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

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.

Explanation

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.

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 .

$6\sqrt{ 7K }$

$\frac{ \sqrt{ 7K } }{24}$

$\frac{ \sqrt{ 7K } }{2}$

$\frac{7K} {48}$

Explanation

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:

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

Parallelogram

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 .

Explanation

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.

Rhombus

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 .

Box

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

$2x^{2} + 80x$

$20 x^{2}$

Explanation

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

Box

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

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

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?

$14,700 \textrm{ cm}^{2}$

$29,400 \textrm{ cm}^{2}$

$58,800 \textrm{ cm}^{2}$

$19,600\textrm{ cm}^{2}$

$11,025 \textrm{ cm}^{2}$

Explanation

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.

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.

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

${Area_{(Square)}}=4\cdot {Area_{(Parallelogram)}}$

${Area_{(Square)}}={Area_{(Parallelogram)}}$

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

${Area_{(Square)}}=3\cdot {Area_{(Parallelogram)}}$

Explanation

The area of a parallelogram is given by:

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

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

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:

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

Box

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

Explanation

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

Box 2

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

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?

$37,500 \textrm{ in}^{2}$

$75,000\textrm{ in}^{2}$

$50,000\textrm{ in}^{2}$

$13,500\textrm{ in}^{2}$

$15,000\textrm{ in}^{2}$

Explanation

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.

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.

$20\ inches$

$16\ inches$

$12\ inches$

$10\ inches$

$8\ inches$

Explanation

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.

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?