## Monday, June 4, 2012

### answer to last week's MGRE Math Beast Challenge

The answer to last week's MGRE Math Beast Challenge was indeed (B). But MGRE shows how to arrive at this answer without using Heron's Formula. To wit:

To find the area of triangle PQR we need a base and a height. If we consider side PR the base of the triangle, then QS, which is at a right angle to PR, is the height. We know that both triangle PQS and triangle SQR are right triangles, but to find the height QS we’ll need to know the length of either PS or SR. Unfortunately we don’t know either one, so we’ll have to name variables for several legs of the triangle.

In this case we’ll let x represent the length of PS and y represent the height QS. Note that if PS has length x, then SR has length 21 – x, so we do not need to (and should not) name a variable for length SR. Updating our diagram yields the following:

Apply the Pythagorean Theorem to each right triangle.
From triangle PQS: x2 + y2 = 100
From triangle SQR: (21 – x)2 + y2 = 289

We solve each in terms of y2:
y2 = 100 – x2
y2 = 289 – (21 – x)2

Then set the two expressions equal to y2 equal to each other:
289 – (21 – x)2 = 100 – x2

Now simplify:
189 – (21 – x)2 = -x2
189 = (21 – x)2 – x2
189 = (441 – 42x + x2) – x2
189 = 441 – 42x
42x = 252
x = 6

Now that the value of x is known, solve for y:
100 = 36 + y2
64 = y2
8 = y
(Or just recognize that PQS is a 6 – 8 – 10 right triangle.)

Finally, we can solve for the area of triangle PQR:

(1/2)bh = (1/2)(21)(8) = 84