties in solving problems?
Answer:
To solve problems:

• Identify the similar triangles in the given figure.

• Use proportionality (i.e., corresponding sides are proportional).

• Apply similarity criteria such as AA (Angle-Angle), SAS (Side-Angle-Side), and SSS (Side-Side-Side).

Q5. How do NCERT Solutions help with exam preparation?
Answer:
These solutions provide step-by-step solutions and clear explanations to help students understand how to apply similarity theorems in geometry problems. This improves their problem-solving abilities and helps them prepare well for board exams.