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?

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#

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.!")#