How to find the area of a rectangle - SSAT Upper Level Quantitative

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Question

Mark wants to seed his lawn, which measures 225 feet by 245 feet. The grass seed he wants to use gets 400 square feet of coverage to the pound; a fifty-pound bag sells for $45.00, and a ten-pound bag sells for $13.00. What is the least amount of money Mark should expect to spend on grass seed?

Answer

The area of Mark's lawn is $225 \cdot245 = 55,125$ . The amount of grass seed he needs is $55,125 \div 400 \approx 137.8$ pounds.

He has two options.

Option 1: he can buy three fifty-pound bags for $$45 \cdot3 = $135$

Option 2: he can buy two fifty-pound bags and four ten-pound bags for $$45 \cdot 2 + $13 \cdot 4 = $142$

The first option is the more economical.

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Question

The base length of a parallelogram is equal to the side length of a square. The base length of the parallelogram is two times longer than its corresponding altitude. Compare the area of the parallelogram with the area of the square.

Answer

The area of a parallelogram is given by:

Where is the base length and is the corresponding altitude. In this problem we have:

or

$a=\frac{b}{2}$

So the area of the parallelogram would be:

$Area_{(Parallelogram)}=ba=b\times \frac{b}{2}=\frac{b^2}{2}$

The area of a square is given by:

weher is the side length of a square. In this problem we have , so we can write:

$Area _{(Square)}=x^2=b^2$

Then:

$\frac{Area_{(Parallelogram)}}{Area_{(Square)}}=\frac{\frac{b^2}{2}}{b^2}=\frac{1}{2}$

or:

${Area_{(Square)}}=2\cdot {Area_{(Parallelogram)}}$

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Question

How many squares with the side length of 2 inches can be fitted in a rectangle with the width of 10 inches and height of 4 inches?

Answer

Solution 1:

We can divide the rectangle width and height by the square side length and multiply the results:

rectangle width $\div$ square length = $10\div 2=5$

rectangle height $\div$ square length = $4\div 2=2$

$5\times 2=10$

Solution 2:

As the results of the division of rectangle width and height by the square length are integers and do not have a residual, we can say that the squares can be perfectly fitted in the rectangle. Now in order to find the number of squares we can divide the rectangle area by the square area:

Rectangle area = $4\times 10=40$ square inches

Square area = $2\times 2=4$ square inches

So we can get:

$40\div 4=10$

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Question

The width and height of a rectangle are and , respectively. Give the area of the rectangle in terms of .

Answer

The area of a rectangle is given by multiplying the width times the height. As a formula:

Where:

is the width and is the height. So we can get:

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Question

A rectangle has the area of 80 square inches. The width of the rectangle is 2 inches longer that its height. Give the height of the rectangle.

Answer

The area of a rectangle is given by multiplying the width times the height. That means:

where:

width and height.

We know that: $w=h+2\Rightarrow h=w-2$ . Substitube the in the area formula:

$Area=80=wh=(h+2)\times h\Rightarrow 80=h^2+2h\Rightarrow h^2+2h-80=0$

Now we should solve the equation for :

$h^2+2h-80=0\Rightarrow h=-1\pm \sqrt{1^2+1\times 80}=-1\pm \sqrt{81}=-1\pm 9$

The equation has two answers, one positive and one negative . As the length is always positive, the correct answer is inches.

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Question

A rectangle with a width of 6 inches has an area of 48 square inches. Give the sum of the lengths of the rectangle's diagonals.

Answer

A rectangle has two congruent diagonals. A diagonal of a rectangle divides it into two identical right triangles. The diagonal of the rectangle is the hypotenuse of these triangles. We can use the Pythagorean Theorem to find the length of the diagonal if we know the width and height of the rectangle.

$Diagonal=\sqrt{w^2+h^2}$

where:

is the width of the rectangle
is the height of the rectangle

First, we find the height of the rectangle:

$h=\frac{area}{w}=\frac{48}{6}=8$

So we can write:

$Diagonal=\sqrt{w^2+h^2}=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ inches

As a rectangle has two diagonals with the same length, the sum of the diagonals is inches.

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Question

A rectangle has the width of and the diagonal length of . Give the area of the rectangle in terms of .

Answer

First we need to find the height of the rectangle. Since the width and the diagonal lengths are known, we can use the Pythagorean Theorem to find the height of the rectangle:

$Diagonal=\sqrt{w^2+h^2}$

$t+1=\sqrt{w^{2}+(t-1)^{2}}$

$\left ( t+1 \right )^{2}=w^{2}+\left ( t-1 \right )^{2}$

So we have:

$w^2=(t+1)^2-(t-1)^{2}$

$\Rightarrow w^2=(t^2+2t+1)-(t^2-2t+1)=4t$

$\Rightarrow w=\sqrt{4t}=2\sqrt{t}$

So we can get:

$Area=wh=2\sqrt{t}(t-1)$

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Question

The perimeter of a rectangle is 800 inches. The width of the rectangle is 60% of its length. What is the area of the rectangle?

