Question

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Answer 1

From image we can see That :-

→ AC = AD = Let x (Given)

→ DB = Let y

→ ∠AED = ∠DBE (Given).

→ AE = 4 unit (Given).

→ ∆ABc is a Right Angle ∆ . (Given).

Solution :-

in ∆ABE & ∆AED, we Have :-

→ ∠AED = ∠DBE , or = ∠ABE (Given).

→ ∠EAD = ∠BAE (Common Angle).

Hence,

→ ∆AED ≈ ∆ABE ( By AA Similarity.)

Therefore,

→ (AD/AE) = (AE/AB)

Putting values Now, we get :-

→ x/4 = 4/(x+y)

→ x(x+y) = 4*4

→ x(x + y) = 16 .

______________________

So,

☛Area [∆ABC] = (1/2) * Base * Height .

☛Area [∆ABC] = (1/2) * AB * AC

☛Area [∆ABC] = (1/2) * (x + y) * x

Putting value from above , we get,

☛Area [∆ABC] = (1/2) * 16 = 8unit². (Ans).

Hence, Area of ∆ABC will be 8unit².

Answer 2

Hope this helps uh......!