How to find the length of the diagonal of a rectangle

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Math › How to find the length of the diagonal of a rectangle

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A rectangle has perimeter 140 inches and area 1,200 square inches. Which is the greater quantity?

(A) The length of a diagonal of the rectangle.

(B) 4 feet

(A) is greater

(B) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Explanation

Let and be the dimensions of the rectangle. Then

and, subsequently,

Since the product of the length and width is the area, we are looking for two numbers whose sum is 70 and whose product is 1,200; through trial and error, they are found to be 30 and 40. We can assign either to be and the other to be since the result is the same.

The length of a diagonal of the rectangle can be found by applying the Pythagorean Theorem:

$D = \sqrt{L^{2}+ W^{2}}$

$D = \sqrt{30^{2}+ 40^{2}} = \sqrt{900+1,600} = \sqrt{2,500} = 50$

A diagonal is 50 inches long; since 4 feet are equivalent to 48 inches, (A) is the greater quantity.

The sides of rectangle ABCD are 4 in and 13 in.

How long is the diagonal of rectangle ABCD?

$\sqrt{185}$

$\sqrt{125}$

Explanation

A diagonal of a rectangle cuts the rectangle into 2 right triangles with sides equal to the sides of the rectangle and with a hypotenuse that is the diagonal. All you need to do is use the pythagorean theorem:

where a and b are the sides of the rectangle and c is the length of the diagonal.

$\sqrt{4^2+16^2}=\sqrt{185}=c$

If a rectangle is inscribed in a circle with a circumference of $16\pi$ , what is the length of the diagonal of the rectangle?

$32\pi$

Explanation

You should notice that the diagonal of the rectangle is the same as the diameter of the circle.

Now, recall how to find the circumference of a circle.

$\text{Circumference}=\text{diameter}\times\pi$

Since we are given the circumference and need the diameter, rerwite the equation to solve for the diameter.

$\text{Diameter}=\frac{\text{Circumference}}{\pi}$

Plug in the given circumference to find the diameter.

$\text{Diameter}=\frac{16\pi}{\pi}=16$

Now, recall the relationship between the diameter of the circle and the diagonal of the rectangle:

$\text{Diameter}=\text{Diagonal}=16$

Find the length of the diagonal of a rectangle that has a length of and a width of .

$3\sqrt{34}$

$6\sqrt{2}$

$12\sqrt{35}$

$4\sqrt3$

Explanation

The diagonal of a rectangle is also the hypotenuse of a right triangle that has the length and the width of the rectangle as its legs.

We can then use the Pythgorean Theorem to find the diagonal.

$\text{Diagonal}=\sqrt{\text{length}^2+\text{width}^2}$

For the given rectangle,

$\text{Diagonal}=\sqrt{9^2+15^2}=\sqrt{306}=3\sqrt{34}$

Find the length of the diagonal of a rectangle with side lengths 6 and 7.

$\sqrt{85}$

Explanation

To find the diagonal of a rectangle recall that the diagonal will create a triangle where the width and length are legs of the triangle and the diagonal is the hypotenuse.

To solve, simple use the Pythagorean Theorem and solve for the hypotenuse, which will be the diagonal of the rectangle.

Thus,

$c=\sqrt{a^2+b^2}=\sqrt{36+49}=\sqrt{85}$

A rectangle is inscribed in a circle. If the circumference of the circle is $4\pi$ , what is the length of the diagonal of the rectangle?

Explanation

You should notice that the diagonal of the rectangle is the same as the diameter of the circle.

Now, recall how to find the circumference of a circle.

$\text{Circumference}=\text{diameter}\times\pi$

Since we are given the circumference and need the diameter, rerwite the equation to solve for the diameter.

$\text{Diameter}=\frac{\text{Circumference}}{\pi}$

Plug in the given circumference to find the diameter.

$\text{Diameter}=\frac{4\pi}{\pi}=4$

Now, recall the relationship between the diameter of the circle and the diagonal of the rectangle:

$\text{Diameter}=\text{Diagonal}=4$

A rectangle has sides of length and . What is the length of its diagonal?

$3\sqrt{10} cm$

$3\sqrt{5}cm$

$\sqrt{30}cm$

Explanation

The diagonal of a rectangle forms the hypotenuse of a right triangle with one long and one short side of the rectangle. Therefore, to find the diagonal, we use the Pythagorean Theorem using the two side lengths and solving for the hypotenuse:

$\sqrt{90}=\sqrt{c^2}$

$c=\sqrt{90}$

Then, all we have to do is simplify the radical, as follows:

$\sqrt{90}=\sqrt{9*10}$

$\sqrt{9*10}=\sqrt{9}*\sqrt{10}$

$\sqrt{9}*\sqrt{10}=3*\sqrt{10}=3\sqrt{10}$

Therefore, $c=3\sqrt{10}$ .

Which is the greater quantity?

(a) The length of a diagonal of a square with sidelength 20 inches

(b) The length of a diagonal of a rectangle with length 25 inches and width less than 10 inches

(a) is greater

(b) is greater

(a) and (b) are equal

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

Explanation

The lengths of the diagonals of these rectangles can be computed using the Pythagorean Theorem:

(a) $c = \sqrt{20^{2} +20^{2}} = \sqrt{400+400} = \sqrt{800}$

(b) $c < \sqrt{25^{2} +10^{2}} = \sqrt{625+100} = \sqrt{725}$

so $\sqrt{725 }<\sqrt{ 800}$ . Since the diagonal of the rectangle in (b) measures less than $\sqrt{725 }$ , it must also measure less than that of the square in (a)

Find the length of the diagonal of a rectangle that has a length of and a width of .

$2\sqrt{17}$

$2\sqrt{15}$

$2\sqrt{19}$

$2\sqrt{13}$

Explanation

The diagonal of a rectangle is also the hypotenuse of a right triangle that has the length and the width of the rectangle as its legs.

We can then use the Pythgorean Theorem to find the diagonal.

$\text{Diagonal}=\sqrt{\text{length}^2+\text{width}^2}$

For the given rectangle,

$\text{Diagonal}=\sqrt{8^2+2^2}=\sqrt{68}=2\sqrt{17}$

Rectangle has length 60 inches and width 80 inches. The two diagonals of the rectangle intersect at point . Which is the greater quantity?

(a)

(b)

(a) is greater.

(b) is greater.

(a) and (b) are equal.

It is impossible to tell from the information given.

Explanation

Two consecutive sides of a rectangle and a diagonal form a right triangle, so the length of any diagonal can be determined using the Pythagorean Theorem, substituting :