AMC 8 2007 Problem 14

AMC8 2007 年第 14 题

三角形★★★☆☆

The base of isosceles

\triangle ABC

is 24 and its area is 60 . What is the length of one of the congruent sides?

等腰三角形△ABC 的底是 24，面积为 60。那么它的一条腰长是多少？

A.5
B.8
C.13
D.14
E.18

正确答案 · Correct Answer

解析 · Solution

Solution 1

The area of a triangle is shown by

\displaystyle \frac{1}{2} bh

. We set the base equal to 24, and the area equal to 60, and we get the triangle's height, or altitude, to be 5. In this isosceles triangle, the height bisects the base, so by using the Pythagorean Theorem,

a^2+b^2=c^2

, we can solve for one of the legs of the triangle (it will be the hypotenuse,

c

). a = 12, b = 5, c = 13. The answer is 13. According to Heron's Formula, setting side

a

as 24, we have

\sqrt{s(s-24)(s-b)(s-c)}=60

where

s

is the triangle's semiperimeter (i.e.

\displaystyle \frac{a+b+c}{2}

). Since the triangle is isosceles, b=c, so we can rewrite

s

\displaystyle \frac{24+2b}{2} = 12+b

. Substituting and solving the equation and taking the positive solution for

b

\sqrt{(12+b)(-12+b)(12)(12)}=60

\sqrt{144(144-b^2)}=60

144(144-b^2)=3600

-b^2=-169

b=13

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