The areas of two similar triangles are 81 cm² and 49 cm² respectively. If the altitude of the smaller triangle is 3.5 cm, then the corresponding altitude of the bigger triangle is
(a) 9 cm
(b) 7 cm
(c) 6 cm
(d) 4.5 cm
Solution:
Given ∆ABC ~ ∆PQR, area of ∆ABC = 54 cm² and area of ∆PQR = 24 cm². If AD and PM are medians of ∆’s ABC and PQR respectively, and length of PM is 10 cm, then length of AD is
(a) 
cm
(b) 
cm
(c) 15 cm
(d) 22.5 cm
Solution:
More Solutions:
- CM and RN are respectively the medians of ∆ABC and ∆PQR.
- Medians AD and BE of ∆ABC meet at the point G.
- In the figure given below, AB, EF and CD are parallel lines.
- ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm.
- A 15-meter high tower casts a shadow of 24 meters.
- In the figure, DE parallel BG, AD = 3 cm, BD = 4 cm and BC = 5 cm.