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

Question 1. Find the area of each of the following parallelograms:
Answer:
(a) Here, Base \( (b) = 7 \) cm and Height \( (h) = 4 \) cm.
The area of the parallelogram is calculated as:
\( \text{Area} = b \times h = 7 \times 4 \text{ cm}^2 = 28 \text{ cm}^2 \).
(b) Here, Base \( (b) = 5 \) cm and Height \( (h) = 3 \) cm.
The area of the parallelogram is found by:
\( \text{Area} = b \times h = 5 \times 3 \text{ cm}^2 = 15 \text{ cm}^2 \).
(c) Here, Base \( (b) = 2.5 \) cm and Height \( (h) = 3.5 \) cm.
The area of the parallelogram is determined as:
\( \text{Area} = b \times h = 2.5 \times 3.5 \text{ cm}^2 \)
\( = \frac { 25 }{10} \times \frac {35}{ 10 } \text{ cm}^2 \)
\( = \frac { 875 }{ 100 } \text{ cm}^2 \)
\( = 8.75 \text{ cm}^2 \).
(d) Here, Base \( (b) = 5 \) cm and Height \( (h) = 4.8 \) cm.
The area of the parallelogram is calculated using:
\( \text{Area} = b \times h = 5 \times 4.8 \text{ cm}^2 = 24 \text{ cm}^2 \).
(e) Here, Base \( (b) = 2 \) cm and Height \( (h) = 4.4 \) cm.
The area of the parallelogram is obtained by:
\( \text{Area} = b \times h = 2 \times 4.4 \text{ cm}^2 = 8.8 \text{ cm}^2 \).
In simple words: For each parallelogram, we multiply the base measurement by its corresponding height measurement to find its area. This calculation gives us the space each figure covers in square centimeters.

Exam Tip: Carefully identify the base and the perpendicular height for each parallelogram from the given diagrams to avoid errors in calculation.

Question 2. Find the area of each of the following triangles:
Answer:
(a) Here, Base \( (b) = 4 \) cm and Height \( (h) = 3 \) cm.
The area of the triangle is calculated as:
\( \text{Area} = \frac {1}{2} \times b \times h = \frac { 1 }{ 2 } \times 4 \times 3 \text{ cm}^2 = 6 \text{ cm}^2 \).
(b) Here, Base \( (b) = 5 \) cm and Height \( (h) = 3.2 \) cm.
The area of the triangle is found by:
\( \text{Area} = \frac {1}{2} \times b \times h = \frac { 1 }{ 2 } \times 5 \times 3.2 \text{ cm}^2 = 8 \text{ cm}^2 \).
(c) Here, Base \( (b) = 3 \) cm and Height \( (h) = 4 \) cm.
The area of the triangle is determined as:
\( \text{Area} = \frac {1}{2} \times b \times h = \frac { 1 }{ 2 } \times 3 \times 4 \text{ cm}^2 = 6 \text{ cm}^2 \).
(d) Here, Base \( (b) = 3 \) cm and Height \( (h) = 2 \) cm.
The area of the triangle is obtained by:
\( \text{Area} = \frac {1}{2} \times b \times h = \frac { 1 }{ 2 } \times 3 \times 2 \text{ cm}^2 = 3 \text{ cm}^2 \).
In simple words: To find the area of each triangle, we take half of the product of its base and its height. This formula gives us the space covered by the triangle in square centimeters.

Exam Tip: Remember the formula for the area of a triangle \( (\frac{1}{2} \times \text{base} \times \text{height}) \) and ensure you use the perpendicular height for the calculation.

Question 3. Find the missing values:

