Parallelograms

Exercise 9.4

Part 2

Question 8: In this figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AC⊥DE meets BC at Y. Show that

(a) ΔMBC ≅ ΔABD

Answer: MB = AB (Sides of a square)

BC = BD (Sides of a square)

∠MBA = ∠DBC = 90° (angles of squares)

So, ∠MBA + ∠ABC = ∠DBC + ∠ABC

So, ∠MBC = ∠ABD

So, ΔMBC ≅ ΔABD (SAS Theorem)

(b) ar(BYXD) = 2 ar(MBC)

Answer: As BYXD and ΔABD are on the same base BD and between same parallels BD and AX

So, ar(ΔABD) = `1/2` ar(BYXD)

In previous question we proved ΔMBC ≅ ΔABD

So, ar(ΔMBC) = `1/2` ar(BYXD)

Or, ar(BYXD) = 2 ar(ΔMBC)

Answer: FC = AC (sides of a square)

CB = CE (sides of a square)

∠ACF = ∠BCE = 90°

So, ∠ACF + ∠ACB = ∠BCE + ∠ACB

Or, ∠FCB = ∠ACE

So, ΔFCB ≅ ΔACE

(d) ar(CYXE) = 2 ar(FCB)

Answer: CYXE and ΔACE are on same base CE and between same parallels AX and CE

So, ar(CYXE) = 2 ar(ACE)

In previous question we proved ΔACE ≅ ΔFCB

So, ar(CYXE) = 2 ar(FCB)

(e) ar(CYXE) = ar(ACFG)

Answer: ACFG and Δ FCB are on same base FC and between same parallels GB and FC

So, ar(ACFG) = 2 ar(FCB)

In previous question, we proved ar(CYXE) = 2 ar(FCB)

So, ar(CYXE) = ar(ACFG)

(f) ar(BCED) = ar(ABMN) + ar(ACFG)

Answer: Ar(BCED) = BC²

Ar(ACFG) = AC²

Ar(ABMN) = BA²

Here, BC is hypotenuse, and AC and BC are remaining two sides of right angle triangle ABC

According to Pythagoras theorem,

`h^2=p^2+b^2`

So, ar(BCED) = ar(ABMN) + ar(ACFG)