A trapezium BCDE is formed by drawing a line ED parallel to the base BC in the triangle ΔABC, such that the ratio of the lengths of sides AB and AE are 5 ∶ 2. If the area of the trapezium so formed is 105 cm², find the area of the triangle ABC.

Question

Question

Answer 1

Given:

A triangle ABC is given

The ratio of length of sides AB and AE are 5 ∶ 2

The area of the trapezium is 105 cm²

Calculations:

F1 Sharad.C 03-03-2020 Savita D3

Given, AB/AE = 5/2

The triangle ABC and the triangle AED so formed will be similar triangles

As we know, the ratio of areas of two similar triangles is equal to the ratio of the squares of their corresponding sides

⇒ (Area of ΔABC)/(Area of ΔAED) = (AB/AE)² = 25/4

⇒ (Area of ΔAED) = 4/25 × (Area of ΔABC) ----(1)

Now,

Area of trapezium BCDE = Area of ΔABC – Area of ΔAED

Substituting from (1), we get,

⇒ Area of trapezium BCDE = (1 – 4/25) × (Area of ΔABC) = (21/25) × (Area of ΔABC)

But, Area of trapezium BCDE = 105 cm²

∴ Area of ΔABC = (25/21) × 105 = 125 cm²

∴ The area of ΔABC is 125cm².

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