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In the figure, if $PR=12 cm, QR = 6 cm$ and $PL=8cm$, then find $QM$.
"
Given :
In the given figure, if $PR=12 cm, QR = 6 cm$ and $PL=8cm$.
To do :
We have to find the length of $QM$.
Solution :
$\triangle PLR$ is a right-angled triangle.
By Pythagoras theorem,
$PL^2+LR^2=PR^2$
$8^2+LR^2=12^2$
$64+LR^2=144$
$LR^2=144-64=80$
$LR=\sqrt{80}$
$LR=4\sqrt{5}$
$LR = LQ+QR$
$LQ=LR-QR$
$LQ=4\sqrt{5}-6$
We know that area of the triangle is $\frac{1}{2} \times b \times h$
So, the area of $\triangle PLR = \frac{1}{2} \times LR \times PL$
$ = \frac{1}{2} \times 4\sqrt{5} \times 8$
$ = 16\sqrt{5} cm^2$.
The area of $\triangle PLQ = \frac{1}{2} \times LQ \times PL$
$ = \frac{1}{2} \times 4\sqrt{5}-6 \times 8$
$= 4(4\sqrt{5}-6)$
$ = 16\sqrt{5} -24 cm^2$.
Area of $\triangle PQR = \frac{1}{2} \times PR \times QM$
Here, the area of $\triangle PQR = area of \triangle PLR-area of \triangle PLQ$
Area of $\triangle PQR = 16\sqrt{5} cm^2 - (16\sqrt{5} -24 )cm^2$
$ = 16\sqrt{5} cm^2 - 16\sqrt{5} +24 cm^2$
Area of $\triangle PQR =24 cm^2$.
$\frac{1}{2} \times PR \times QM = 24$
$\frac{1}{2} \times 12 \times QM = 24$
$6 \times QM = 24 $
$QM = \frac{24}{6} = 4$
Therefore, the length of $QM$ is 4 cm.
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