Question

Hey Brainly User,

I will mark brainliest for the correct answer.

I am having some problems with this two Quad. sums (Q.7 & Q.8).

Hope help me out by solving these Questions.

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Answer 1

.................................sorry for late reply

Answer 2

(7) Given, ABCD is a square.

So, AB = BC = CD = AD.

Also, it's given that ∠PQR = 90° and PB = QC = DR.

*(i)* We know ABCD is a *square* which means BC = CD. Also given that QC = RD.

Now, BC-QC = CD-RD

Or, QB = RC `(proved)`

*(ii)* In ∆PBQ and ∆RQC, we get :

PB = QC ; QB = RC ; ∠PBQ = ∠RCQ (each angle is 90°, i.e., a right angle)

So, ∆PBQ ≅ RQC (S-A-S rule)

Hence, PQ = QR (C.P.C.T.) `(proved)`

*(iii)* In ∆PQR, PQ = QR

So, ∠PRQ = ∠RPQ (Angles opposite to equal sides of a trianglebare also equal)

Now, ∠PQR +∠QPR + ∠PRQ = 180°

Or, 90° + 2∠QPR = 180°

Or, 2∠QPR = 90°

Or, ∠QPR = 45° `(proved)`