Question 33
$$\triangle ABC$$ and $$\triangle DBC$$ are on the same BC but on opposite sides of it. AD and BC intersect each other at O.If AO = a cm, DO = b cm and the area of $$\triangle ABC = x cm^2$$, then what is the area(in $$cm^2$$) of $$\triangle DBC$$Â
Solution
Area of $$\triangle$$ ABC :Â area of $$\triangle$$ DBC = $$\frac{1}{2} \times a \times BC :Â \frac{1}{2} \times b \times BC$$
x :Â area of $$\triangle$$ DBC = $$\frac{a}{b}$$
Area of $$\triangle$$ DBC =Â $$\frac{ax}{b}$$
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