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

# 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?

A) 1:2

B) 2:1

C) 1:2

D) 3:1

### 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

## 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