Rectangles - GMAT Quantitative Reasoning

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Question

Find the area of a rectangle whose width is and length is .

Answer

To find area, simply multiply length times width. Thus

$2x\cdot2y=4xy$

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Question

The area of a square that has sides with a length of 12 inches is equal to the area of a rectangle. If the rectangle has a width of 3 inches, what is the length of the rectangle?

Answer

If the area of the rectangle is equal to the area of the square, then it must have an area of $\dpi{100} \small 144in^{2}$ $\dpi{100} \small (12\times 12)$ . If the rectangle has an area of $\dpi{100} \small 144 in^{2}$ and a side with a lenth of 3 inches, then the equation to solve the problem would be $\dpi{100} \small 144=3x$ , where $\dpi{100} \small x$ is the length of the rectangle. The solution:

$\frac{144}{3} = 48$ .

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Question

A rectangle twice as long as it is wide has perimeter . Write its area in terms of .

Answer

Let be the width of the rectangle; then its length is , and its perimeter is

Set this equal to and solve for :

$\frac{6W}{6} =\frac{ 6N-12}{6}$

The width is and the length is $L = 2W = 2 \left (N - 2 \right ) = 2N - 4$ , so multiply these expressions to get the area:

$A = LW = (N-2) \cdot \left (2N-4 \right ) = 2N^{2} -8N + 8$

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Question

A rectangle has its vertices at . What part, in percent, of the rectangle is located in Quadrant III?

Answer

A rectangle with vertices has width and height , thereby having area $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant III is a rectangle with vertices

.

It has width and height , thereby having area $4 \times 3 = 12$ .

Therefore, $\frac{12}{50 }$ of the rectangle is in Quadrant III; this is equal to

$\frac{12}{50 } \times 100 = 24 %$

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Question

A rectangle has its vertices at . What percentage of the rectangle is located in Quadrant IV?

Answer

A rectangle with vertices has width and height ; it follows that its area is $10 \times 5 = 50$ .

The portion of the rectangle in Quadrant IV has vertices . Its width is , and its height is , so its area is $1 \times 3 = 3$ .

Therefore, $\frac{3}{50 }$ , or $\frac{3}{50} \times 100 % = 6%$ , of this rectangle is in Quadrant IV.

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Question

What is the area of a rectangle given the length of and width of ?

Answer

To find the area of a rectangle, you must use the following formula:

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Question

What polynomial represents the area of a rectangle with length and width ?

Answer

The area of a rectangle is the product of the length and the width. The expression can be multplied by noting that this is the product of the sum and the difference of the same two terms; its product is the difference of the squares of the terms, or

$(A+4)(A-4) = A^{2}-4^{2}= A^{2}-16$

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Question

The perimeter of a rectangle is $104\textup{ inches}$ and its length is times the width. What is the area?

Answer

The perimeter of a rectangle is the sum of all four sides, that is:

since , we can rewrite the equation as:

$w=13\textup{ in}$

We are being asked for the area so we still aren't done. The area of a rectangle is the product of the width and length. We know what the width is so we can find the length and then take their product.

$l=3w=3(13)=39\textup{ in}$

$A= l \cdot w = 39 \cdot 13$

$A = 507\textup{ in}^2$

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Question

Find the area of a rectangle whose side lengths are $7\textup{ and }9$ .

Answer

To calculate area, multiply width times height. Thus,

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Question

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

Answer

Mark is building a garden with raised beds. One side of the garden will be 10 feet long and the other will be 5 less than three times the first side. What area of will Mark's garden be?

This problem asks us to find the area of a rectangle. We are given one side and asked to find the other. To find the other, we need to use the provided clues.

"...five less..."

"...three times the first side..." or

So put it together:

Next, find the area via the following formula:

So our answer is:

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Question

Note: figure NOT drawn to scale

Give the area of the above rectangle.

Answer

The area of a rectangle is the product of its length and width;

$A = 3^{x+1} \cdot 3^{x} = 3 ^{x+1+x} = 3 ^{2x+1}$

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Question

A rectangle has a length of and a width of . What is the length of the diagonal of the rectangle?

