Question

n the given figure angle EDB = angle ACB . if BE=6cm ,BD=5cm and EC=4cm and the area of triangle BED=9cmsq find the area of triangle ABC​

Answer 1

Given,

∆ABC and ∆EBD,

angle ACB is equal to angle EDB,

common angle both the triangles is angle EBD and angle ABC ,

area of triangle BED = 9cm^2

∆ABC is similar to ∆EBD (by AA Similarly)

Step 1:-

AB/BE = BC/BD

AB = BC×BE/BD

(Given in the question that BC = 10cm, BE = 6cm and BD = 5cm)

AB = 10×6/5

By applying the values we get that

AB =12cm

Step 2:-

area of ∆ABC/area of ∆BED = (AB/BE) ^2

area of ∆ABC = (AB/BE) ^2 × area of ∆ BED

area of ∆ABC = (12/6) ^2 × 9

area of ∆ABC = 4 × 9

area of ∆ABC = 36cm^ 2

I hope that you can understand the problem