# NCERT Solutions for Class 7 Maths Exercise 11.2

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NCERT solutions for Maths Perimeter and Area

## NCERT Solutions for Class 7 Maths Perimeter and Area

###### Question 1.Find the area of each of the following parallelograms:

We know that the area of parallelogram = base x height

(a) Here base = 7 cm and height = 4 cm

{tex}\therefore {/tex} Area of parallelogram = 7 x 4 = 28{tex}c{m^2}{/tex}

(b) Here base = 5 cm and height = 3 cm

{tex}\therefore {/tex} Area of parallelogram = 5 x 3 = 15 {tex}c{m^2}{/tex}

(c) Here base = 2.5 cm and height = 3.5 cm

{tex}\therefore {/tex} Area of parallelogram = 2.5 x 3.5 = 8.75 {tex}c{m^2}{/tex}

(d) Here base = 5 cm and height = 4.8 cm

{tex}\therefore {/tex} Area of parallelogram = 5 x 4.8 = 24 {tex}c{m^2}{/tex}

(e) Here base = 2 cm and height = 4.4 cm

{tex}\therefore {/tex} Area of parallelogram = 2 x 4.4 = 8.8 {tex}c{m^2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 2.Find the area of each of the following triangles:

We know that the area of triangle = {tex}\frac{1}{2}{/tex} x base x height

(a) Here, base = 4 cm and height = 3 cm

{tex}\therefore {/tex} Area of triangle = {tex}\frac{1}{2}{/tex} x 4 x 3 = 6 {tex}c{m^2}{/tex}

(b) Here, base = 5 cm and height = 3.2 cm

{tex}\therefore {/tex} Area of triangle = {tex}\frac{1}{2}{/tex} x 5 x 3.2 = 8 {tex}c{m^2}{/tex}

(c) Here, base = 3 cm and height = 4 cm

{tex}\therefore {/tex} Area of triangle = {tex}\frac{1}{2}{/tex} x 3 x 4 = 6{tex}c{m^2}{/tex}

(d) Here, base = 3 cm and height = 2 cm

{tex}\therefore {/tex} Area of triangle = {tex}\frac{1}{2}{/tex} x 3 x 2 = 3{tex}c{m^2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 3.Find the missing values:
 S. No. Base Hieght Area of the parallelogram a. 20 cm 246 {tex}c{m^2}{/tex} b. 15 cm 154.5{tex}c{m^2}{/tex} c. 84 cm 48.72 {tex}c{m^2}{/tex} d. 15.6 cm 16.38{tex}{m^2}{/tex}

We know that the area of parallelogram = base x height

(a) Here, base = 20 cm and area = 246 {tex}c{m^2}{/tex}

{tex}\therefore {/tex} Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 246 = 20 x height {tex} \Rightarrow {/tex} height = {tex}\frac{{246}}{{20}}{/tex} = 12.3 cm

(b) Here, height = 15 cm and area = 154.5 cm2

{tex}\therefore {/tex} Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 154.5 = base x 15 {tex} \Rightarrow {/tex} base = {tex}\frac{{154.5}}{{15}}{/tex} = 10.3 cm

(c) Here, height = 8.4 cm and area = 48.72{tex}c{m^2}{/tex}

{tex}\therefore {/tex} Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 48.72 = base x 8.4 {tex} \Rightarrow {/tex} base = {tex}\frac{{48.72}}{{8.4}}{/tex} = 5.8 cm

(d) Here, base = 15.6 cm and area = 16.38 {tex}c{m^2}{/tex}

{tex}\therefore {/tex} Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 16.38 = 15.6 x height {tex} \Rightarrow {/tex} height = {tex}\frac{{16.38}}{{15.6}}{/tex} = 1.05 cm

Thus, the missing values are:

 S. No. Base Hieght Area of the parallelogram a. 20 cm 12.3 cm 246 {tex}c{m^2}{/tex} b. 10.3 cm 15 cm 154.5{tex}c{m^2}{/tex} c. 5.8 cm 84 cm 48.72{tex}c{m^2}{/tex} d. 15.6 cm 1.05 16.38 {tex}c{m^2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 4.Find the missing values:
 Base Hieght Area of triangle 15 cm — 87 {tex}c{m^2}{/tex} — 31.4 mm 1256 {tex}m{m^2}{/tex} 22 cm — 170.5{tex}c{m^2}{/tex}

