DAV Class 7 Maths Chapter 11 Practice Worksheet
Perimeter and Area Practice Worksheet
1. Find area of parallelogram whose base is 5 m and height is 20 cm.
Solution
\[ \begin{aligned} b &= 5 \ \text{m} \implies 500 \ cm \\ h &= 20 \ \text{cm} \\[6pt] \color{green}\text{Area of parallelogram} &= \color{green} b \times h \\ &= 500 \times 20 \\ &= 10000 \ \text{cm}^2 \ (or) \ 1 \ m^2 \end{aligned} \]Answer Area of parallelogram \(= \color{red}{10000 \ cm^2 \ (or) \ 1 \ \text{m}^2}\)
2. Find the area of a rhombus, if the length of each side be 2.4 dm and the altitude be 32 cm.
Solution
\[ \begin{aligned} base &= 2.4 \ \text{dm} \implies 24 \ cm \\ height &= 32 \ \text{cm} \\[6pt] \color{green}\text{Area of rhombus} &= \color{green} \text{base} \times \text{height} \\ &= 24 \times 32 \\ &= 768 \ \text{cm}^2 \end{aligned} \]Answer Area of rhombus \(= \color{red}{768 \ \text{cm}^2}\)
3. Find the altitude of rhombus whose area is 420 sq cm and the base is 14 cm.
Solution
\[ \begin{aligned} b &= 14 \ \text{cm} \\ h &= \ ? \\ \color{green}\text{Area of rhombus} &= 420 \ \text{cm}^2 \\[6pt] \color{green} b \times h &= 420 \\[6pt] 14 \times h &= 420 \\[6pt] h &= \frac{\cancel{420}^{30}}{\cancel{14}_1} \\[6pt] h &= 30 \ \text{cm} \end{aligned} \]Answer Altitude \(= \color{red}{30 \ \text{cm}}\)
4. Two sides of a parallelogram are 80 cm and 100 cm. If the altitude corresponding to the side of length 100 cm is 40 cm, find the altitude corresponding to the other pair of sides.
Solution
\[ \begin{align*} \text{Area of (ABCD)} &= \text{Area of (ABCD)} \\ \text{Base} \times \text{Height} &= \text{Base} \times \text{Height} \\ \text{DA} \times \text{BF} & = \text{AB} \times \text{DE} \\ 80 \times \text{BF} & = 100 \times 40 \\[6pt] \text{BF} & = \frac{4000}{80} \\[6pt] \text{BF} & = 50 \ cm \end{align*} \]
Answer Required altitude \(= \color{red}{50 \ \text{cm}}\)
5. A 88 cm long wire is bent to form a circle. Find the diameter of the circle.
Solution
\[ \begin{aligned} \text{Circumference } &= 88 \ \text{cm} \\[4pt] \color{green}\pi d &= 88 \ \text{cm} \\[4pt] \frac{22}{7} \times d &= 88 \ \text{cm} \\[4pt] d &= \cancel{88}^{ \ 4} \times \frac{7}{\cancel{22}_1} \\[6pt] d &= 28 \ \text{cm} \end{aligned} \]Answer Diameter \(= \color{red}{28 \ \text{cm}}\)
6. A wire is bent to form a square of side 22 cm. If the same wire is bent in the form of a circle, then find its radius and circumference.
Solution
\[ \begin{aligned} \text{Side of square} &= 22 \ \text{cm} \\ \text{Perimeter} &= 4 \times 22 \\ &= 88 \ \text{cm} \\[6pt] \color{green}\text{Circumference} &= 88 \ \text{cm} \\[6pt] \color{green} 2\pi r &= 88 \\[6pt] 2 \times \frac{22}{7} \times r &= 88 \\[6pt] \frac{44}{7} \times r &= 88 \\[6pt] r &= \frac{\cancel{88}^2 \times 7}{\cancel{44}_1} \\[6pt] r &= 14 \ \text{cm} \end{aligned} \]Answer Radius \(= \color{red}{14 \ \text{cm}}\), Circumference \(=\ \color{red}{88 \ \text{cm}}\)
7. The base of a triangular field is three times its height. If the cost of cultivating the field at ₹ 36 per hectare is ₹ 486, find its base and height.
