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Bunuel
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Bunuel
What is the area of a rectangle if its width is half its length?

(1) The perimeter of the rectangle is 30.
(2) The diagonal of the rectangle has length \(5\sqrt{5}\).


Length = 2x

Width = x

Statement 1: perimeter = 30

2(2x + x ) = 30

x = 5.

2x = 10

Area: 5*10 = 50

Statement 2: Diagonal = 5\(\sqrt{5}\)

As it's a rectangle, its diagonal is basically the hypotenuse of right angle.

So, we have our chance to apply Pythagorean Theorem.

\((2x)^2\) + \(x^2\)= \((5\sqrt{5})^2\)
Solving the equation we will get the value of x.

So, it's possible to determine the value of the area of rectangle.

The best answer is D.
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Solution


Given:
    • Width of rectangle = 1/2 * length of rectangle

To find:
    • Area of the rectangle

Approach and Working:
    • Area of rectangle = l * b, and we know that, l = 2b
    • Thus, we can find the area, if we know the value of l or b

Analysing Statement 1
    • Perimeter = 30, implies, 2(l+b) = 30
    • l+b = 15, implies, 3b = 15
    • Thus, b = 5
    • Hence, we can find the area of the rectangle
Therefore, Statement 1 alone is sufficient to answer this question

Analysing Statement 2
    • Diagonal = 5√5, implies, \(√ (l^2 + b^2) = 5√5\)
    • Squaring on both sides gives, \((l^2 + b^2) = 25 * 5\)
    • Implies, \(5b^2 = 125\), which gives the value of b as 5
Therefore, Statement 2 alone is sufficient to answer this question.

Hence, the correct answer is option D.

Answer: D

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Let length of rectangle= 2x

.............Width...........= x

Area= 2\(x^2\)

1) Perimeter= 2(L+B)= 2(2x+x)= 30. Value of x can be found. [Suff]

2) \(x^2\)+4\(x^2\)= \((5\sqrt{5})^2\). Value of x can be found. [Suff]


Answer: D.
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