AMC8 2011

AMC8 2011 · Q16

AMC8 2011 · Q16. It mainly tests Triangles (properties), Pythagorean theorem.

Let $A$ be the area of a triangle with sides of length 25, 25, and 30. Let $B$ be the area of a triangle with sides of length 25, 25, and 40. What is the relationship between $A$ and $B$?

设$A$为边长分别为25、25和30的三角形的面积。设$B$为边长分别为25、25和40的三角形的面积。$A$和$B$之间是什么关系？

(A) $A = \frac{9}{16}B$ $A = \frac{9}{16}B$

(B) $A = \frac{3}{4}B$ $A = \frac{3}{4}B$

(D) $A = \frac{4}{3}B$ $A = \frac{4}{3}B$

(E) $A = \frac{16}{9}B$ $A = \frac{16}{9}B$

Answer

Correct choice: (C)

正确答案：（C）

Solution

25-25-30 We can draw the altitude for the side with length 30. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 30 into two segments with length 15. By the Pythagorean Theorem, we have \[15^2 + x^2 =25^2\] \[x^2 = 25^2 - 15^2\] \[x^2 = (25 + 15)(25-15)\] \[x^2= 40\cdot 10\] \[x^2= 400\] \[x = \sqrt{400}\] \[x= 20\] Thus we have two 15-20-25 right triangles. 25-25-40 We can draw the altitude for the side with length 40. By HL Congruence, the two triangles formed are congruent. Thus the altitude splits the side with length 40 into two segments with length 20. From the 25-25-30 case, we know that the other side length is 15, so we have two 15-20-25 right triangles. Let the area of a 15-20-25 right triangle be $x$. \[a = 2x\] \[b = 2x\] \[\boxed{\textbf{(C) } A = B}\]

25-25-30 我们可以对于边长30的那一边作高线。由HL全等，两构成的三角形全等。因此高线将边长30的一边分成两个长度为15的线段。由勾股定理，有 \[15^2 + x^2 =25^2\] \[x^2 = 25^2 - 15^2\] \[x^2 = (25 + 15)(25-15)\] \[x^2= 40\cdot 10\] \[x^2= 400\] \[x = \sqrt{400}\] \[x= 20\] 因此我们有两个15-20-25直角三角形。 25-25-40 我们可以对于边长40的那一边作高线。由HL全等，两构成的三角形全等。因此高线将边长40的一边分成两个长度为20的线段。从25-25-30的情况，我们知道另一边长为15，因此我们有两个15-20-25直角三角形。设15-20-25直角三角形的面积为$x$。 \[a = 2x\] \[b = 2x\] \[\boxed{\textbf{(C) } A = B}\]

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