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?

2 Answers

26+4\sqrt1326+413

Explanation:

The area of isosceles \triangle PQRPQR having sides PQ=PR=13PQ=PR=13 & included angle \angle QPRQPR or haing base PR=13PR=13 & altitude from Q to the side PR is 1212 hence the area of \triangle PQRPQR is given as

1/2\text{(base PR)}\times \text{(altitude on PR)}=1/2(PQ)(PR)\sin \angle QPR12(base PR)×(altitude on PR)=12(PQ)(PR)sinQPR

1/2(13)(12)=1/2(13)(13)\sin\angle QPR12(13)(12)=12(13)(13)sinQPR

\sin\angle QPR=12/13sinQPR=1213

\angle QPR=\sin^{-1}(12/13)QPR=sin1(1213)

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

angle PQR+\angle PRQ+\angle QPR=\piPQR+PRQ+QPR=π

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

Jul 20, 2018

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!

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