# NCERT Solutions for Class 7 Maths Exercise 6.5

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NCERT solutions for Maths Triangles and its Properties

## NCERT Solutions for Class 7 Maths Triangles and its Properties

###### Question 1.PQR is a triangle, right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Given: PQ = 10 cm, PR = 24 cm

Let QR be {tex}x{/tex} cm.

In right angled triangle QPR,

{tex}{\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}{/tex}

[By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {QR} \right)^2} = {\text{ }}{\left( {PQ} \right)^2} + {\text{ }}{\left( {PR} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{x^2} = {\left( {10} \right)^2} + {\left( {24} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{x^2}{/tex} = 100 + 576 = 676

{tex} \Rightarrow {/tex} {tex}x = \sqrt {676} {/tex} = 26 cm

Thus, the length of QR is 26 cm.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 2.ABC is a triangle, right angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Given: AB = 25 cm, AC = 7 cm

Let BC be {tex}x{/tex} cm.

In right angled triangle ACB,

{tex}{\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}{/tex}

[By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {AB} \right)^2} = {\text{ }}{\left( {AC} \right)^2} + {\text{ }}{\left( {BC} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {25} \right)^2} = {\left( 7 \right)^2} + {x^2}{/tex}

{tex} \Rightarrow {/tex} 625 = 49 + {tex}{x^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{x^2}{/tex} = 625 – 49 = 576

{tex} \Rightarrow {/tex} {tex}x = \sqrt {576} {/tex} = 24 cm

Thus, the length of BC is 24 cm.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 3.A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance {tex}a.{/tex} Find the distance of the foot of the ladder from the wall.

Let AC be the ladder and A be the window.

Given: AC = 15 m, AB = 12 m, CB = {tex}a{/tex} m

In right angled triangle ACB,

{tex}{\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}{/tex}

[By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {AC} \right)^2} = {\text{ }}{\left( {CB} \right)^2} + {\text{ }}{\left( {AB} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {15} \right)^2} = {\left( a \right)^2} + {\left( {12} \right)^2}{/tex}

{tex} \Rightarrow {/tex} 225 = {tex}{a^2}{/tex} + 144

{tex} \Rightarrow {/tex} {tex}{a^2}{/tex} = 225 – 144 = 81

{tex} \Rightarrow {/tex} {tex}a = \sqrt {81} {/tex} = 9 cm

Thus, the distance of the foot of the ladder from the wall is 9 m.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 4.Which of the following can be the sides of a right triangle?
1. 2.5 cm, 6.5 cm, 6 cm
2. 2 cm, 2 cm, 5 cm
3. 1.5 cm, 2 cm, 2.5 cm

In the case of right angled triangles, identify the right angles.

Let us consider, the larger side be the hypotenuse and also using Pythagoras theorem,

{tex}{\left( {Hypotenuse} \right)^2} = {\text{ }}{\left( {Base} \right)^2} + {\text{ }}{\left( {Perpendicular} \right)^2}{/tex}

(i) 2.5 cm, 6.5 cm, 6 cm

In {tex}\Delta {\text{ABC}},{/tex} {tex}{\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}{/tex}

L.H.S. = {tex}{\left( {6.5} \right)^2}{/tex} = 42.25 cm

R.H.S. = {tex}{\left( 6 \right)^2} + {\left( {2.5} \right)^2}{/tex} = 36 + 6.25 = 42.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 6.5 cm, i.e., at B.

(ii) 2 cm, 2 cm, 5 cm

{tex}{\left( 5 \right)^2} = {\left( 2 \right)^2} + {\left( 2 \right)^2}{/tex}

L.H.S. = {tex}{\left( 5 \right)^2}{/tex} = 25

R.H.S. = {tex}{\left( 2 \right)^2} + {\left( 2 \right)^2}{/tex} = 4 + 4 = 8

Since, L.H.S. {tex} \ne {/tex} R.H.S.

Therefore, the given sides are not of the right angled triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

In {tex}\Delta {/tex}PQR, {tex}{\left( {{\text{PR}}} \right)^2} = {\left( {{\text{PQ}}} \right)^2} + {\left( {{\text{RQ}}} \right)^2}{/tex}

L.H.S. = {tex}{\left( {2.5} \right)^2}{/tex} = 6.25 cm

R.H.S. = {tex}{\left( {1.5} \right)^2} + {\left( 2 \right)^2}{/tex} = 2.25 + 4 = 6.25 cm

Since, L.H.S. = R.H.S.

Therefore, the given sides are of the right angled triangle.

Right angle lies on the opposite to the greater side 2.5 cm, i.e., at Q.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 5.A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Let A’CB represents the tree before it broken at the point C and let the top A’ touches the ground at A after it broke. Then {tex}\Delta {\text{ABC}}{/tex} is a right angled triangle, right angled at B.

