Question

In the given figure, PR and QS are two diameters of the circle.
If $PR=28cm$ and $PS=14\sqrt{3}$ cm, then find area of triangle $ORQ$.

• A. $87.87$ c$m^2$
• B. $84.87$ c$m^2$
• C. $86.87$ c$m^2$
• D. $85.87$ c$m^2$
  

Answer 1

Answer: (B)

$84.87$ c$m^2$

In the given figure,

Drop a perpendicular from $O$ to side $QR$.

It is given that $PR=28cm$

$\therefore OR=\frac{PR}{2}=\frac{28}{2}=14cm$

So, the radius of a circle is $14cm$

Also,

$PS=QR=14\sqrt{3}$

Now, after construction, in$△RTO$

$\angle OTR={90}^0,TR=\frac{14\sqrt{3}}{2}=7\sqrt{3},OR=14cm$

Using Pythagorus theorem,

$OT=\sqrt{{14}^2-(7\sqrt{3}{)}^2}=\sqrt{196-147}=7cm$

Area of triangle $ROQ$

$=\frac{1}{2}\times OT\times QR=\frac{1}{2}\times 7\times 14\sqrt{3}{cm}^2$

$=84.87{cm}^2$

Hence option ${}^{\prime }{B}^{\prime }$ is the answer.