Ratio of triangle A and B if the length of the sides are A:25,25, 30 and b:25,25, 40

Question

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

Answer 1

$A$'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 $Area_A:Area_B = 1:1$

Answer 2

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?