Question

please solve question 10
if I get the proper answer I will give 100points and mark as brainliest

Answer 1

Step-by-step explanation:

From figure:

(i)

⇒ ∠EAC = ∠EAB + ∠BAC

⇒ ∠EAC = 90° + ∠BAC    ----- (1)

⇒ ∠BAF = ∠FAC + ∠BAC

⇒ ∠BAF = 90° + ∠BAC    ---- (2)

From (1) & (2), we get

⇒ ∠EAC = ∠BAF

In ΔEAC and ΔBAF, EA = EB

⇒ In ∠EAC = ∠BAF, AC = AF

∴ ΔEAC ≅ ΔBAF

(ii)

ar(ΔEAC) = ar(ΔBAF)

⇒ ∠ABD + ∠ABC = 90° + 90°

⇒ ∠ABD + ∠ABC = 180°

(a)

ΔEAC and square ABDE are on the same base AE and between same parallel lines AF and BH.

∴ ar(ΔEAC) = (1/2) * ar(square ABDE)   ---- (3)

(b)

ΔBAF and rectangle ARHF are on the same base AF and between the same parallel lines AF and BH.

∴ ar(ΔBAF) = (1/2) * ar(rectangle ARHF)  ---- (4)

Since, ar(ΔEAC) = ar(ΔBAF)

From (3) & (4), we get

⇒ (1/2) * ar(square. ABDE) = (1/2) * ar(rectangle. ARHF)

⇒ ar(square. ABDE) = ar(rectangle. ARHF)

Hope it helps!

Answer 2

Refer to the attachment

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