Question

In the given figure, AC is the angle bisector and AB = 5.2 cm, AD = 4.5 cm. What will be ratios of areas of ∆ABC and ∆ACD?

(a) 131 : 90

(b) 131 : 91

(c) 121 : 81

(d) 120 : 80

This question was posed to me by my school teacher while I was bunking the class.

I need to ask this question from Area of Similar Triangle topic in portion Triangles of Mathematics – Class 10

Answer 1

Correct choice is (c) 121 : 81

Best explanation: We know that the ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding altitudes.

Here, AB=5.2cm and AD=4.5 cm

Now, In ∆ACD & ∆ACB

∠ACD = ∠ACB      (Both 90°)

AC = AC       (Common side)

∠BAC = ∠DAC      ( AC is the angle bisector)

Hence, ∆ACD ∼ ∆ACB      (By, ASA similarity test)

According to the theorem,

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle ADC}=(\frac {AB}{AD}) \)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle ADC}=(\frac {5.5}{4.5}) \)^2

\(\frac {area \, of \, \triangle ABC}{area \, of \, \triangle PQR}=\frac {30.25}{20.25}=\frac {121}{81} \)