In ΔABC, a line through A cuts the side BC at D such that BD : DC = 4 :5. If the area of ΔABD = 60 cm , then the area of ΔADC is
Let BD = 4$$x$$ and DC = 5$$x$$
Let height of the triangle be $$h$$ which will be same for both triangles ABD and ADC.
Now, area of $$\triangle$$ABD = $$\frac{1}{2} * BD * h$$ = 60
=> $$\frac{1}{2} * 4x * h$$ = 60
=> $$xh$$ = 30
Now, area of $$\triangle$$ADC = $$\frac{1}{2} * DC * h$$
= $$\frac{1}{2} * 5x * h = \frac{1}{2} * 5 * 30$$
= 75 $$cm^2$$
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