If hatQ is a Right-angle in DeltaPQR and PR-PQ=9 , PR-QR=18 then what is the perimeter of the triangle?

Mar 18, 2018

Perimeter is $108$

Explanation:

As in right angled triangle $P Q R$, $m \angle Q = {90}^{\circ}$, $\overline{P R}$ is hyptenuse and te largest side. Further, according to Pythagoras theorem

$P {R}^{2} = Q {R}^{2} + P {Q}^{2}$ .......................(A)

Now let $P R = x$, then as $P R - P Q = 9$, $P Q = x - 9$

and as $P R - Q R = 18$ and therefore $Q R = x - 18$

Putting these values in (A), we have

${x}^{2} = {\left(x - 9\right)}^{2} + {\left(x - 18\right)}^{2}$

or ${x}^{2} = {x}^{2} - 18 x + 81 + {x}^{2} - 36 x + 324$

or ${x}^{2} - 54 x + 405 = 0$

or ${x}^{2} - 45 x - 9 x + 405 = 0$

or $x \left(x - 45\right) - 9 \left(x - 45\right) = 0$

or $\left(x - 9\right) \left(x - 45\right) = 0$

Observe that as all sides are positive, we must have $x > 18$

and hence only solution is $x = 45$

and sides are $45$, $36$ and $27$

and perimeter is $45 + 36 + 27 = 108$

Mar 18, 2018

Perimeter = $27 + 36 + 45 = 108$

Explanation:

If $\hat{Q}$ is 90°, then $P R$ is the hypotenuse.

$P Q \mathmr{and} R Q$ can be written in terms of the hypotenuse.

Let $P R = x$

Then $P Q = x - 9 \mathmr{and} R Q = x - 18$

Write an equation using Pythagoras' Theorem:

${\left(x - 9\right)}^{2} + {\left(x - 18\right)}^{2} = {x}^{2}$

${x}^{2} - 18 x + 81 + {x}^{2} - 36 x + 324 = {x}^{2}$

${x}^{2} - 54 x + 405 = 0$

$\left(x - 45\right) \left(x - 9\right) = 0$

$\therefore x = 45 \mathmr{and} x = 9 \rightarrow$ reject $9$ as being too short

If the hypotenuse is $45$, then the sides are

$45 - 9 = 36 \mathmr{and} 45 - 18 = 27$

Perimeter = $27 + 36 + 45 = 108$