Question

In the given figure, ΔPQR is right-angled at P. PABQ and QRST are squares on the side PQ and hypotenuse QR. If PN ⊥ TS, show that:
(a) ΔQRB ≅ ΔPQT
(b) Area of square PABQ = area of rectangle QTNM.

Answer 1

∠BQR = ∠BQP + ∠PQR
⇒ ∠BQR = 90° + ∠PQR
∠PQT = ∠TQR + ∠PQR
⇒ ∠PQT = 90° + ∠PQR   ....(i)
⇒ ∠BQR = ∠PQT 

(a) In ΔQRB and ΔPQT,
BQ = PQ              ....(sides of a square PABQ)
QR = QT             ....(sides of a square QRST)
∠BQR = ∠PQT  ...[From (i)]
∴ ΔQRB ≅ ΔPQT ...(by SAS congruence criterion)
⇒  A(ΔBQR) = A(ΔPQT)   ....(ii)

(b) ΔPQT and rect. QTNM are on the same base QT
and between the same parallel lines QT and PN.
∴ A(ΔPQT) = `(1)/(2)"A"("rect. QTNM")`

⇒ A(rect. QTMN) =  x A(ΔPQT)
⇒ A(rect. QTMN) =  x ar(ΔBQR)    [From (ii)]....(iii)
ΔBQR and sq. PABQ are on the same base BQ
and between the same parallel lines BQ and AR.
∴ 2 x A(ΔBQR) = A(sq. PABQ)    ....(iv)
From (i) and (iv),
A(sq. PABQ) = A(rect. QTNM)