Question

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

Q 12 em Gem R

• A. $6cm$
• B. $9cm$
• C. $4cm$
• D. $2cm$

Answer 1

The correct option is C

$4cm$

Explanation for the correct option:

Step 1: Applying Pythagoras theorem.

$PR=12cm$

$QR=6cm$

$PL=8cm$

$Hypotenus{e}^2=Perpendicula{r}^2+bas{e}^2$

From the figure we observe

$P{R}^2=P{L}^2+L{R}^2$

$L{R}^2=P{R}^2-P{L}^2$

$L{R}^2={12}^2-8^2$

$L{R}^2=144-64$

$L{R}^2=80$

Step 2: Transferring the square.

It becomes,$LR=\sqrt{80}$

Therefore, $LR=4\sqrt{5}$

Then,

$LR=LQ+QR$

$LQ=LR-QR$

$LQ=4\sqrt{5}-6$

Step 3: Find the area of triangle $PLR$.

We know that,

$Areaof△PLR=\frac{1}{2}\times LR\times PL$

$=\frac{1}{2}\times 4\sqrt{5}\times 8$

$=1\times 4\sqrt{5}\times 4$

$=16\sqrt{5}$

Step 4: Find the area of triangle $PLQ$.

$Areaof△PLQ=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times LQ\times PL$

$=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times (4\sqrt{5}-6)\times 8$

$=4(4\sqrt{5}-6)$

$=16\sqrt{5}-24c{m}^2$

So, $AreaofPLR=AreaofPLQ+AreaofPQR$

$16\sqrt{5}=(16\sqrt{5}-24)+AreaofPQR$

Step 4: Compute the required value:

Therefore,

$AreaofPQR=24c{m}^2$

$\frac{1}{2}\times PR\times QM=24$

$\frac{1}{2}\times 12\times QM=24$

$6\times QM=24$

$QM=\frac{24}{6}$

$QM=4cm$

Hence option $C$ is the correct answer.