We know that the area of triangle = {tex}\frac{1}{2}{/tex} x base x height

In first row, base = 15 cm and area = 87 {tex}c{m^2}{/tex}

{tex}\therefore {/tex} 87 = {tex}\frac{1}{2}{/tex} x 15 x height {tex} \Rightarrow {/tex} height = {tex}\frac{{87 \times 2}}{{15}}{/tex} 11.6 cm

In second row, height = 31.4 mm and area = 1256 {tex}m{m^2}{/tex}

{tex}\therefore {/tex} 1256 = {tex}\frac{1}{2}{/tex} x base x 31.4 {tex} \Rightarrow {/tex} base = {tex}\frac{{1256 \times 2}}{{31.4}}{/tex} 80 mm

In third row, base = 22 cm and area = 170.5 {tex}c{m^2}{/tex}

{tex}\therefore {/tex} 170.5 = {tex}\frac{1}{2}{/tex} x 22 x height {tex} \Rightarrow {/tex} height = {tex}\frac{{170.5 \times 2}}{{22}}{/tex} 15.5 cm

Thus, the missing values are:

 Base Hieght Area of triangle 15 cm 11.6 cm 87 {tex}c{m^2}{/tex} 80 mm 31.4 mm 1256 {tex}m{m^2}{/tex} 22 cm 15.5 cm 170.5 {tex}c{m^2}{/tex}

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 5.PQRS is a parallelogram. QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find:

(a) the area of the parallelogram PRS

(b) QN, if PS = 8 cm

Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.

(a) Area of parallelogram = base x height

= 12 x 7.6 = 91.2 {tex}c{m^2}{/tex}

(b) Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 91.2 = 8 x QN {tex} \Rightarrow {/tex} QN = {tex}\frac{{91.2}}{8}{/tex} = 11.4 cm

Question 6.DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD. If the area of the parallelogram is 1470{tex}c{m^2}{/tex}, AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Given: Area of parallelogram = 1470{tex}c{m^2}{/tex}

Base (AB) = 35 cm and base (AD) = 49 cm

Since Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 1470 = 35 x DL {tex} \Rightarrow {/tex} DL = {tex}\frac{{1470}}{{35}}{/tex}

{tex} \Rightarrow {/tex} DL = 42 cm

Again, Area of parallelogram = base x height

{tex} \Rightarrow {/tex} 1470 = 49 x BM {tex} \Rightarrow {/tex} BM = {tex}\frac{{1470}}{{49}}{/tex}

{tex} \Rightarrow {/tex} BM = 30 cm

Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 7.{tex}\Delta {/tex}ABC is right angled at A. AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of {tex}\Delta {/tex}ABC. Also, find the length of AD.

In right angles triangle BAC, AB = 5 cm and AC = 12 cm

Area of triangle = {tex}\frac{1}{2}{/tex} x base x height = {tex}\frac{1}{2}{/tex} x AB x AC

= {tex}\frac{1}{2}{/tex} x 5 x 12 = 30{tex}c{m^2}{/tex}

Now, in {tex}\Delta {/tex}ABC,

Area of triangle ABC = {tex}\frac{1}{2}{/tex} x BC x AD

{tex} \Rightarrow {/tex} 30 = {tex}\frac{1}{2}{/tex} x 13 x AD {tex} \Rightarrow {/tex} AD = {tex}\frac{{30 \times 2}}{{13}}{/tex} = {tex}\frac{{60}}{{13}}{/tex} cm

NCERT Solutions for Class 7 Maths Exercise 11.2

###### Question 8.{tex}\Delta {/tex}ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm. The height AD from A to BC, is 6 cm. Find the area of {tex}\Delta {/tex}ABC. What will be the height from C to AB i.e., CE?

In {tex}\Delta {/tex}ABC, AD = 6 cm and BC = 9 cm

Area of triangle = {tex}\frac{1}{2}{/tex} x base x height = {tex}\frac{1}{2}{/tex} x BC x AD

= {tex}\frac{1}{2}{/tex} x 9 x 6 = 27{tex}c{m^2}{/tex}

Again, Area of triangle = {tex}\frac{1}{2}{/tex} x base x height = {tex}\frac{1}{2}{/tex} x AB x CE

{tex} \Rightarrow {/tex} 27 = {tex}\frac{1}{2}{/tex} x 7.5 x CE {tex} \Rightarrow {/tex} CE = {tex}\frac{{27 \times 2}}{{7.5}}{/tex}

{tex} \Rightarrow {/tex} CE = 7.2 cm

Thus, height from C to AB i.e., CE is 7.2 cm.

## NCERT Solutions for Class 7 Maths Exercise 11.2

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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