Q)the triangles ABC and EBD are similar (AC and DE are not parallel). If AB=8 cm, BE=4 cm and the area of DBE=6 cm2. find the area of ABC in cm2 A)18 B)24 C)36 D)48 E)none

Since its given AC and DE are not parallel, the only way the triangles could be similar is when Angle A = Angle E (and) Angle C = Angle D. Therefore using properties of similar triangles, we can write AB/BE = BC/BD 8/4 = BC/3 BC = 6. Area of ABC = 1/2 * AB * BC = 1/2 * 8* 6 = 24 B IMO

using S(DBE)=6 and BE=4 we find the height, h=2S/BE=12/4=3 now turn triangle DBE inside of triangle ABC, the common angle DBE=ABC won't change, but the side BE will stand in place of BD. let's use similarity of two triangles BE=4 and AB=8, the ratio between the sides is 1:2; the height drawn to BE is 3 and this must be in ratio of 2 x (multiplied) to the side AB. Thus, S(ABC)=(8*6)/2=24 arifaisal wrote:Q)the triangles ABC and EBD are similar (AC and DE are not parallel). If AB=8 cm, BE=4 cm and the area of DBE=6 cm2. find the area of ABC in cm2 A)18 B)24 C)36 D)48 E)none

arifaisal wrote:Q)the triangles ABC and EBD are similar (AC and DE are not parallel). If AB=8 cm, BE=4 cm and the area of DBE=6 cm2. find the area of ABC in cm2 A)18 B)24 C)36 D)48 E)none Solution: We can also use the relationship that if two triangles are similar, ratio of their areas is the square of the ratio of their corresponding sides. This means that (4/8)^2 = 6/(area of ABC). Or, area of ABC is 4*6 = 24 square cms.

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