Question

IN the adjoining figure $AQ=2,QB=4,BP=3,PC=5,CR=6$ and $RA=4$. Find the area of triangle $PQR$

• A. $4.8$
• B. $5.2$
• C. $5.8$
• D. $6$

Answer 1

The correct option is C $5.8$

$AQ=2,QB=4,BP=3,PC=5,CR=6,AR=4$

$AB=AQ+QB=2+4=6$

$AC=AR+CR=4+6=10$

$CB=BP+CP=5+3=8$

$6^2+8^2=36+64=100={10}^2$

$△ABC$ is $\perp △$

Using by pythagoras theorem

$A{B}^2+C{B}^2=A{C}^2,\angle B={90}^o$

In, $△PBQ,$

$B{Q}^2+B{P}^2=P{Q}^2$

$4^2+3^2=$

$9+16=25=P{Q}^2$

$PQ=5$

$ar\left(△ABC\right)=\frac{1}{2}\times BC\times BC=24$ square units

$ar\left(PQR\right)=24-(9+6+\frac{16}{5})=5.8$ sq. units

$\sin C=\frac{P}{H}=\frac{6}{10}=\frac{1}{3}$

$\sin A=\frac{8}{10}=\frac{4}{5}$

$\sin B=1$

Since,

$ar\left(△PBQ\right)=\frac{1}{2}\times 3\times 4=6sq.unit$

$ar\left(△AQR\right)=\frac{1}{2}\times AQ\times QR\times \sin \angle QAR$

$ar\left(△PCR\right)=\frac{1}{2}\times PC\times CR\times \sin \angle OCR=\frac{1}{2}\times 5\times 6\times \frac{3}{5}=9$