1.

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 : 80This 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»

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

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


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