Question

Question 11

Ratio of areas of $\Delta MNO,\Delta MOPand\Delta MPQ$ in the given figure is



(a) 2 : 1 : 3
(b) 1 : 3 : 2
(c) 2 : 3 : 1
(d) 1 : 2 : 3

Answer 1

From the given figure,
Area of $\Delta MNO=\frac{1}{2}\times NO\times MO=\frac{1}{2}\times 4\times 5=10c{m}^2$
$[\because areaoftriangle=\frac{1}{2}\times base\times height]$
Area of $\Delta MOP=\frac{1}{2}\times OP\times MO=\frac{1}{2}\times 2\times 5=5c{m}^2$
Area of $\Delta MPQ=\frac{1}{2}\times PQ\times MO=\frac{1}{2}\times 6\times 5=15c{m}^2$
Hence, required ratio = 10 : 5 : 15 = 2 : 1 : 3