Question

In ΔABC, P divides the side AB such that AP : PB = 1 : 2. Q is a point in AC such that PQ||BC. Find the ratio of the areas of ΔAPQ and trapezium BPQC.

Answer 1

Ratio of the areas of ΔAPQ and trapezium BPQC is 1:2

• In triangle APQ and triangle ABC

• <A = <A (Common)

• <P = <B ( Corresponding angles)

• <Q = <C ( Corresponding angles)

• triangle APQ is similar to triangle

ABC

• AP:AB = AP/(AP+PB)

• AP:AB = AP/(AP+2AP)

• AP:AB = AP/(3AP) = 1:3

• As AP:AB = 1:3

ar(APQ):ar(ABC) = 1:3

• ar(ABC) = 3ar(APQ)

• ar(APQ) + ar(BPQC) = 3ar(APQ)

• ar(BPQC)=2ar(APQ)

• ar(APQ):ar(BPQC) =1:2

Answer 2

The ratio of the areas of ΔAPQ and trapezium BPQC is 1:8

Step-by-step explanation:

• Here it is given that

        $\mathbf{\frac{PB}{AP}=\frac{2}{1}}$

        On adding both side 1

        $\mathbf{\frac{PB}{AP}+1=\frac{2}{1}+1}$

        $\mathbf{\frac{PB+AP}{AP}=\frac{2+1}{1}}$

        $\mathbf{\frac{AB}{AP}=\frac{3}{1}}$        ...1)

• From given condition, we can say that

        Area of trapezium BPQC = arΔABC -arΔAPQ       ...2)

• Here it is also given that  PQ||BC , we can say that

        $\mathbf{\Delta ABC\sim \Delta APQ}$

• From theorem of area of similar triangle Its state that ratio of area of similar triangle is equal to ratio of square of corresponding side.

        $\mathbf{\frac{\Delta ABC}{\Delta APQ}=\left ( \frac{AB}{AP} \right )^{2}}$

        $\mathbf{\frac{\Delta ABC}{\Delta APQ}=\left ( \frac{3}{1} \right )^{2}}$

        $\mathbf{\frac{\Delta ABC}{\Delta APQ}=\frac{9}{1}}$

• On subtracting both side 1, we get

        $\mathbf{\frac{\Delta ABC}{\Delta APQ} -1 =\frac{9}{1} - 1}$

        $\mathbf{\frac{\Delta ABC-\Delta APQ}{\Delta APQ}=\frac{9-1}{1}}$               ...3)

• From equation 2) and equation 3)

        $\mathbf{\frac{Area\ of\ trapezium\ BPQC}{\Delta APQ}=\frac{8}{1}}$

        We can also write above

        $\mathbf{\frac{\Delta APQ}{Area\ of\ trapezium\ BPQC}=\frac{1}{8}}$