ABCD is trapezium with A B parallel to C D and the diagonals intersect at the point P - The area- of - ABP is 32 cm 2 and the area of CDP is 50 cm 2 - The area of the trapezium is -

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Area of Trapezium by Finding the Area of a Triangle of Same Area

ABCD is trape...

Question

$A B C D$ is trapezium with $¯ ¯¯¯¯¯¯ ¯ A B$ parallel to $¯ ¯¯¯¯¯¯¯ ¯ C D$ and the diagonals intersect at the point $P$ . The area, of $Δ$ $A B P$ is $32$ $c m^{2}$ and the area of $△$ $C D P$ is $50$ $c m^{2} .$ The area of the trapezium is :

162

c m^{2}

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364

c m^{2}

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324

c m^{2}

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164

c m^{2}

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Solution

The correct option is A $162$ $c m^{2}$
ar $Δ$ APD $=$ ar $Δ$ BPC
[Since triangles having same base between same parallels are equal in area
$∴ a r Δ A C D = a r Δ B C D$ ]
$\frac{D P}{P B} = \frac{Δ A D P}{Δ A B P} = \frac{x}{32}$
Also $\frac{D P}{P B} = \frac{Δ C P D}{Δ B P C} = \frac{50}{x}$
$\frac{x}{32} = \frac{50}{x}$
$x^{2} = 50 \times 32 = 100 \times 16$
$x = 10 \times 4 = 40$
Area of the trapezium $= 32 + 50 + 40 + 40$
$= 162 c m^{2}$ .

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Question 16

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ABCD is trapezium with ¯¯¯¯¯¯¯¯AB parallel to ¯¯¯¯¯¯¯¯¯CD and the diagonals intersect at the point P . The area, of Δ ABP is 32 cm2 and the area of △ CDP is 50 cm2. The area of the trapezium is :

The correct option is A 162 cm2ar ΔAPD=ar ΔBPC[Since triangles having same base between same parallels are equal in area∴arΔACD=arΔBCD]DPPB=ΔADPΔABP=x32Also DPPB=ΔCPDΔBPC=50xx32=50xx2=50×32=100×16x=10×4=40Area of the trapezium=32+50+40+40=162cm2.

$A B C D$ is trapezium with $¯ ¯¯¯¯¯¯ ¯ A B$ parallel to $¯ ¯¯¯¯¯¯¯ ¯ C D$ and the diagonals intersect at the point $P$ . The area, of $Δ$ $A B P$ is $32$ $c m^{2}$ and the area of $△$ $C D P$ is $50$ $c m^{2} .$ The area of the trapezium is :