Answer

Let be the length of the rectangle. Then its width is 60% of this, or . The perimeter is the sum of the lengths of its sides, or

; we set this equal to 800 inches and solve for :

$3.20 L \div 3.20 = 800 \div 3.20$

The width is therefore

$0.60 \times 250 = 150$ .

The product of the length and width is the area:

$250 \times 150 = 37,500$ square inches.

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Question

The perimeter of a rectangle is 490 centimeters. The width of the rectangle is three-fourths of its length. What is the area of the rectangle?

Answer

Let be the length of the rectangle. Then its width is three-fourths of this, or $\frac{3}{4} L$ . The perimeter is the sum of the lengths of its sides, or

$L + \frac{3}{4} L + L + \frac{3}{4} L = \frac{7}{2} L$ .

Set this equal to 490 centimeters and solve for :

$\frac{7}{2} L = 490$

$\frac{2} {7} \cdot \frac{7}{2} L =\frac{2} {7} \cdot 490$

The length of the rectangle is 140 centimeters; the width is three-fourths of this, or

$\frac{3}{4} \times 140 = 105$ centimeters.

The area is the product of the length and the width:

$140 \times 105 = 14,700$ square centimeters.

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Question

The area of a rectangle is square feet. The width of the rectangle is four-sevenths of its length. Give the length of the rectangle in inches in terms of .

Answer

Let be the length in feet. Then the width of the rectangle in feet is four-sevenths of this, or $\frac{4}{7}L$ . The area is equal to the product of the length and the width, so set up this equation and solve for :

$\frac{4}{7}L \cdot L = K$

$\frac{4}{7}L ^{2}= K$

$\frac{7} {4}\cdot \frac{4}{7}L ^{2}= \frac{7} {4}\cdot K$

$L ^{2}= \frac{7K} {4}$

$L =\sqrt{ \frac{7K} {4}} =\frac{ \sqrt{ 7K } }{\sqrt{4}}=\frac{ \sqrt{ 7K } }{2}$

Since this is the length in feet, we multiply this by 12 to get the length in inches:

$\frac{ \sqrt{ 7K } }{2} \cdot 12 = 6\sqrt{ 7K }$

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Question

Rectangle A has length 40 inches and height 24 inches. Rectangle B has length 30 inches and height 28 inches. Rectangle C has length 72 inches, and its area is the mean of the areas of the other two rectangles. What is the height of Rectangle C?

Answer

The area of a rectangle is the product of the length and its height, Rectangle A has area $40 \times 24 = 960$ square inches; Rectangle B has area $30 \times 28 = 840$ square inches.

The area of Rectangle C is the mean of these areas, or

$\frac{960+840}{2} = 900$ square inches, so its height is this area divided by its length:

$900 \div 72 = 12 \frac{1}{2}$ inches.

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Question

Rectangle A has length 40 inches and height inches; Rectangle B has length 30 inches and height inches; Rectangle C has height inches, and its area is the sum of those of the other two rectangles. What is its length?

Answer

The area of a rectangle is the product of the length and its height.

Rectangle A has area square inches, and Rectangle B has an area of $30 \cdot 2H = 60 H$ square inches, so the sum of their areas is square inches. This is the area of Rectangle C; divide it by height inches to get a length of

$\frac{100 H}{4H} = \frac{100}{4} = 25$ inches. This answer is not among the given choices.

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Question

A basketball team wants to paint a 4-foot wide border around its court to make sure fans don't get too close to the action. If the court is 94 by 50 feet, and one can of paint can cover 300 square feet, how many cans of paint does the team need to ensure that the entire border is painted?

(Assume that you cannot buy partial cans of paint.)

Answer

We begin this problem by finding the difference of two areas: the larger rectangle bounded by the outer edge of the border and the smaller rectangle that is the court itself.

The larger rectangle is square feet $\left(94+4+4\right\times\left(50+4+4 \right )$ , and the court is square feet $(94\times 50)$ .

The difference, , repesents the area of the border.

Now we divide this by , which is just a bit over . But since we can't leave square feet unpainted, we have to round up to cans of paint.

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Question

Figure NOT drawn to scale

The above figure shows Rhombus ; and are midpoints of their respective sides. Rhombus has area 900.

Give the area of Rectangle .