AB = 12 m and BC = 5 m

Using Pythagoras theorem, In {tex}\Delta {\text{ABC}}{/tex}

{tex}{\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{BC}}} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{AC}}} \right)^2} = {\left( {12} \right)^2} + {\left( 5 \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{AC}}} \right)^2} = 144 + 25{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{AC}}} \right)^2} = 169{/tex}

{tex} \Rightarrow {/tex} AC = 13 m

Hence, the total height of the tree = AC + CB = 13 + 5 = 18 m.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 6.Angles Q and R of a {tex}\Delta {/tex}PQR are {tex}25^\circ {/tex} and {tex}65^\circ .{/tex}

Write which of the following is true:

1. {tex}P{Q^2} + {\text{ }}Q{R^2} = {\text{ }}R{P^2}{/tex}
2. {tex}P{Q^2} + {\text{ }}R{P^2} = {\text{ }}Q{R^2}{/tex}
3. {tex}R{P^2} + {\text{ }}Q{R^2} = {\text{ }}P{Q^2}{/tex}

{tex}25^\circ {/tex} {tex}65^\circ {/tex}

In {tex}\Delta {/tex}PQR,

{tex}\angle {/tex}PQR + {tex}\angle {/tex}QRP + {tex}\angle {/tex}RPQ = {tex}180^\circ {/tex}

[By Angle sum property of a {tex}\Delta {/tex} ]

{tex} \Rightarrow {/tex} {tex}25^\circ + 65^\circ + \angle {\text{RPQ = 180}}^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}90^\circ + \angle {\text{RPQ = 180}}^\circ {/tex}

{tex} \Rightarrow {/tex} {tex}\angle {/tex}RPQ = {tex}180^\circ – 90^\circ = 90^\circ {/tex}

Thus, {tex}\Delta {/tex}PQR is a right angled triangle, right angled at P.

{tex}\therefore {/tex} (Hypotenuse)2 = (Base)2 + (Perpendicular)2 [By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{QR}}} \right)^2} = {\left( {{\text{PR}}} \right)^2} + {\left( {{\text{QP}}} \right)^2}{/tex}

Hence, Option (ii) is correct.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 7.Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Given diagonal (PR) = 41 cm, length (PQ) = 40 cm

Let breadth (QR) be {tex}x{/tex} cm.

Now, in right angled triangle PQR,

{tex}{\left( {{\text{PR}}} \right)^2} = {\left( {{\text{RQ}}} \right)^2} + {\left( {{\text{PQ}}} \right)^2}{/tex}

[By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {41} \right)^2} = {x^2} + {\left( {40} \right)^2}{/tex}

{tex} \Rightarrow {/tex} 1681 = {tex}{x^2}{/tex} + 1600 {tex} \Rightarrow {/tex} {tex}{x^2}{/tex} = 1681 – 1600

{tex} \Rightarrow {/tex} {tex}{x^2}{/tex} = 81 {tex} \Rightarrow {/tex} {tex}x = \sqrt {81} = 9{/tex} cm

Therefore the breadth of the rectangle is 9 cm.

Perimeter of rectangle = 2(length + breadth)

= 2 (9 + 49)

= 2 x 49 = 98 cm

Hence the perimeter of the rectangle is 98 cm.

NCERT Solutions for Class 7 Maths Exercise 6.5

###### Question 8.The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Given: Diagonals AC = 30 cm and DB = 16 cm.

Since the diagonals of the rhombus bisect at right angle to each other.

Therefore, OD = {tex}\frac{{{\text{DB}}}}{2} = \frac{{16}}{2}{/tex} = 8 cm

And OC = {tex}\frac{{{\text{AC}}}}{2} = \frac{{30}}{2}{/tex} = 15 cm

Now, In right angle triangle DOC,

{tex}{\left( {{\text{DC}}} \right)^2} = {\left( {{\text{OD}}} \right)^2} + {\left( {{\text{OC}}} \right)^2}{/tex} [By Pythagoras theorem]

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{DC}}} \right)^2} = {\left( 8 \right)^2} + {\left( {15} \right)^2}{/tex}

{tex} \Rightarrow {/tex} {tex}{\left( {{\text{DC}}} \right)^2}{/tex} = 64 + 225 = 289

{tex} \Rightarrow {/tex} DC = {tex}\sqrt {289} {/tex} = 17 cm

Perimeter of rhombus = 4 x side

= 4 x 17 = 68 cm

Thus, the perimeter of rhombus is 68 cm.

## NCERT Solutions for Class 7 Maths Exercise 6.5

NCERT Solutions Class 7 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 7 Maths includes text book solutions from Class 7 Maths Book . NCERT Solutions for CBSE Class 7 Maths have total 15 chapters. 7 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 7 solutions PDF and Maths ncert class 7 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.

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### 18 thoughts on “NCERT Solutions for Class 7 Maths Exercise 6.5”

1. Thanks

2. Thanks

3. Thanks anamika tyagi

4. thanks

5. In a 7 question you write 2(9+49) but it is 2(9+40)

6. In question 7 how it come 49cm

7. Very good

8. Thanks????????????????????????????????????

9. Thanks????????????????????????????????????

10. 7th qns confused me
But it is ok ?

11. THANK YOU

12. Very useful and we can understand very easily.