If $A$ triangle's side length are $25,25$ and $30$ and $B$ triangle's length are $25,25$ and $40$ what is the ratio between the two areas of the triangles? #math
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$\begingroup$$A$'s side lengths are: $25, 25 \text{, and }30$
Let $S = \dfrac{25+25+30}{2}=40$
Using Heron's Formula we know: $$Area_A = \sqrt{40(40-25)(40-25)(40-30)}$$ $$=\sqrt{40(15)(15)(10)}$$ $$=\sqrt{90000}$$ $$=300$$
$B$'s side lengths are: $25, 25 \text{, and }40$
Let $S = \dfrac{25+25+40}{2}=45$
Using Heron's Formula we know: $$Area_B = \sqrt{45(45-25)(45-25)(45-40)}$$ $$ =\sqrt{45(20)(20)(5)}$$ $$=\sqrt{90000}$$ $$=300$$
So $Area_A:Area_B = 1:1$
$\endgroup$ $\begingroup$Hint: These two triangles are isosceles, so the altitude on the base bisects the base.
That means the $25-25-30$ triangle is split in to two right triangles with a hypotenuse of $25$ and one leg of $15$. What is the altitude of the $25-25-30$ triangle?
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The $25-25-40$ triangle is split in to two right triangles with a hypotenuse of $25$ and one leg of $20$. What is the altitude of the $25-25-40$ triangle?
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