Answer

A rhombus, by definition, has four sides of equal length. Therefore, , and, by the Multiplication Property, $\frac{1}{2 }AB=\frac{1}{2 } CD$ . Also, since and are the midpoints of their respective sides, $AX = XB = \frac{1}{2 }AB$ and $\frac{1}{2 } CD = CY = YD$ . Combining these statements, and letting :

$n = AX = XB = \frac{1}{2 }AB=\frac{1}{2 } CD = CY = YD$

Also, both $\overline{AY}$ and $\overline{XC}$ are altitudes of the rhombus; they are congruent, and we will call their common length (height).

The figure, with the lengths, is below.

The area of the entire Rhombus is the product of its height and the length of a base , so

$A_{1} = h \cdot 2n = 2hn =900$ .

Rectangle has as its length and width and , so its area is their product , Since

,

From the Division Property, it follows that

$2 hn \div 2 = 900 \div 2$ ,

and

.

This makes 450 the area of Rectangle .

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Question

The above diagram shows a rectangular solid. The shaded side is a square. Give the total surface area of the solid.

Answer

A square has four sides of equal length, as seen in the diagram below.

All six sides are rectangles, so their areas are equal to the products of their dimensions:

Top, bottom, front, back (four surfaces): $20 \times 10 = 200$

Left, right (two surfaces): $10 \times 10 = 100$

The total area: $4 \times 200 + 2 \times 100 = 800 + 200 = 1,000$

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Question

The above diagram shows a rectangular solid. The shaded side is a square. In terms of , give the surface area of the solid.

Answer

Since a square has four sides of equal length, the solid looks like this:

The areas of each of the individual surfaces, each of which is a rectangle, are the product of their dimensions:

Front, back, top, bottom (four surfaces):

Left, right (two surfaces): $x \cdot x = x^{2}$

The total surface area is therefore

$2 \cdot x^{2} + 20x \cdot 4 = 2 x^{2} + 80x$

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Question

Rectangle A has length 40 inches and height 24 inches. Rectangle B has length 30 inches and height 28 inches. Rectangle C has length 72 inches, and its area is the mean of the areas of the other two rectangles. What is the height of Rectangle C?

Answer

The area of a rectangle is the product of the length and its height, Rectangle A has area $40 \times 24 = 960$ square inches; Rectangle B has area $30 \times 28 = 840$ square inches.

The area of Rectangle C is the mean of these areas, or

$\frac{960+840}{2} = 900$ square inches, so its height is this area divided by its length:

$900 \div 72 = 12 \frac{1}{2}$ inches.

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Question

Mark wants to seed his lawn, which measures 225 feet by 245 feet. The grass seed he wants to use gets 400 square feet of coverage to the pound; a fifty-pound bag sells for $45.00, and a ten-pound bag sells for $13.00. What is the least amount of money Mark should expect to spend on grass seed?

Answer

The area of Mark's lawn is $225 \cdot245 = 55,125$ . The amount of grass seed he needs is $55,125 \div 400 \approx 137.8$ pounds.

He has two options.

Option 1: he can buy three fifty-pound bags for $$45 \cdot3 = $135$

Option 2: he can buy two fifty-pound bags and four ten-pound bags for $$45 \cdot 2 + $13 \cdot 4 = $142$

The first option is the more economical.

Compare your answer with the correct one above

Question

The base length of a parallelogram is equal to the side length of a square. The base length of the parallelogram is two times longer than its corresponding altitude. Compare the area of the parallelogram with the area of the square.

Answer

The area of a parallelogram is given by:

Where is the base length and is the corresponding altitude. In this problem we have:

or

$a=\frac{b}{2}$

So the area of the parallelogram would be:

$Area_{(Parallelogram)}=ba=b\times \frac{b}{2}=\frac{b^2}{2}$

The area of a square is given by:

weher is the side length of a square. In this problem we have , so we can write:

$Area _{(Square)}=x^2=b^2$

Then:

$\frac{Area_{(Parallelogram)}}{Area_{(Square)}}=\frac{\frac{b^2}{2}}{b^2}=\frac{1}{2}$

or:

${Area_{(Square)}}=2\cdot {Area_{(Parallelogram)}}$

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Question

How many squares with the side length of 2 inches can be fitted in a rectangle with the width of 10 inches and height of 4 inches?

Answer

Solution 1:

We can divide the rectangle width and height by the square side length and multiply the results:

rectangle width $\div$ square length = $10\div 2=5$

rectangle height $\div$ square length = $4\div 2=2$

$5\times 2=10$

Solution 2:

As the results of the division of rectangle width and height by the square length are integers and do not have a residual, we can say that the squares can be perfectly fitted in the rectangle. Now in order to find the number of squares we can divide the rectangle area by the square area:

Rectangle area = $4\times 10=40$ square inches

Square area = $2\times 2=4$ square inches

So we can get:

$40\div 4=10$

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