In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If $∠ D A B = 90^{\circ}$ . The area of field is :

309 m^{2}

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307 m^{2}

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308 m^{2}

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306 m^{2}

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Solution

The correct option is D $306 m^{2}$
Draw a perpendicular CM on AB from point C.
From the figure,
BM = AB –AM = (40 -28) m =12 m
Area of the field = area of rectangle AMCD + area of right angled triangle MBC
$= 28 \times 9 + \frac{1}{2} \times 12 \times 9 = 306 m^{2}$

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In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If $∠ D A B = 90^{\circ}$ . The area of field is :

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In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If ∠DAB=90∘. The area of field is :

The correct option is D 306 m2Draw a perpendicular CM on AB from point C. From the figure, BM = AB –AM = (40 -28) m =12 m Area of the field = area of rectangle AMCD + area of right angled triangle MBC =28×9+12×12×9=306m2

In the given figure, ABCD is a field in the form of a quadrilateral whose sides are indicated. If $∠ D A B = 90^{\circ}$ . The area of field is :

The correct option is D $306 m^{2}$
Draw a perpendicular CM on AB from point C.
From the figure,
BM = AB –AM = (40 -28) m =12 m
Area of the field = area of rectangle AMCD + area of right angled triangle MBC
$= 28 \times 9 + \frac{1}{2} \times 12 \times 9 = 306 m^{2}$