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Geometry Problem URGENTTTTTT

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by shankar.ashwin » Sun Nov 20, 2011 7:31 am
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
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by pemdas » Sun Nov 20, 2011 12:56 pm
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
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by Anurag@Gurome » Sun Nov 20, 2011 7:05 pm
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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