Answer

If the rectangle has a length of and a width of , we can imagine the diagonal as being the hypotenuse of a right triangle. The length and width are the other two sides to this triangle, so we can use the Pythagorean Theorem to calculate the length of the diagonal of the rectangle:

$d=\sqrt{45}=\sqrt{5\cdot 9}=3\sqrt{5}$

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Question

is a rectangle, and $\widehat{CAD}=60^{\circ}$ . What is the length of the diagonal of the rectangle?

Answer

Here, we also have a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle, indeed both ADC and ABD are $30^{\circ}-60^{\circ}-90^{\circ}$ triangles. We can see that by calculating the missing angles. in each triangle we find that this angle to be 30 degrees. Since , we also know that the other sides of the triangles will be $\sqrt{3}$ and , since a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle have its sides in ratio $n-\sqrt{3}n-2n$ , where is a constant. In this case . Therefore the hypotenuse will be , which is also the length of the diagonal of the rectangle.

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Question

A rectangle has a length of and a width of . What is the length of the diagonal of the rectangle?

Answer

The diagonal of a rectangle can be thought of as the hypotenuse of a right triangle with the length and width of the rectangle as the other two sides. This means we can use the Pythagorean theorem to solve for the diagonal of a rectangle if we are given its length and its width:

$d^2=34\rightarrow d=\sqrt{34}$

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Question

Rectangle has an area of and . What is the length of the diagonal ?

Answer

Firstly, before we try to set up an equation for the length of the sides and the area, we should notice that the area is a perfect square. Indeed $13^{2}=169$ . Now let's try to see whether 13 could be the length of two consecutive sides of the rectangle, Indeed, we are told that , therefore ABDC is a square with side 13 and with diagonal $s\sqrt{2}$ , where is the length of the side of the square. Therefore, the final answer is $13\sqrt{2}$ .

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Question

Rectangle , has area and . What is the length of diagonal ?

Answer

To find the length of the sides to calculate the length of the diagonal with the Pythagorean Theorem, we would need to set up two equations for our variables. However, trial and error, in my opinion in most GMAT problems is faster than trying to solve a quadratic equation. The way we should test values for our sides is firstly by finding the different possible factors of the area, that way we can see possible factors.

As follows has possible factors . From these values we should find the two that will give us 8, the length of the two consecutive sides.

We find that and are the values for the two sides.

Now we just need to apply the Pythagorean Theorem to find the length of the diagonal: $\sqrt{3^{2}+5^{2}}$ , or $\sqrt{34}$ .

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Question

Find the length of the diagonal of a rectangle whose sides are lengths $4\textup{ and }7$ .

Answer

To find the diagonal, you must use the pythaorean theorem. Thus:

$c=\sqrt{65}$

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Question

Calculate the length of the diagonal of a rectangle whose width is and length is .

Answer

To solve, simply use the Pythagorean theorem.

$c^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}$

$c=\sqrt{9^2+4^2}=\sqrt{81+16}=\sqrt{97}$

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Question

If a garden bed will have side lengths of 9 meters and 12 meters, what will the distance be across its diagonal?

Answer

If a rectangular garden bed will have side lengths of 9 meters and 12 meters, what will the distance be across its diagonal?

This question is a rectangle question, but it could also be seen as a triangle question. If we have a rectangle with two side lengths, we can find the diagonal by using Pythagorean theorem:

$c=\sqrt{a^2+b^2}$

$c=\sqrt{9^2+12^2}=\sqrt{225}=15 meters$

If you are feeling really observant, you may have seen that we have a Pythagorean triple. In this case, a 3-4-5 triangle. You could have skipped using Pythagoran theorem and simply done three times five to get fifteen meters.

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Question

Craig is building a fence around his rectangular back yard. He knows that the yard is feet longer than twice the width. What is the width of the yard if Craig needs feet of fencing to completely enclose the yard?

Answer

The length is 5 more twice the width. Let y be the length of the yard and w the width.

The perimeter of the yard is 160, so we can write:

The yard is 25 feet wide and 55 feet long.

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