Question

Areas of $△$ADC and $△$CDB are in ratio 3:2
and if AB = 10 cm and Area of $△$ABC is 20 $c{m}^2$, find the length of AD, BD.

• A. 3 cm, 2 cm
• B. 2 cm, 3 cm
• C. 6 cm, 4 cm
• D. 4 cm, 6 cm

Answer 1

The correct option is C 6 cm, 4 cm

Area of triangle = $\frac{1}{2}$$\times$base$\times$height

Area of $△$ABC = $\frac{1}{2}$$\times$AB$\times$CD
20 = $\frac{1}{2}$$\times$10$\times$CD
CD = $\frac{20\times 2}{10}$ = 2$\times$2 = 4 cm

Let common multiple of area of $△$ADC and $△$CDB be y.
According to given condition
Area of $△$ADC +Area of $△$ CDB = Area of $△$ABC
3y+2y=20
5y = 20
y = 4 $c{m}^2$

Area of $△$ADC = 3y = 3$\times$4 = 12
12 = $\frac{1}{2}$$\times$base$\times$height
12 = $\frac{1}{2}$$\times$AD$\times$CD
AD = $\frac{12\times 2}{CD}$ = $\frac{12\times 2}{4}$ = 3$\times$2 = 6 cm

AB= AD+BD
10 = 6 + BD
BD = 10-6 = 4 cm