GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

   

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

GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of each of the following parallelograms:
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 1
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 2
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

GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 2.
Find the area of each of the following triangles:
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 3
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:
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 4
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

GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

Question 4.
Find the missing values:
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 4a
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.
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 5a
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

GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

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.
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 5
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.
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 6
Solution:
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 7

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?
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 8
Solution:
Here, base, BC = 9 cm
Corresponding height, AD = 6 cm
GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2 9
Thus, height CE = 7.2 cm

GSEB Solutions Class 7 Maths Chapter 11 Perimeter and Area Ex 11.2

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