Answer:
(a) Here, Base \( (b) = 20 \) cm and Area \( = 246 \text{ cm}^2 \).
Let the Height be \( h \) cm.
We know that \( b \times h = \text{Area of the parallelogram} \).
So, \( 20 \times h = 246 \)
\( h = \frac { 246 }{ 20 } \text{ cm} \)
\( h = \frac { 123 }{ 10 } \text{ cm} = 12.3 \text{ cm} \).
Thus, the missing value (height) is \( 12.3 \) cm.
(b) Here, Height \( (h) = 15 \) cm and Area of the parallelogram \( = 154.5 \text{ cm}^2 \).
Let the base of the parallelogram be \( b \) cm.
So, \( b \times h = 154.5 \)
\( b \times 15 = 154.5 \)
\( b = \frac { 154.5 }{ 15 } = 10.3 \text{ cm} \).
Thus, the missing value (base) is \( 10.3 \) cm.
(c) Here, Height \( (h) = 8.4 \) cm and Area of the parallelogram \( = 48.72 \text{ cm}^2 \).
Let the base of the parallelogram be \( b \) cm.
So, \( b \times h = 48.72 \)
\( b \times 8.4 = 48.72 \)
\( b = \frac { 48.72 }{ 8.4 } \)
\( b = 5.8 \text{ cm} \).
Thus, the missing value (base) is \( 5.8 \) cm.
(d) Here, Base \( (b) = 15.6 \) cm and Area of the parallelogram \( = 16.38 \text{ cm}^2 \).
Let the height be \( h \) cm.
So, \( b \times h = 16.38 \)
\( 15.6 \times h = 16.38 \)
\( h = \frac { 16.38 }{ 15.6 } = 1.05 \text{ cm} \).
Thus, the missing value (height) is \( 1.05 \) cm.
In simple words: We used the formula for the area of a parallelogram (base multiplied by height) to calculate each missing dimension. By rearranging the formula, we could find either the base or the height when the other two values were provided.

Exam Tip: When finding a missing dimension, always make sure to correctly rearrange the area formula. Division is used when finding a missing side, while multiplication finds the area.

Question 4. Find the missing values:

Answer:
(i) Here, Base \( (b) = 15 \) cm and Area of the triangle \( = 87 \text{ cm}^2 \).
Let the height be \( h \) cm.
We know that \( \frac { 1 }{ 2 } \times b \times h = \text{Area} \).
So, \( \frac { 1 }{ 2 } \times 15 \times h = 87 \)
\( h = \frac { 87 \times 2 }{ 15 } \)
\( h = \frac { 174 }{ 15 } = 11.6 \text{ cm} \).
Thus, the missing value (height) is \( 11.6 \) cm.
(ii) Here, Height \( (h) = 31.4 \) mm and Area of the triangle \( = 1256 \text{ mm}^2 \).
Let the base be \( b \) mm.
So, \( \frac { 1 }{ 2 } \times b \times h = 1256 \)
\( \frac { 1 }{ 2 } \times b \times 31.4 = 1256 \)
\( b = \frac { 1256 \times 2 }{ 31.4 } \)
\( b = \frac { 2512 }{ 31.4 } = 80 \text{ mm} \).
Thus, the missing value (base) is \( 80 \) mm.
(iii) Here, Base \( (b) = 22 \) cm and Area of the triangle \( = 170.5 \text{ cm}^2 \).
Let the height be \( h \) cm.
So, \( \frac { 1 }{ 2 } \times b \times h = 170.5 \)
\( \frac { 1 }{ 2 } \times 22 \times h = 170.5 \)
\( 11 \times h = 170.5 \)
\( h = \frac { 170.5 }{ 11 } = 15.5 \text{ cm} \).
The missing value (height) is \( 15.5 \) cm.
In simple words: We used the formula for the area of a triangle (half times base times height) to figure out the missing measurements. We rearranged the formula to calculate either the base or the height when the area and the other dimension were already known.

Exam Tip: Pay close attention to the units (cm vs. mm) and ensure consistency throughout your calculations and final answers for each part.

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.
(a) Find the area of the parallelogram PQRS.
(b) QN, if PS = 8 cm.
Answer:
Here, Base \( (\text{SR}) = 12 \) cm and Corresponding height \( (\text{QM}) = 7.6 \) cm.
(a) The area of the parallelogram PQRS is given by \( \text{Base} \times \text{Height} \).
\( \text{Area} = \text{SR} \times \text{QM} \)
\( = 12 \times 7.6 \text{ cm}^2 \)
\( = 91.2 \text{ cm}^2 \).
(b) Now, the area of the parallelogram is \( 91.2 \text{ cm}^2 \). The base \( (\text{PS}) = 8 \) cm.
Let the corresponding height \( (\text{QN}) \) be \( h \) cm.
We know that \( b \times h = \text{Area} \).
So, \( 8 \times h = 91.2 \)
\( h = \frac { 91.2 }{ 8 } \)
\( \text{QN} = 11.4 \text{ cm} \).
In simple words: For part (a), we multiply the base SR by height QM to find the parallelogram's area. For part (b), we use this calculated area and the new base PS to divide and find the other height QN.

