Question 1209158: In the diagram equal angles are marked. Given that
EB=6 cm, CE 18 cm and DB=4 cm. Find the area of Triangle ABC.
https://ibb.co/vxJcWkc
Answer by greenestamps(13203)   (Show Source):
You can put this solution on YOUR website!
In triangles ABC and EBD, we are given that angles BAC and BED are congruent. Then, since the two triangles share angle B, those two triangles are similar.
Corresponding sides BC and BD have lengths 24 and 4, so the ratio of similarity between the two triangles is 6:1.
The ratio of the areas of the two triangles is the square of the ratio of similarity, which is 36:1.
The diagram shows that angle EBD and EDB are congruent, so triangle EBD is isosceles. The altitude of triangle EDB from vertex E to side DB then cuts DB in half. From that, the Pythagorean Theorem tells us that that altitude of triangle EDB has length 4*sqrt(2); and from that we can find the area of triangle EDB is 8*sqrt(2).
And since the ratio of the areas of the two triangles is 36:1, the area of triangle ABC is 36*(8*sqrt(2)) = 288*sqrt(2).
ANSWER: 288*sqrt(2)
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