Solution
\[ \begin{align*} \text{Let Height} & = x \\ \text{Let Base} & = 3x \\ \text{Total cost for cultivation} & = \text{₹} 486 \\ \text{Cost per hectare} & = \text{₹} 36 \\[6pt] \text{Area of the triangular field} & = \frac{486}{36} \text{hectare} \\[6pt] & = \frac{27}{2} \text{hectare} \\[6pt] & = \frac{27}{2} \times 10000 \text{m}^2 \\[6pt] & = 135000 \text{m}^2 \\ \\ \text{Area of the triangular field} & = 135000 \text{m}^2 \\[6pt] \frac{1}{2} \times \text{b} \times \text{h} & = 135000 \text{m}^2 \\[6pt] \frac{1}{2} \times 3x \times x & = 135000 \\[6pt] \frac{3x^2}{2} & = 135000 \\[6pt] \ x^2 & = \cancel{135000}^{ \ 45000} \times \frac{2}{\cancel3_1} \\[6pt] \ x^2 & = 45000 \times \frac{2}{1} \\[6pt] \ x^2 & = 90000 \\[6pt] \ x^2 & = 300 \times 300 \\ \ x^2 & = 300^2 \\ x & = 300 \, \text{m} \, (\text{Height of the field}) \\ \\ \text{Base} &= 3 \times x \\ &= 3 \times 300 \\ \text{Base} &= 900 \, \text{m} \\ \end{align*} \]
Answer Height \(= \color{red}{300 \ \text{m}}\), Base \(=\ \color{red}{900 \ \text{m}}\)
8. The radius of the wheel of a bus is 42 cm. How much distance will it cover in 500 revolutions?
Solution
\[ \begin{aligned} r &= 42 \ \text{cm} \\ \color{green}\text{Distance in 1 revolution} &= \color{green}2\pi r \\[6pt] &= 2 \times \frac{22}{\cancel7_1} \times \cancel{42}^6 \\[6pt] &= 44 \times 6 \\[6pt] &= 264 \ \text{cm} \\[6pt] \text{Distance in 500 revolutions} &= 500 \times 264 \\ &= 132000 \ \text{cm} \\[6pt] 1 \ cm & = \frac{1}{100000} \ km \\[6pt] & = \frac{132000}{100000} \\[6pt] \implies & 1.32 \ \text{km} \end{aligned} \]Answer Distance covered \(= \color{red}{1.32 \ \text{km}}\)
9. Find the diameter of a circle whose circumference is 13.2 m.
Solution
\[ \begin{aligned} \color{green} \text{Circumference} &= 13.2 \ \text{m} \\[4pt] \color{green} \pi d &= 13.2 \\[6pt] \frac{22}{7} \times d &= 13.2 \\[6pt] d &= \cancel{13.2}^{0.6} \times \frac{7}{\cancel{22}_1} \\[6pt] d &= 0.6 \times 7 \\[6pt] d &= 4.2 \ \text{m} \end{aligned} \]Answer Diameter \(= \color{red}{4.2 \ \text{m}}\)
10. Find the area of a rhombus whose diagonals are 7.7 m and 21 dm respectively.
Solution
\[ \begin{aligned} d_1 &= 7.7 \ \text{m} \\ d_2 &= 21 \ \text{dm} \implies 2.1 \ \text{m} \\[6pt] \color{green}\text{Area of rhombus} &= \color{green}\dfrac{1}{2} \times d_1 \times d_2 \\[6pt] &= \frac{1}{\cancel2_1} \times 7.7 \times \cancel{2.1}^{1.05} \\[6pt] &= 7.7 \times 1.05 \\[6pt] &= 8.085 \ \text{m}^2 \end{aligned} \]Answer Area of rhombus \(= \color{red}{8.085 \ \text{m}^2}\)