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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