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 :

$306 m^{2}$

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$307 m^{2}$

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$308 m^{2}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$309 m^{2}$

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Solution

The correct option is A
$306 m^{2}$

Draw the perpendicular CM on AB from C.
From the figure,
BM = AB – AM = 40 - 28 = 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 ∠DAB=90∘. The area of field is :

The correct option is A 306 m2 Draw the perpendicular CM on AB from C. From the figure, BM = AB – AM = 40 - 28 = 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 A
$306 m^{2}$

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