# In PQR triangle , PQ=PR=13. If the length of the altitude drawn from Q perpendicular to side PR is 12 what is the perimeter of PQR?

$26 + 4 \setminus \sqrt{13}$

#### Explanation:

The area of isosceles $\setminus \triangle P Q R$ having sides $P Q = P R = 13$ & included angle $\setminus \angle Q P R$ or haing base $P R = 13$ & altitude from Q to the side PR is $12$ hence the area of $\setminus \triangle P Q R$ is given as

$\frac{1}{2} \setminus \textrm{\left(b a s e P R\right)} \setminus \times \setminus \textrm{\left(a < i t u \mathrm{de} o n P R\right)} = \frac{1}{2} \left(P Q\right) \left(P R\right) \setminus \sin \setminus \angle Q P R$

$\frac{1}{2} \left(13\right) \left(12\right) = \frac{1}{2} \left(13\right) \left(13\right) \setminus \sin \setminus \angle Q P R$

$\setminus \sin \setminus \angle Q P R = \frac{12}{13}$

$\setminus \angle Q P R = \setminus {\sin}^{- 1} \left(\frac{12}{13}\right)$

Now, in isosceles $\setminus \triangle P Q R$, sum of interior angles is zero hence

$\angle P Q R + \setminus \angle P R Q + \setminus \angle Q P R = \setminus \pi$

$2 \setminus \angle P Q R + \setminus {\sin}^{- 1} \left(\frac{12}{13}\right) = \setminus \pi \setminus \quad \left(\setminus \because \setminus \angle P R Q = \setminus \angle P Q R\right)$

$\setminus \angle P Q R = \setminus \frac{\pi}{2} - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{12}{13}\right)$

Now, applying sine rule in isosceles $\setminus \triangle P Q R$

$\setminus \frac{Q R}{\setminus \sin \setminus \angle Q P R} = \setminus \frac{P R}{\setminus \sin \setminus \angle P Q R}$

$\setminus \frac{Q R}{\setminus \sin \left(\setminus {\sin}^{- 1} \left(\frac{12}{13}\right)\right)} = \setminus \frac{13}{\setminus \sin \left(\setminus \frac{\pi}{2} - \frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{12}{13}\right)\right)}$

$\setminus \frac{Q R}{\frac{12}{13}} = \setminus \frac{13}{\setminus \cos \left(\frac{1}{2} \setminus {\sin}^{- 1} \left(\frac{12}{13}\right)\right)}$

$Q R = \frac{12}{\setminus \cos \left(\setminus {\cos}^{- 1} \left(\setminus \sqrt{\frac{1 + \frac{5}{13}}{2}}\right)\right)}$

$= \frac{12}{\setminus \sqrt{\frac{9}{13}}}$

$= \frac{12 \setminus \sqrt{13}}{3}$

$= 4 \setminus \sqrt{13}$

hence the perimeter of isosceles $\setminus \triangle P Q R$

$= P Q + Q R + P R$

$= 13 + 4 \setminus \sqrt{13} + 13$

$= 26 + 4 \setminus \sqrt{13}$

Jul 20, 2018

$26 + 4 \sqrt{13}$.

#### Explanation:

Let $S$ be the foot of the $\bot$ from $Q$ on the side $P R$.

$\therefore Q S = 12. \ldots \ldots . . \text{[Given]}$.

Also, suppose that, the area of $\Delta P Q R$ is $A$.

Then, from Trigo., $A = \frac{1}{2} \cdot P Q \cdot P R \cdot \sin \angle Q P R , \mathmr{and} ,$

by Geom., $A = \frac{1}{2} \cdot Q S \cdot P R$.

$\therefore \frac{1}{2} \cdot P Q \cdot P R \cdot \sin \angle Q P R = \frac{1}{2} \cdot Q S \cdot P R$.

$\therefore 13 \cdot 13 \cdot \sin \angle Q P R = 12 \cdot 13 \Rightarrow \sin \angle Q P R = \frac{12}{13.}$

Now in the right- $\Delta Q S P ,$

$\angle Q S P = {90}^{\circ} , Q P = 13 , Q S = 12.$

$\therefore P {S}^{2} = Q {P}^{2} - Q {S}^{2} = {13}^{2} - {12}^{2} = 25$.

$\therefore P S = 5 , \therefore , S R = P R - P S = 13 - 5 = 8.$

Now in the right- $\Delta Q S R , \angle Q S R = {90}^{\circ} , Q S = 12 , S R = 8.$

$\therefore Q {R}^{2} = Q {S}^{2} + S {R}^{2} = {12}^{2} + {8}^{2} = 208$.

$\therefore Q R = 4 \sqrt{13}$.

$\therefore \text{The Reqd. Perimeter of } \Delta P Q R = P Q + Q R + R P ,$

$= 13 + 4 \sqrt{13} + 13$,

$= 26 + 4 \sqrt{13}$, as Respected Harish Chandra Rajpoot has

$\textcolor{b l u e}{\text{Enjoy Maths.!}}$