Question

In the given figure, $AD$ is perpendicular bisector of $BC$ and the perimeter of the triangle is $36$ cm and $BC=10$ cm. Then value of $2AD+BC$ is

• A. $17$ cm
• B. $32$ cm
• C. $22$ cm
• D. $34$ cm

Answer 1

The correct option is D $34$ cm

In $△ABD\text{and}△ACD$
$AD=AD\text{(common side in both triangles)}$
$BD=CD\text{(AD bisect BC)}$
$\angle ADB=\angle ADC\text{(AD perpendicular to BC so both angles are equal to}{90}^o)$
So, $△ABD\cong △ACD\text{(SAS property)}$

We know that corresponding sides of congruent triangles are equal.
$\Rightarrow AB=AC$

As perimeter of triangle is $36$ cm,
So, $AB+BC+AC=36$
$2AB+10=36(\because \text{Put AB=AC, BC=10})AB=\frac{36-10}{2}AB=\frac{26}{2}AB=13\text{cm}$

Now, In $△ABD\text{as}\angle ADB={90}^o$
applying pythagoras theorem,
$A{B}^2=B{D}^2+A{D}^2$
$A{D}^2=A{B}^2-B{D}^2$
$A{D}^2={13}^2-5^2(\because BD=\frac{BC}{2}\text{as AD bisect BC})$
$A{D}^2=169-25$
$A{D}^2=144$
$AD=\sqrt{144}$
$AD=12$ cm

Now, $2AD+BC=2\times 12+10$
$2AD+BC=24+10$
$2AD+BC=34$ cm

So, option $\left(d\right)$ correct.