Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 1.

Find the area of each of the following parallelograms:

Solution:

(a) Here, Base (b) = 7 cm, Height (h) = 4 cm

∴ Area of the parallelogram

= b x h = 7 x 4 cm² = 28 cm²

(b) Here, Base =5 cm

Height (h) = 3 cm

∴ Area of the parallelogram = b x h

= 5 x 3 cm² = 15 cm²

(c) Here, Base(b) = 2.5 cm

Height (h) =3.5 cm

∴ Area of the parallelogram = b x h

= 2.5 x 3.5 cm²

= \(\frac { 25 }{ 10 }\) x \(\frac { 35 }{ 10 }\) cm²

= \(\frac { 875 }{ 100 }\)cm²

= 8.75 cm²

(d) Here, Base (b) =5 cm

Height (h) = 4.8 cm

∴ Area of the parallelogram = b x h

= 5 x 4.8 cm² = 24 cm²

(e) Here, Base (b) =2 cm

Height (h) = 4.4 cm

∴ Area of the parallelogram = b x h

= 2 x 4.4 cm² = 8.8 cm²

Question 2.

Find the area of each of the following triangles:

Solution:

(a) Here, Base (b) =4 cm

Height (h) = 3 cm

∴ Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h

= \(\frac { 1 }{ 2 }\) x 4 x 3 cm² = 6 cm²

(b) Base (b) = 5 cm

Height (h) =3.2 cm

∴ Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h

= \(\frac { 1 }{ 2 }\) x 5 x 3.2 cm²

(c) Here

Base (b) = 3 cm

Height (h) = 4 cm

∴ Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h

= \(\frac { 1 }{ 2 }\) x 3 x 4 cm² = 6 cm²

(d) Here, Base (b) = 3 cm

Height (h) = 2 cm

∴ Area of the triangle = \(\frac { 1 }{ 2 }\) x b x h

= \(\frac { 1 }{ 2 }\) x 3 x 2cm²

= 3 cm²

Question 3.

Find the missing values:

Solution:

(a) Here,

Base (b) = 20 cm

Area = 246 cm²

Consider Height = h cm

∴ b x h = Area of the parallelogram

or 20 x h = 246 ⇒ h = \(\frac { 246 }{ 20 }\) cm

= \(\frac { 123 }{ 10 }\) cm = 12.3 cm

Thus, the missing value (height) = 12.3 cm

(b) Here, Height (h) = 15 cm

Area of the parallelogram = 154.5 cm²

Suppose the base of the parallelgoram = b cm

∴ b x h = 154.5 ⇒ b x 15 = 154.5

or b = \(\frac { 154.5 }{ 15 }\) = 10.3 cm

Thus, the missing value (base) = 10.3 cm

(c) Here, Height (h) = 8.4 cm

Area of the parallelogram = 48.72 cm²

Suppose the base of the parallelogram = b cm

∴ b x h = 48.72

b x 8.4 = 48.72

or b = \(\frac { 48.72 }{ 8.4 }\)

Thus, the missing value (base) = 5.8 cm

(d) Here, Base (b) = 15.6 cm

Area of the parallelogram = 16.38 cm²

Let the height be h cm.

∴ b x h = 6.38

or 15.6 x h = 16.38

or h = \(\frac { 16.38 }{ 15.6 }\) = 1.05 cm

Thus, the missing value (height) = 1.05 cm

Question 4.

Find the missing values:

Solution:

(i) Here, Base (b) = 15 cm

Area of the triangle = 87 cm² Let the height be h cm

∴ \(\frac { 1 }{ 2 }\) x b x h = 87

or \(\frac { 1 }{ 2 }\) x 15 x h = 87

or h = \(\frac { 87×2 }{ 15 }\)

= \(\frac { 174 }{ 15 }\) = 11.6 cm

∴ The missing value (height) = 11.6 cm

(ii) Here, Height (h) =31.4 mm

Area of the triangle = 1256 mm²

Let the base be b mm

∴\(\frac { 1 }{ 2 }\) x b x h = 1256

or \(\frac { 1 }{ 2 }\) x b x 31.4 = 1256

or b = \(\frac { 1256×2 }{ 31.4 }\)

= \(\frac { 2512 }{ 31.4 }\) = 80 mm

Thus, the missing value (base) = 80 mm.

(iii) Here, Base (b) = 22 cm

Area of the triangle = 170.5 cm²

Let the height be h cm

∴ \(\frac { 1 }{ 2 }\) x b x h = 170.5

or \(\frac { 1 }{ 2 }\) x 22 x h = 170.5

∴ h = \(\frac { 170.5×2 }{ 22 }\)

= \(\frac { 170.5 }{ 11 }\) = 15.5 mm

∴The missing value (height) = 15.5 cm.

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 PQRS.

(b) QN, if PS = 8 cm.

Solution:

Here, Base (SR) = 12 cm

Corresponding height (QM) = 7.6 cm

(a) ∴ Area of the parallelogram PQRS = b x h

= SR x QM

= 12 x 7.6 cm²

= 91.2 cm²

(b) Now, Area of the parallelogram = 91.2 cm²

Base (PS) = 8 cm

Let the corresponding height (QN) be h cm

∴ b x h = 91.2

or 8 x h = 91.2

or h = \(\frac { 91.2 }{ 8 }\)

∴ QN = 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 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.

Solution:

Since, Area of the parallelogram ABCD = Base x height

= AD x BM

∴ AD x BM = 1470

[∵ Area = 1470 cm²]

or 49 x BM = 1470

or BM = \(\frac { 1470 }{ 49 }\) = 30 cm

Again, Area of parallelogram ABCD = Base x Height = AB x DL

∴ AB X DL = 1470

or 35 x DL = 1470

or DL = \(\frac { 1470 }{ 49 }\) cm = 42 cm

Question 7.

∆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 ∆ABC. Also find the length of AD.

Solution:

Question 8.

∆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 ∆ABC. What will be the height from C to AB, i.e. CE?

Solution:

Here, base, BC = 9 cm

Corresponding height, AD = 6 cm

Thus, height CE = 7.2 cm