What is the area of the given figure? $A B C D$ is a rectangle and $B D E$ is an isosceles right triangle.

a b

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a b^{2}

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c a b

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b (a + \frac{b}{2})

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Solution

The correct option is B $b (a + \frac{b}{2})$
Area of figure $= a r □ A B C D + a r △ B D E = a \times b + \frac{1}{2} \times b \times b = b (a + \frac{b}{2})$

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What is the area of the given figure? ABCD is a rectangle and BDE is an isosceles right triangle.

The correct option is B b(a+b2)Area of figure=ar□ABCD+ar△BDE=a×b+12×b×b=b(a+b2)

What is the area of the given figure? $A B C D$ is a rectangle and $B D E$ is an isosceles right triangle.

The correct option is B $b (a + \frac{b}{2})$
Area of figure $= a r □ A B C D + a r △ B D E = a \times b + \frac{1}{2} \times b \times b = b (a + \frac{b}{2})$