Area of triangles

Question

Triangles ABC and CDE have a common vertex C with side AB of triangle ABC being parallel to side DE of triangle CDE. If length of side AB = 2 cm and length of side DE = 7 cm and perpendicular distance between sides AB and DE is 7.2 cm, then find the sum of areas of triangle ABC and triangle CDE.

Answer 1

Proof ∆ABC ≂∆CDE (alternate angle and vertical angle are same ,so using AA property both are similar) therefore,

Area ∆ABC/Area ∆CDE =(AB/DE)^2 =(2/7)^2 ,Now we know that Area ∆ABC = 1/2 *h1*AB and  Area ∆CDE = 1/2 *h2*DE where h1 and h2 are height of triangle ABC and CDE respectively

4/49 =(h1*2)/(h2*7) --> h1/h2 =2/7 Now h1+h2 =7.2 we get h1 = 1.6 and  h2= 5.6 and sum of both triangle area =21.2