A triangular corner is cut from a rectangular piece of paper and the resulting pentagon has sides 5, 6, 8, 9, 12 in some order. The ratio of the area of the pentagon to the area of the rectangle is

$\frac{11}{18}$

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$\frac{13}{18}$

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$\frac{15}{18}$

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$\frac{17}{18}$

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Solution

The correct option is D
$\frac{17}{18}$

Angle of rectangle = $9 \times 12$

Area of pentagon = $12 \times 9 - \frac{1}{2} \times 3 \times 4$

$= 12 \times 9 - 6$

Required = $\frac{12 \times 9 - 6}{12 \times 9} = \frac{17}{18}$

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A triangular corner is cut from a rectangular piece of paper and the resulting pentagon has sides 5, 6, 8, 9, 12 in some order. The ratio of the area of the pentagon to the area of the rectangle is

The correct option is D 1718 Angle of rectangle = 9×12 Area of pentagon = 12×9−12×3×4 =12×9−6 Required = 12×9−612×9=1718

The correct option is D
$\frac{17}{18}$

Angle of rectangle = $9 \times 12$

Area of pentagon = $12 \times 9 - \frac{1}{2} \times 3 \times 4$

$= 12 \times 9 - 6$

Required = $\frac{12 \times 9 - 6}{12 \times 9} = \frac{17}{18}$