Question

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?

Answer 1

`26+4\sqrt13`

Explanation:

The area of isosceles `\triangle PQR` having sides `PQ=PR=13` & included angle `\angle QPR` or haing base `PR=13` & altitude from Q to the side PR is `12` hence the area of `\triangle PQR` is given as

`1/2\text{(base PR)}\times \text{(altitude on PR)}=1/2(PQ)(PR)\sin \angle QPR`

`1/2(13)(12)=1/2(13)(13)\sin\angle QPR`

`\sin\angle QPR=12/13`

`\angle QPR=\sin^{-1}(12/13)`

Now, in isosceles `\triangle PQR`, sum of interior angles is zero hence

`angle PQR+\angle PRQ+\angle QPR=\pi`

`2\angle PQR+\sin^{-1}(12/13)=\pi\quad(\because \angle PRQ=\angle PQR)`

`\angle PQR=\pi/2-1/2\sin^{-1}(12/13)`

Now, applying sine rule in isosceles `\triangle PQR`

`\frac{QR}{\sin\angle QPR}=\frac{PR}{\sin\angle PQR}`

`\frac{QR}{\sin(\sin^{-1}(12/13))}=\frac{13}{\sin(\pi/2-1/2\sin^{-1}(12/13))}`

`\frac{QR}{12/13}=\frac{13}{\cos(1/2\sin^{-1}(12/13))}`

`QR=12/{\cos(\cos^{-1}(\sqrt{{1+5/13}/{2}}))}`

`=12/{\sqrt{9/13}}`

`={12\sqrt13}/3`

`=4\sqrt13`

hence the perimeter of isosceles `\triangle PQR`

`=PQ+QR+PR`

`=13+4\sqrt13+13`

`=26+4\sqrt13`

Answer 2

`26+4sqrt13`.

Explanation:

Let `S` be the foot of the `bot` from `Q` on the side `PR`.

`:. QS=12........."[Given]"`.

Also, suppose that, the area of `DeltaPQR` is `A`.

Then, from Trigo., `A=1/2*PQ*PR*sin/_QPR, and,`

by Geom., `A=1/2*QS*PR`.

`:. 1/2*PQ*PR*sin/_QPR=1/2*QS*PR`.

`:. 13*13*sin/_QPR=12*13 rArr sin/_QPR=12/13.`

Now in the right- `DeltaQSP,`

`/_QSP=90^@, QP=13, QS=12.`

`:. PS^2=QP^2-QS^2=13^2-12^2=25`.

`:. PS=5, :., SR=PR-PS=13-5=8.`

Now in the right- `DeltaQSR, /_QSR=90^@, QS=12, SR=8.`

`:. QR^2=QS^2+SR^2=12^2+8^2=208`.

`:. QR=4sqrt13`.

`:."The Reqd. Perimeter of "DeltaPQR=PQ+QR+RP,`

`=13+4sqrt13+13`,

`=26+4sqrt13`, as Respected Harish Chandra Rajpoot has

readily derived!

`color(blue)("Enjoy Maths.!")`