Quadrilaterals

Help Questions

Math › Quadrilaterals

Questions 1 - 10

Find the perimeter of a square inscribed in a circle that has a diameter of .

$24\sqrt2$

$48\sqrt2$

$36\sqrt2$

Explanation

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

$\text{diameter}=\text{diagonal}$

Now, use the Pythagorean theorem to find the length of the sides of the square.

$\text{Diagonal}^2=\text{side}^2+\text{side}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}$

Now, substitute in the value of the diagonal to find the length of a side of the square.

$\text{side}=\frac{12(\sqrt2)}{2}$

Simplify.

$\text{side}=6\sqrt2$

Now, recall how to find the perimeter of a square:

$\text{Perimeter}=4(\text{side})$

Substitute in the value of the side to find the perimeter of the square.

$\text{Perimeter}=4(6\sqrt2)$

Solve.

$\text{Perimeter}=24\sqrt2$

The perimeter of a square is 48. What is the length of its diagonal?

$12\sqrt{2}$

$\sqrt{12}$

$24\sqrt{2}$

$48\sqrt{2}$

$2\sqrt{12}$

Explanation

Perimeter = side * 4

48 = side * 4

Side = 12

We can break up the square into two equal right triangles. The diagonal of the sqaure is then the hypotenuse of these two triangles.

Therefore, we can use the Pythagorean Theorem to solve for the diagonal:

$\sqrt{144+144}=\sqrt{hypotenuse^2}$

$hypotenuse=\sqrt{2\times 144}=\sqrt{2\times 12\times 12}=12\sqrt{2}$

The lengths of the sides of ten squares form an arithmetic sequence. One side of the smallest square measures sixty centimeters; one side of the second-smallest square measures one meter.

Give the area of the largest square, rounded to the nearest square meter.

18 square meters

16 square meters

20 square meters

22 square meters

24 square meters

Explanation

Let $a_{n}$ be the lengths of the sides of the squares in meters. $a_{1} = 60 \div 100 = 0.6$ and $a_{2} = 1$ , so their common difference is

The arithmetic sequence formula is

$a_{n} = a_{1} + (n-1)d$

The length of a side of the largest square - square 10 - can be found by substituting $a_{1} = 0.6, n= 10, d = 0.4$ :

$a_{10} =0.6+ (10-1) 0.4 = 0.6 + 9 (0.4) = 0.6 + 3.6 = 4.2$

The largest square has sides of length 4.2 meters, so its area is the square of this, or $A = 4.2^{2} = 17.64$ square meters.

Of the choices, 18 square meters is closest.

The perimeter of a square is one yard. Which is the greater quantity?

(a) The area of the square

(b) $\frac{1}{2}$ square foot

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell form the information given.

Explanation

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\frac{3}{4}$ feet. The area of the square is $\frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\frac{9}{16} > \frac{1}{2}$ , making (a) greater.

If a rectangle has a width of and a length that is double the width, what would be the area of the rectangle? Round to the nearest tenth.

Explanation

To calculate the area of a triangle, we want to multiply the length by the width. Since the length is twice that of the width, which is , we can determine length as such:

$2.3;cm\cdot2=4.6;cm$

Now that we know the values for length and width, we can calculate the area of the triangle:

$Area=2.3;cm\cdot4.6;cm=10.6;cm^2$

If the perimeter of a rectangle is , and the width of the rectangle is , what is the area of a rectangle?

Explanation

Recall how to find the perimeter of a rectangle:

$\text{Perimeter}=2(\text{length}+\text{width})$

Since we are given the width and the perimeter, we can solve for the length.

$\text{length}+\text{width}=\frac{\text{Perimeter}}{2}$

$\text{length}=\frac{\text{Perimeter}}{2}-\text{width}$

Substitute in the given values for the width and perimeter to find the length.

$\text{length}=\frac{36}{2}-17$

Simplify.

$\text{length}=18-17$

Solve.

$\text{length}=1$

Now, recall how to find the area of a rectangle.

$\text{Area}=\text{length}\times\text{width}$

Substitute in the values of the length and width to find the area.

$\text{Area}=17 \times 1$

Solve.

$\text{Area}=17$

A rectangle has an area of $96 ; cm^{2}$ . The width is four less than the length. What is the perimeter?

Explanation

For a rectangle, area is and perimeter is , where is the length and is the width.

Let = length and = width.

The area equation to solve becomes , or $x^{2}-4x-96=0$ .

To factor, find two numbers the sum to -4 and multiply to -96. -12 and 8 will work:

x2 + 8x - 12x - 96 = 0

x(x + 8) - 12(x + 8) = 0

(x - 12)(x + 8) = 0

Set each factor equal to zero and solve:

or .

Therefore the length is and the width is , giving a perimeter of .

Which is the greater quantity?

(a) The length of a diagonal of a square with sidelength 10 inches

(b) The hypotenuse of an isosceles right triangle with legs 10 inches each

(a) and (b) are equal.

(a) is greater.

(b) is greater.

It is impossible to tell which is greater from the information given.

Explanation

A diagonal of a square cuts the square into two isosceles right triangles, of which the diagonal is the common hypotenuse. Therefore, each figure is the hypotenuse of an isosceles right triangle with legs 10 inches, making them equal in length.

Find the length of the square's diagonal.

Square_8