Areas within a right triangle

Geometry Level 3

△ A B C \triangle ABC is a right triangle at B B , with A B ‾ = 3 \overline{AB} = 3 and B C ‾ = 4 \overline{BC} = 4 . We take point D D on B C BC , such that B D ‾ = 1 \overline{BD} = 1 . We construct the perpendicular at D D meeting A C AC at E E . Find the ratio of areas of the yellow region to that of the blue region. That is, find [ A B E ] [ B D E ] \dfrac{[ABE]}{[BDE]} .


The answer is 1.3333.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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2 solutions

The altitude of the yellow triangle that corresponds to its side A B AB is congruent to the altitude of the blue triangle that corresponds to its side E D ED .
They both have length 1 1 .

Therefore, [ E A B ] [ B E D ] = A B E D = B C D C = 4 3 ≈ 1.33 \dfrac{\left[ EAB \right]}{\left[ BED \right]}=\dfrac{AB}{ED}=\dfrac{BC}{DC}=\dfrac{4}{3}\approx \boxed{1.33}

2 Helpful 1 Interesting 2 Brilliant 0 Confused
Hosam Hajjir
Sep 5, 2020

We note that △ A B E \triangle ABE and △ B D E \triangle BDE have the same altitude of 1 1 . Therefore, the ratio of their areas is the ratio of A B ‾ D E ‾ \dfrac{\overline{AB}}{\overline{DE}} and this ratio is (by similar triangles) the same as the ratio B C ‾ D C ‾ = 4 3 \dfrac{\overline{BC}}{\overline{DC}} = \dfrac{4}{3} .

Hence, [ A B E ] [ B D E ] = 4 3 = 1.3333 \dfrac{[ABE]}{ [BDE]} = \dfrac{4}{3} = 1.3333

2 Helpful 1 Interesting 1 Brilliant 0 Confused

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