Question

In a triangle abc p divides the side ab such that ap :pb=2:3 . q is a point on ac such that pq parallel to bc . the ratio of the area of triangle apq and trapezium bpqc is?

Answer 1

The ratio of the areas of Δ APQ and the trapezium BPQC 2 : 21  

Given :  

In ΔABC , P divides the side AB such that AP : PB = 2 : 3,

Q is a point on AC on such that PQ || BC.

To prove : The ratio of the areas of Δ APQ and the trapezium BPQC.

Proof :  

In Δ APQ  and Δ ABC

∠ APQ = ∠ B         [corresponding angles]

∠ PAQ = ∠ BAC    [common]

Δ APQ ∼ Δ ABC  [By AA Similarity criterion]

By Using the Theorem, the ratio of the areas of two triangles is equal to the square of the ratio of their corresponding sides.

ar(Δ APQ) / ar(Δ ABC) = (AP/AB)²

ar(Δ APQ) / ar(Δ ABC) = AP²/AB²

Let AP = 2x and PB = 3x

AB = AP + PB  

AB = 2x + 3x

AB = 5x

ar (Δ APQ) / ar (Δ ABC) = (2x)²/ (5x)²

ar (Δ APQ) / ar (Δ ABC) = 4x² / 25x²

ar (Δ APQ) / ar (Δ ABC) = 4 / 25

Let Area of ΔAPQ  = 4x sq. units and Area of ΔABC = 25x sq.units

ar [trap. BPQC] = ar (Δ ABC  ) – ar (Δ APQ)

ar [trap. BPQC] = 25x - 4x

ar [trap. BPQC] = 21x sq units

Now,

ar Δ APQ / ar (trap. BCED) = 2x sq.units / 21x sq.units

ar Δ APQ / ar (trap. BCED) = 2 / 21

ar Δ APQ : ar (trap. BCED) = 2 : 21  

Hence, the ratio of the areas of ΔAPQ and the trapezium BPQC 2 : 21  

Answer 2

The ratio of the area of triangle ΔAPQ and trapezium BCQP is 4:21

Step-by-step explanation:

• Given data

        $\mathbf{\frac{AP}{PB}=\frac{2}{3}}$

        we can also write this

        $\mathbf{\frac{PB}{AP}=\frac{3}{2}}$

        On adding both side 1 ,we get

        $\mathbf{\frac{PB}{AP}+1=\frac{3}{2}+1}$

        $\mathbf{\frac{PB+AP}{AP}=\frac{3+2}{2}}$

        $\mathbf{\frac{AB}{AP}=\frac{5}{2}}$      ...1)

• From given condition it is clear that

        Area o trapezium BCQP =Area ofΔABC-Area ofΔAPQ

        Or

        Area o trapezium BCQP =arΔABC - arΔAPQ      ...2)

• Here it is given that $\mathbf{PQ\parallel BC}$ ,so

        $\mathbf{\Delta ABC\sim \Delta APQ}$       ...3)

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

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

       On putting respective value, we get

        $\mathbf{\frac{\Delta ABC }{\Delta APQ}=\left ( \frac{5}{2} \right )^{2}}$

        $\mathbf{\frac{\Delta ABC }{\Delta APQ}=\frac{25}{4} }$

• On subtracting 1 on both side

        $\mathbf{\frac{\Delta ABC }{\Delta APQ}-1=\frac{25}{4}-1 }$

        $\mathbf{\frac{\Delta ABC-\Delta APQ }{\Delta APQ}=\frac{25-4}{4} }$

        $\mathbf{\frac{\Delta ABC-\Delta APQ }{\Delta APQ}=\frac{21}{4} }$        ...4)

• From equation 2) , equation 4) can be written as

        $\mathbf{\frac{Area\ of\ trapezium\ BCQP }{\Delta APQ}=\frac{21}{4} }$

        We can also write this

        $\mathbf{\frac{\Delta APQ }{Area\ of\ trapezium\ BCQP}=\frac{4}{21} }$