Question

(iii) In ∆ABC, DE ||AB, AD=3DC,
A([]ABED)=90 cm².
Find A(∆ABC)
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Answer 1

The value of Area (Δ ABC) = 96 cm².

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Let's understand a few concepts:

To find Area (triangle ABC) we must use the Theorem of Ratio of Areas of Similar Triangles.

What are similar triangles?

Two triangles are said to be similar if their corresponding angles are equal and their corresponding sides are proportional to each other.

What is the Theorem of Areas of Similar Triangles?

The theorem states that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.

For example: if ΔABC and ΔPQR are two similar triangles then we can say that,

$\boxed{\bold{\frac{A(\triangle ABC)}{A (\triangle PQR)} = \bigg(\frac{AB}{PQ}\bigg)^2 = \bigg(\frac{BC}{QR}\bigg)^2 = \bigg(\frac{AC}{PR}\bigg)^2 }}$

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Let's solve the given problem:

In Δ CDE and Δ CAB, we have

∠CDE = ∠CAB . . . [∵ DE // AB, ∴ ∠CDE & ∠CAB corresponding angles]

∠DCE = ∠ACB . . . [common angles]

∴ Δ CDE ~ Δ CAB . . . [By AA similarity]

By using the above theorem of Ratio of Areas of Similar Triangles, we can say

$\frac{A(\triangle CDE)}{A (\triangle CAB)} = \bigg(\frac{CD}{CA}\bigg)^2$

$\implies \frac{A(\triangle CDE)}{A (\triangle CAB)} = \bigg(\frac{CD}{CD + AD}\bigg)^2$

• On substituting AD = 3DC (given)

$\implies \frac{A(\triangle CDE)}{A (\triangle CAB)} = \bigg(\frac{CD}{CD + 3CD}\bigg)^2$

$\implies \frac{A(\triangle CDE)}{A (\triangle CAB)} = \bigg(\frac{CD}{4CD }\bigg)^2$

$\implies \frac{A(\triangle CDE)}{A (\triangle CAB)} = \bigg(\frac{1}{4 }\bigg)^2$

$\implies \frac{A(\triangle CDE)}{A (\triangle CAB)} = \frac{1}{16 }$

$\implies A (\triangle CAB) = 16 A(\triangle CDE)$ . . . (1)

From the figure, we get

$A (\triangle CAB)$ $=$ A([] ABED) $+ A (\triangle CDE)$

• on substituting from (1) and A ([]ABED) = 90 cm²

$\implies 16 A (\triangle CDE) = 90 + A (\triangle CDE)$

$\implies 16 A (\triangle CDE) - A (\triangle CDE) = 90$

$\implies 15 A (\trinagle CDE) = 90$

$\implies A (\triangle CDE) = \frac{90}{15}$

$\implies \bold{A (\triangle CDE) =6 \:cm^2}$

On substituting A (Δ CDE) = 6 cm² in equation (1), we get

$A (\triangle ABC) = A (\triangle CAB) = 16 \times 6 = \bold{96 \:cm^2}$

Thus, Area (∆ ABC) = 96 cm².

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