Question

Question 16

In the given figure, if PR = 12 cm, QR = 6 cm and PL = 8 cm, then QM is



(a) 6 cm
(b) 9 cm
(c) 4 cm
(d) 2 cm

Answer 1

Given that, PR = 12 cm, QR = 6 cm and PL = 8cm
Now, in right angled $\Delta PLR$, using Pythagoras theorem,
$(Hypotenuse{)}^2=(Perpendicular{)}^2+(Base{)}^2$
$\Rightarrow P{R}^2=P{L}^2+L{R}^2$
$\Rightarrow L{R}^2=P{R}^2-P{L}^2=(12{)}^2-(8{)}^2$
$\Rightarrow L{R}^2=144-64=80$
$\Rightarrow LR=\sqrt{80}=4\sqrt{5}cm$
$\because LR=LQ+QR\Rightarrow LQ=LR-QR=(4\sqrt{5}-6)cm$
Now, area of $\Delta PLR,$
$A_1=\frac{1}{2}\times LR\times PL$
$=\frac{1}{2}\times \left(4\sqrt{5}\right)\times 8$
$=16\sqrt{5}c{m}^2$
Again, area of $\Delta PLQ$,
$A_2=\frac{1}{2}\times LQ\times PL$
$=\frac{1}{2}\times (4\sqrt{5}-6)\times 8$
$=(16\sqrt{5}-24)c{m}^2$
$\therefore$ Area of $\Delta PLQ$ + Area of $\Delta PQR$
$\Rightarrow 16\sqrt{5}=(16\sqrt{5}-24)+Areaof\Delta PQR$
$\Rightarrow$ Area of $\Delta PQR=24c{m}^2$
$\Rightarrow \frac{1}{2}\times PR\times QM=24$
$\Rightarrow \frac{1}{2}\times 12\times QM=24$
$\therefore QM=4cm$