Question

In the given figure, PQRT is a square with each side of length 16 cm. If 2OR = TS, then ratio of area of TORS and area ∆PQR is equal to

• A.
  $1:2$
• B.
  $4:5$
• C.
  $3:2$
• D.
  $5:3$

Answer 1

The correct option is C

$3:2$
Given that PQRT is a square.
Since diagonals of a square bisects each other at ${90}^{\circ }.$

$\therefore \angle TOR={90}^{\circ }andOR=OT\left(sayx\right)Now,in\Delta TOR,\text{by Pythagoras theorem,}O{R}^2+O{T}^2=R{T}^2\Rightarrow x^2+x^2=\left(16{)}^2\right(\because RT=16cm)\Rightarrow 2x^2=256\Rightarrow x=8\sqrt{2}cm\dots ..(i)Now,TS=2\times QR(Given)=2\times 8\sqrt{2}=16\sqrt{2}cm...(ii)Thus,AR(\Delta PQR)=\frac{1}{2}\times OQ\times PR=\frac{1}{2}\times x\times 2x(\because OQ=OR=OT=x)=x^2=128c{m}^2...(iii)Also,Ar(TORS)=Ar(\Delta TOR)+Ar(\Delta TRS)=\frac{1}{2}\times TO\times OR+\frac{1}{2}\times TO\times TS=\frac{1}{2}\times 8\sqrt{2}\times 8\sqrt{2}+\frac{1}{2}\times 8\sqrt{2}\times 16\sqrt{2}[From\left(i\right)and\left(ii\right)]=192c{m}^2....(iv)\therefore \frac{Ar\left(TORS\right)}{Ar\left(\Delta PQS\right)}=\frac{192}{128}=\frac{3}{2}From(iii\left)and\right(iv\left)\right]$

Hence, the correct answer is option (c).