If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?
(A) 40
(B) 60
(C) 80
(D) 100
(E) 120
[spoiler]OA=B[/spoiler]
Source: Official Guide
If a square mirror has a 20-inch diagonal, what is the approximate perimeter of the mirror, in inches?
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Let x = the length of one side of the square
The diagonal creates a right triangle with each leg having length x and the hypotenuse with length 20.
Plug these values into the Pythagorean Theorem to get: x² + x² = 20²
Simplify: 2x² = 400
Divide both sides by 2 to get x² = 200
So, x = √200
Rewrite as x = 10√2
NOTE: For the GMAT, you should memorize approximations for √2, √3 and √5
√2 ≈ 1.4
So, 10√2 ≈ (10)1.4 ≈ 14
If each side of the square has length 14, then the perimeter of the square = (4)(14) = 56
The best approximation is B (60)
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Brent
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Solution:
We are given that a square has a 20-inch diagonal, and we must solve for the perimeter of the square. This means we first need to know the length of a side of the square. To determine the length of a side we can use the diagonal formula for a square. We know that:
diagonal of a square = side√2 = s√2
20 = s√2
s = 20/√2
To rationalize the denominator, we multiply 20/√2 by √2/√2. This gives us:
s = 20/√2 × √2/√2 = 20√2/√4
s = 20√2/2 = 10√2
We also should have memorized that the approximate value of √2 is 1.4. Thus, one side of the square is approximately 10 x 1.4 = 14 inches.
Finally, the perimeter of the square is approximately 4 x 14 = 56 inches. Since we are asked to approximate, the closest answer is 60.
Alternate Solution:
Since the diagonal is 20 and since the diagonal of a square is √2 times the side of a square, we have
s√2 = 20
s = 20/√2 = 10√2
where s denotes the side of the square. Thus, the perimeter of the square is 4 x 10√2 = 40√2.
Instead of estimating the side of a square, let’s compare the square of the perimeter to the possible answers. We see that the square of the perimeter is (40√2)^2 = 40^2 x (√2)^2 = 1600 x 2 = 3200.
Now, let’s square the possible perimeters:
A) perimeter = 40, perimeter^2 = 1600
B) perimeter = 60, perimeter^2 = 3600
C) perimeter = 80, perimeter^2 = 6400
At this point, we don’t even need to check answer choices D and E because the square of those numbers will be even greater than 6400. The choice whose square is closest to 3200 is B.
Answer: B
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