The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.
Step-by-step solution
To find: Area of the trapezium
Idea: Drop perpendiculars from the ends of the short side. They cut off two equal right triangles with hypotenuse 26 cm and base 10 cm, which gives the height.
- Drop perpendiculars from the ends of the 20 cm side to the 40 cm side. The middle part is 20 cm, so each end piece is (40 − 20) ÷ 2 = 10 cm.1 mark
- Each end is a right triangle with hypotenuse 26 cm and base 10 cm: height h = √(262 − 102) = √(676 − 100) = √576 = 24 cm.1 mark
- Area = 12 × (sum of parallel sides) × height = 12 × (40 + 20) × 24 = 30 × 24 = 720 cm2.1 mark
Check: Rectangle 20 × 24 = 480 plus two triangles 12 × 10 × 24 = 120 each: 480 + 240 = 720 ✓.
Answer to write in the exam
Overhang on each side = 40 − 202 = 10 cm
h = √(262 − 102) = √576 = 24 cm
Area = 12(40 + 20) × 24
∴ Area = 720 cm2
Common mistakes that cost marks
- Taking the overhang as 20 cm (forgetting it is shared between two ends).
- Using the slant side 26 cm as the height.
- Forgetting the 12 in the trapezium formula.
How this can come in the exam
An isosceles trapezium has parallel sides 18 cm and 8 cm and slant sides 13 cm. Its height is
- 5 cm
- 10 cm
- 12 cm
- 13 cm
Show answer
(C) 12 cm
Overhang = (18 − 8) ÷ 2 = 5; height = √(169 − 25) = 12 cm.
A farmer’s field is shaped like an isosceles trapezium. The side along the road is 75 m and the opposite parallel side is 45 m; the two other sides are 25 m each.
(i) Find the overhang at each end. (ii) Find the distance between the parallel sides. (iii) Find the area of the field. (iv) If seed costs ₹2 per m2, find the cost of sowing the field.
Show answer
(i) (75 − 45) ÷ 2 = 15 m (1 mark). (ii) √(252 − 152) = 20 m (1 mark). (iii) 12(75 + 45) × 20 = 1200 m2 (1 mark). (iv) 1200 × 2 = ₹2400 (1 mark).Try one yourself
The parallel sides of an isosceles trapezium are 25 cm and 11 cm, and each slant side is 25 cm. Find its area.
Show answer
Overhang 7, height √(625 − 49) = 24, area 12(36)(24) = 432 cm2.
More questions like this
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