Exam Tip: Remember that a parallelogram has two pairs of base-height combinations. The area remains constant regardless of which base and corresponding height pair you use.

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.
Answer:
Since, the area of the parallelogram ABCD \( = \text{Base} \times \text{Height} \).
Using base AD and height BM:
\( \text{AD} \times \text{BM} = 1470 \text{ cm}^2 \).
Given \( \text{AD} = 49 \) cm.
So, \( 49 \times \text{BM} = 1470 \)
\( \text{BM} = \frac { 1470 }{ 49 } = 30 \text{ cm} \).
Again, using base AB and height DL:
\( \text{AB} \times \text{DL} = 1470 \text{ cm}^2 \).
Given \( \text{AB} = 35 \) cm.
So, \( 35 \times \text{DL} = 1470 \)
\( \text{DL} = \frac { 1470 }{ 35 } = 42 \text{ cm} \).
In simple words: We used the given area and each base length (AD and AB) to find the two different heights (BM and DL) of the parallelogram by dividing the area by the base.

Exam Tip: For a parallelogram, two different height values can exist depending on which side is considered the base. Always match the height to its corresponding base.

Question 7. ΔABC 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.
Answer:
The area of \( \triangle\text{ABC} \) is calculated as:
\( \text{Area} = \frac { 1 }{ 2 } \times \text{Base} \times \text{Height} \).
Since it's a right-angled triangle at A, AB and AC are the perpendicular sides.
\( \text{Area} = \frac { 1 }{ 2 } \times \text{AB} \times \text{AC} \)
\( = \frac { 1 }{ 2 } \times 5 \times 12 \text{ cm}^2 \)
\( = 30 \text{ cm}^2 \).
Again, using BC as the base and AD as the height:
\( \text{Area of } \triangle\text{ABC} = \frac { 1 }{ 2 } \times \text{BC} \times \text{AD} \).
We know the area is \( 30 \text{ cm}^2 \) and \( \text{BC} = 13 \) cm.
So, \( 30 = \frac { 1 }{ 2 } \times 13 \times \text{AD} \)
\( \text{AD} = \frac { 30 \times 2 }{ 13 } \text{ cm} \)
\( \text{AD} = \frac { 60 }{ 13 } \text{ cm} \).
In simple words: First, we find the area of the right-angled triangle using its two perpendicular sides. Then, we use this area and the hypotenuse (longest side) as the base to figure out the length of the altitude (height) to that side.

Exam Tip: In a right-angled triangle, the two sides forming the right angle can serve as base and height. The altitude to the hypotenuse can also be found using the area formula.

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?
Answer:
Here, Base \( (\text{BC}) = 9 \) cm and Corresponding height \( (\text{AD}) = 6 \) cm.
The area of \( \triangle\text{ABC} \) is calculated as:
\( \text{Area} = \frac { 1 }{ 2 } \times \text{Base} \times \text{Height} \)
\( = \frac { 1 }{ 2 } \times \text{BC} \times \text{AD} \)
\( = \frac { 1 }{ 2 } \times 9 \times 6 \text{ cm}^2 \)
\( = 27 \text{ cm}^2 \).
Now, let the height from C to AB be \( h \) (which is CE).
Using AB as the base and CE as the height:
\( \frac { 1 }{ 2 } \times \text{AB} \times h = \text{Area} \).
We know Area \( = 27 \text{ cm}^2 \) and \( \text{AB} = 7.5 \) cm.
So, \( \frac { 1 }{ 2 } \times 7.5 \times h = 27 \)
\( h = \frac { 27 \times 2 }{ 7.5 } \text{ cm} \)
\( h = \frac { 54 }{ 7.5 } \text{ cm} \)
\( h = 7.2 \text{ cm} \).
Thus, the height CE is \( 7.2 \) cm.
In simple words: We first found the triangle's area using the given base (BC) and height (AD). Then, using this area and another side (AB) as the new base, we calculated the length of the second height (CE).

Exam Tip: For any triangle, the area can be calculated using any side as the base, provided you use the corresponding perpendicular height to that base. This is useful for finding unknown heights.