Question 1116888
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The area of a triangle is given by *[tex \Large \frac{1}{2}bh]


So if the area is 20 when the base is 16, then *[tex \Large 8h\ =\ 20], which is to say *[tex \Large h\ =\ 2.5]


Since the larger (area 45) triangle is similar, all dimensions of the two triangles must be in proportion.  The ratio of the base to the height of the small triangle must be the same as the ratio of the base to the height of the large triangle.  Since *[tex \Large h\ =\ 2.5] when *[tex \Large b\ =\ 16], the desired ratio is *[tex \Large \frac{16}{2.5}\ =\ 6.4].  Hence, regardless of the actual height, the base must be 6.4 times larger, which is to say: *[tex \Large b\ =\ 6.4h]


Now we can create a single variable equation to describe the larger triangle:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{2}(6.4h)(h)\ =\ 45]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3.2h^2\ =\ 45]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h^2\ =\ 14.0625]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ \sqrt{14.0625}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ 3.75]


and finally:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ (6.4)(3.75)\ =\ 24]


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  


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