Question

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Answer 1

Answer:

6.2 sq units

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Answer 2

Solution :-

In ∆AJD and ∆AHI we have,

→ ∠ADJ = ∠AIJ { Each 90° because of square }

→ ∠JAD = ∠HAI { common }

So,

→ ∆AJD ~ ∆AHI { By AA similarity }

then,

→ JD/AD = HI/AI {when ∆'s are similar their corresponding sides are in same ratio. }

→ JD/2 = 5/10 { AI = AD + DE + EI = 2 + 3 + 5 = 10 units }

→ JD/2 = 1/2

→ JD = 2/2

→ JD = 1 unit .

similarly,

→ ∆AKE ~ ∆AHI { By AA similarity }

then,

→ KE/AE = HI/AI

→ KE/5 = 1/2

→ KE = (5/2) unit .

therefore,

→ Area trapezium (DJKE) = (1/2) * (sum of parallel sides) * Height

→ Area trapezium (DJKE) = (1/2) * (JD + KE) * DE

→ Area trapezium (DJKE) = (1/2) * (1 + 5/2) * 3

→ Area trapezium (DJKE) = (1/2) * (7/2) * 3

→ Area trapezium (DJKE) = (21/4)

→ Area trapezium (DJKE) = 5.25 unit² ≈ 5.2 unit² (B) (Ans.)

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