# 1. What are the base and the height of the rectangle if the area is 216 and the perimeter is 66? ...

## Question:

1. What are the base and the height of the rectangle if the area is 216 and the perimeter is 66?

2. The perimeter of an isosceles trapezoid is 40 ft. The bases of the trapezoid are 11ft and 19 ft. Find the area of the trapezoid.

## The Area and Perimeter of a Rectangle:

A rectangle is a closed geometrical figure that has two pairs of parallel sides. The parallel sides in a rectangle are congruent, meaning that they have equal measures. The area of a rectangle is the amount of space covered by the figure. It is given by {eq}A = l\times w {/eq}, where {eq}l {/eq} is the length and {eq}w {/eq} is the width. The perimeter of a rectangle is the length of the line forming the boundary of the figure. This length is twice the sum of the length and the width {eq}P = 2(l + w) {/eq}.

1. What are the base and the height of the rectangle if the area is 216 and the perimeter is 66?

The perimeter of a rectangle is given by:

• {eq}P = 2(l + w) {/eq}

where {eq}l {/eq} is the length and {eq}w {/eq} is the width.

If the perimeter is 216, then:

• {eq}66 = 2(l + w) {/eq}
• {eq}33= l + w {/eq}........................................................................................................(i)

The area of a rectangle is given by:

• {eq}A = l\times w {/eq}

If the area is 216, then:

• {eq}216 = l\times w {/eq}........................................................................................................(ii)

From equation (i):

• {eq}l = 33 - w {/eq}

Substituting this into equation (ii):

• {eq}216 = w(33 - w) {/eq}
• {eq}216 = 33w - w^2 {/eq}
• {eq}w^2 - 33w + 216 = 0 {/eq}

Solving the quadratic equation by factorization:

• {eq}(w - 9)(w - 24) = 0 {/eq}
• {eq}w = 9, \quad w = 24 {/eq}

Thus, if the width is 9, then the length is:

• {eq}l = 33 - 9 = 24 {/eq}

And if the width is 24, then the length is:

• {eq}l = 33 - 24 = 9 {/eq}

2. The perimeter of an isosceles trapezoid is 40 ft. The bases of the trapezoid are 11ft and 19 ft. Find the area of the trapezoid.

An isosceles trapezoid is a trapezoid whose legs or the non-parallel sides have equal measures.

For an isosceles trapezoid, the perimeter is given by:

• {eq}P = 2l + a + b {/eq}

where {eq}l {/eq} is the measure of each of the legs and {eq}\rm a \text{ and } b {/eq} are the measures of the teo bases.

If the bases of the trapezoid in our question are {eq}a = 11\; \rm ft {/eq} and {eq}b = 9\; \rm ft {/eq} and the perimeter is 40, then:

• {eq}40 = 2a + 11 + 19 {/eq}
• {eq}2a = 10 {/eq}
• {eq}a = \dfrac{10}{2} = 5\; \rm ft {/eq}

The area of a trapezoid is given by:

• {eq}A = \dfrac{1}{2}(a + b)h {/eq}

Thus, to calculate the area of the trapezoid in our question, we need to calculate the height.

The height will be equal to:

• {eq}h = \sqrt{5^2 - [0.5(19 - 11)]^2 } {/eq}
• {eq}h = \sqrt{25 -16 } = 3 {/eq}

Therefore, the area of the trapezoid is equal to:

• {eq}A = \dfrac{1}{2}\left (11 + 19\right)\times 3 {/eq}
• {eq}\boxed{\color{blue}{A = 45\; \rm ft^2}} {/eq}