Question

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

Answer 1

Perimeter is `108`

Explanation:

As in right angled triangle `PQR`, `m/_Q=90^@`, `bar(PR)` is hyptenuse and te largest side. Further, according to Pythagoras theorem

`PR^2=QR^2+PQ^2` .......................(A)

Now let `PR=x`, then as `PR-PQ=9`, `PQ=x-9`

and as `PR-QR=18` and therefore `QR=x-18`

Putting these values in (A), we have

`x^2=(x-9)^2+(x-18)^2`

or `x^2=x^2-18x+81+x^2-36x+324`

or `x^2-54x+405=0`

or `x^2-45x-9x+405=0`

or `x(x-45)-9(x-45)=0`

or `(x-9)(x-45)=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`

Answer 2

Perimeter = `27+36+45 = 108`

Explanation:

If `hatQ` is `90°`, then `PR` is the hypotenuse.

`PQ and RQ` can be written in terms of the hypotenuse.

Let `PR = x`

Then `PQ = x-9 and RQ = x-18`

Write an equation using Pythagoras' Theorem:

`(x-9)^2 +(x-18)^2 = x^2`

`x^2 -18x+81 + x^2 -36x+324= x^2`

`x^2 -54x + 405 =0`

`(x-45)(x-9)=0`

`:. x=45 or x =9 rarr` reject `9` as being too short

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

`45-9 = 36 and 45-18 = 27`

Perimeter = `27+36+45 = 108`