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In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to t...


Question

In the figure /PX/ = /XQ/, PQ//YZ and XV//QR. What is the ratio of the area of XYZQ to te area of \(\bigtriangleup\)YZR?

Options

A) 1:2

B) 2:1

C) 1:2

D) 3:1

The correct answer is B.

Explanation:

From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY

= Area of \(\bigtriangleup\)YZR = A2

Area of \(\bigtriangleup\)PQR = A = A1 + 2A2

But from similarity of triangles

\(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

\(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

A = 4A2 But, A = A1 + 2A

A1 = 4A2 - 2A2

A1 = 2A2

\(\frac{A_1}{A_2}\) = 2

A1:A2 = 2:1

Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1


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Dicussion (1)

  • From the diagram, XYZQ is a parallelogram. Thus, |YZ| = |XQ| = |PX|; \(\bigtriangleup\)PXY, Let the area of XYZQ = A1, the area of \(\bigtriangleup\)PXY

    = Area of \(\bigtriangleup\)YZR = A2

    Area of \(\bigtriangleup\)PQR = A = A1 + 2A2

    But from similarity of triangles

    \(\frac{\text{Area of PQR}}{\text{Area of PXY}} = (\frac{PQ}{PX})^2 = (\frac{QR}{XY})^2\)

    \(\frac{A}{A_2} = (\frac{2}{1})^2 = \frac{2}{1}\)

    A = 4A2 But, A = A1 + 2A

    A1 = 4A2 - 2A2

    A1 = 2A2

    \(\frac{A_1}{A_2}\) = 2

    A1:A2 = 2:1

    Area of XYZQ:Area of \(\bigtriangleup\)YZR = 2:1

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