Question

The area of two similar triangles is 25cm^2 and 121cm^2. If the median of the bigger triangle is 10 cm, then what will be the corresponding median of the smaller triangle?

(a) \(\frac {51}{11}\)

(b) \(\frac {10}{11}\)

(c) \(\frac {50}{11}\)

(d) \(\frac {5}{11}\)

The question was posed to me in homework.

My question comes from Area of Similar Triangle in section Triangles of Mathematics – Class 10

Answer 1

The correct answer is (d) \(\frac {5}{11}\)

For explanation I would say: We know that the ratio of areas of similar triangles is equal to the ratio of the squares of their corresponding medians.

Here the area of ∆ABC is 25cm^2 and area of ∆PQR is 121cm^2. Also, PS = 10 cm

According to the theorem,

\(\frac {area \, of \, triangle \, \triangle ABC}{area \, of \, triangle \, \triangle PQR}=(\frac {AD}{PS})\)^2

\(\frac {25}{121}=(\frac {AD}{10})\)^2

\(\sqrt {\frac {25}{121}}=\frac {AD}{10}\)

\(\frac {AD}{10}=\frac {5}{11}\)

PS = \(\frac {5}{11} × 10 = \frac {50}{11}\)