I can think of only one way to solve this. The hint is in the original post's title: "Heron's Triangle." When calculating the area of a triangle whose side lengths you know, you have to use Heron's Formula:

A = √((s)(s-a)(s-b)(s-c)),

where s = (1/2)(a + b + c).

With that in mind, let:

a = 10 b = 17 c = 21

Therefore,

s = (10+17+21)/2 = 24

So,

A = √((24)(24-10)(24-17)(24-21))

=

√(24•14•7•3)

At this point, the best advice I can give you is that it's easier to solve this via factoring than to plug the above into the GRE's on-screen calculator:

√(4•2•3•7•2•7•3)

--right away, you can see the above has plenty of repeated factors. In fact, the above number is a perfect square.

A = 84

Since 84 < 87, the answer must be (B).

Quick. Easy. Delicious.

I do have to confess, though, that I hadn't remembered Heron's Formula, despite having spent the past few months tutoring my goddaughter in geometry. I looked the formula up earlier on Monday.

## 1 comment:

I can think of only one way to solve this. The hint is in the original post's title: "Heron's Triangle." When calculating the area of a triangle whose side lengths you know, you have to use Heron's Formula:

A = √((s)(s-a)(s-b)(s-c)),

where s = (1/2)(a + b + c).

With that in mind, let:

a = 10

b = 17

c = 21

Therefore,

s = (10+17+21)/2 = 24

So,

A = √((24)(24-10)(24-17)(24-21))

=

√(24•14•7•3)

At this point, the best advice I can give you is that it's easier to solve this via factoring than to plug the above into the GRE's on-screen calculator:

√(4•2•3•7•2•7•3)

--right away, you can see the above has plenty of repeated factors. In fact, the above number is a perfect square.

A = 84

Since 84 < 87, the answer must be (B).

Quick. Easy. Delicious.

I do have to confess, though, that I hadn't remembered Heron's Formula, despite having spent the past few months tutoring my goddaughter in geometry. I looked the formula up earlier on Monday.

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