In the figure PX XQ PQ YZ and XV QR What is the... - WAEC Mathematics 2008 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
correct option: b
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
= 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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