Question

10. Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m, respectively, v
perpendicular distance between them is 15 m and the rate of watering is 4 per m²
11. The parallel sides of a trapezium are 32 cm and 20 cm. Its non-parallel sides are equal, each being 10 cm.
the area of the trapezium.
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Answer 1

1. Correct Question :–

Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively. If the distance between them is 15 cm and the rate of watering is rs 4 per m².

Given :–

• Length (l) = 10m.
• Breadth (b) = 25m.
• Height (h) = 15cm.
• Cost of watering = Rs. 4 per m².

To Find :–

• Cost of watering a trapezoidal field.

Solution :–

Area of trapezoid = $\frac{1}{2}$ × (l + b) × h

→ $\frac{1}{2}$ × (10 + 25) × 15

→ 0.5 × 35 × 15

→ 0.5 × 252

→ 262.5m²

Hence, the area of trapezoid = 262.5m²

Now, Cost of watering a trapezoidal field = Area × Cost of watering

→ 262.5 × 4

→ Rs. 1,050

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2. Correct Question :–

The parallel sides of a trapezium are 20cm and 32 cm. The non parallel sides are of the equal length 10 cm find the area of trapezium.

Given :–

• The parallel sides of a trapezium are 20cm and 32 cm and the non parallel sides are of the equal length 10 cm.

To Find :–

• Area of trapezium.

Solution :–

First, we divide the trapezium into 3 parts :–

1. two triangles
2. one rectangle.

Base of the triangle = $\frac{subtract \: the \: paralell \: sides}{2}$

→ $\frac{32 - 20}{2}$

→ $\frac{\cancel{12}}{\cancel{2}}$

→ 6

• height of trianlge = 6cm.
• height of rectangle = 10cm.

Height of triangle & rectangle = (h)²

First, we need to find height of trapezium,

→ (10)² – (6)²

→ 100 – 36

→ h² = 64 cm

→ h = √64

→ h = 8

• height of trapezium = 8cm.

Now,

Area of trapezium = $\frac{1}{2}$ × (sum of parallel sides) × (height)

→ $\frac{1}{2}$ × (32 + 20) × (8)

→ $\frac{1}{\cancel2}$ × 52 × $\cancel 8$

→ 52 × 4

→ 208 cm²

Hence,

Area of the trapezium = 208 cm²