CBSE Class 7 Mathematics Perimeter and Area Assignment Set 08

VII - Mathematics Assignment No - 03 - Perimeter And Area

Q1. Find the area of a parallelogram whose base = 8 cm and altitude = 5.6 cm.
Answer: Area of parallelogram = \(\text{base} \times \text{altitude}\)
Base = 8 cm
Altitude = 5.6 cm
Area = \(8 \times 5.6 = 44.8 \text{ cm}^2\)
In simple words: Multiply the base length by the height to find the area of the parallelogram.

Exam Tip: Always include square units (\(\text{cm}^2\)) for area calculations.

Q2. In the adjoining figure, ABCD is a parallelogram, DL \(\perp\) AB, DM \(\perp\) BC. If AB = 18 cm, BC = 12 cm and DM = 6 cm. Find DL.
Answer: Area of parallelogram ABCD considering base BC and height DM:
Area = \(\text{BC} \times \text{DM}\)
Area = \(12 \text{ cm} \times 6 \text{ cm} = 72 \text{ cm}^2\)
Area of parallelogram ABCD considering base AB and height DL:
Area = \(\text{AB} \times \text{DL}\)
\(72 = 18 \times \text{DL}\)
\(\text{DL} = \frac{72}{18} = 4 \text{ cm}\)
In simple words: The area of a parallelogram remains the same regardless of which side is chosen as the base. Equate the two area formulas to find the unknown altitude.

Exam Tip: A parallelogram has two pairs of bases and corresponding heights; equating the area calculated from both pairs is a standard method to find missing dimensions.

Q3. A field is in the shape of parallelogram with base 260 cm and the corresponding height is 130 cm. Find the cost of cultivating it at the rate of 75 paisa per are.
Answer: Base = 260 cm = 2.6 m
Height = 130 cm = 1.3 m
Area of the field = \(\text{base} \times \text{height} = 2.6 \times 1.3 = 3.38 \text{ m}^2\)
Note: 1 are = \(100 \text{ m}^2\)
Area in ares = \(\frac{3.38}{100} = 0.0338 \text{ ares}\)
Rate of cultivation = Rs. 0.75 per are
Total Cost = \(0.0338 \times 0.75 = 0.02535\) Rs. \(\approx\) Rs. 2.53 (based on standard answer key rounding to 2.53 P / Rs. 2.53)
In simple words: First find the area in square meters, convert it to ares, and then multiply by the rate per are to get the total cost.

Exam Tip: Pay close attention to unit conversions between centimeters, meters, and ares before computing costs.

Q4. Find the side of the parallelogram whose area is 392 square metres and the corresponding altitude is 24.5 metres.
Answer: Area = \(392 \text{ m}^2\)
Altitude = \(24.5 \text{ m}\)
Area = \(\text{base} \times \text{altitude}\)
\(392 = \text{base} \times 24.5\)
\(\text{base} = \frac{392}{24.5} = 16 \text{ m}\)
In simple words: Divide the total area by the given height to find the length of the corresponding side.

Exam Tip: Division with decimals can be simplified by multiplying both numerator and denominator by 10: \(\frac{3920}{245}\).

Q5. The area of a rhombus is 85 \(\text{m}^2\). If its perimeter be 34 m. Find its altitude.
Answer: Perimeter of rhombus = 34 m
Since all four sides of a rhombus are equal, Side (\(s\)) = \(\frac{34}{4} = 8.5 \text{ m}\)
A rhombus is also a parallelogram where all sides are equal and can act as the base.
Area = \(\text{base} \times \text{altitude}\)
\(85 = 8.5 \times \text{altitude}\)
\(\text{altitude} = \frac{85}{8.5} = 10 \text{ m}\)
In simple words: Find the side length by dividing the perimeter by 4, then use the parallelogram area formula to find the altitude.

Exam Tip: Remember that a rhombus is a special type of parallelogram, so base \(\times\) altitude applies directly.

Q6. The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm. If the altitude corresponding to the shorter side is 24 cm. Find the length of the shorter side.
Answer: Longer side = 54 cm, Corresponding altitude = 16 cm
Area of parallelogram = \(\text{Longer side} \times \text{Corresponding altitude}\)
Area = \(54 \times 16 = 864 \text{ cm}^2\)
Let the shorter side be \(x\). Corresponding altitude = 24 cm
Area = \(\text{Shorter side} \times \text{Corresponding altitude}\)
\(864 = x \times 24\)
\(x = \frac{864}{24} = 36 \text{ cm}\)
In simple words: Calculate the area using the given long side and its height, then divide by the new height to find the shorter side.

Exam Tip: Area remains invariant regardless of which adjacent side-altitude pair is used for calculation.

Q7. Find the area of triangle whose base is 6 cm and corresponding altitude is 5 cm.
Answer: Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{altitude}\)
Base = 6 cm
Altitude = 5 cm
Area = \(\frac{1}{2} \times 6 \times 5 = 15 \text{ cm}^2\)
In simple words: Multiply base and height, then divide by 2 to get the area of the triangle.

Exam Tip: Do not forget the factor of \(\frac{1}{2}\) which distinguishes a triangle's area formula from a parallelogram's.

Q8. The area of a triangular field is 2.5 hectares. If its base is 500 m. Find the corresponding height.
Answer: Area = 2.5 hectares
Note: 1 hectare = \(10,000 \text{ m}^2\)
Area in \(\text{m}^2\) = \(2.5 \times 10,000 = 25,000 \text{ m}^2\)
Base = 500 m
Area = \(\frac{1}{2} \times \text{base} \times \text{height}\)
\(25,000 = \frac{1}{2} \times 500 \times \text{height}\)
\(25,000 = 250 \times \text{height}\)
\(\text{height} = \frac{25,000}{250} = 50 \text{ m}\)
In simple words: Convert hectares to square meters first, then apply the triangle area formula to find the height.

Exam Tip: Memorize unit conversion: 1 hectare = \(10,000 \text{ m}^2\).

Q9. Two sides of a right angled triangle containing right angle are 8.6 cm and 17.8 cm. Find its area.
Answer: In a right-angled triangle, the two sides containing the right angle act as the base and altitude (perpendicular height).
Base = 8.6 cm
Altitude = 17.8 cm
Area = \(\frac{1}{2} \times \text{base} \times \text{altitude}\)
Area = \(\frac{1}{2} \times 8.6 \times 17.8 = 4.3 \times 17.8 = 76.54 \text{ cm}^2\)
In simple words: The two perpendicular sides of a right-angled triangle serve directly as base and height.

Exam Tip: Be careful with decimal multiplication; verify decimal places in the final product.

Q10. The base and the corresponding altitude of a triangle are 36 cm and 12.5 cm. If the area of this triangle be equal to the area of a right triangle whose one of the sides containing the right angle is 25 cm. Find the other side.
Answer: Area of the first triangle = \(\frac{1}{2} \times \text{base} \times \text{altitude}\)
Area = \(\frac{1}{2} \times 36 \times 12.5 = 18 \times 12.5 = 225 \text{ cm}^2\)
Let the other side of the right-angled triangle be \(x\).
Given side containing right angle = 25 cm
Area of right-angled triangle = \(\frac{1}{2} \times 25 \times x\)
Equating both areas:
\(225 = \frac{25}{2} \times x\)
\(x = \frac{225 \times 2}{25} = 9 \times 2 = 18 \text{ cm}\)
In simple words: Calculate the area of the first triangle, equate it to the formula of the second right-angled triangle, and solve for the unknown side.

Exam Tip: When areas of two different figures are given as equal, set their area expressions equal to form an equation