Trapezoids

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Geometry › Trapezoids

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Which of the following shapes is a trapezoid?

Shapes

Explanation

A trapezoid is a four-sided shape with straight sides that has a pair of opposite parallel sides. The other sides may or may not be parallel. A square and a rectangle are both considered trapezoids.

Find the area of the figure.

Explanation

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

$\text{Hypotenuse}=\sqrt{6^2+12^2}=\sqrt{180}$

Next, use this value to find the area of the trapezoid.

$\text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})$

Plug in the given and found values to find the area.

$\text{Area of Trapezoid}=\frac{1}{2}(10+\sqrt{180})(4)=20+2\sqrt{180}$

Next, find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(6)(12)=36$

To find the area of the figure, add the two areas together.

$\text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}$

$\text{Area of Figure}=20+2\sqrt{180}+36=82.83$

Make sure to round to places after the decimal.

Which of the following shapes is a trapezoid?

Shapes

Explanation

A trapezoid is a four-sided shape with straight sides that has a pair of opposite parallel sides. The other sides may or may not be parallel. A square and a rectangle are both considered trapezoids.

Find the area of the figure.

Explanation

From the figure, you should notice that it is made up of a right triangle and a trapezoid. The lower base of the trapezoid is also the hypotenuse of the right triangle.

First, find the length of the hypotenuse of the right triangle using the Pythagorean Theorem.

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

$\text{Hypotenuse}=\sqrt{6^2+12^2}=\sqrt{180}$

Next, use this value to find the area of the trapezoid.

$\text{Area of Trapezoid}=\frac{1}{2}(\text{base 1}+\text{base 2})(\text{height})$

Plug in the given and found values to find the area.

$\text{Area of Trapezoid}=\frac{1}{2}(10+\sqrt{180})(4)=20+2\sqrt{180}$

Next, find the area of the triangle.

$\text{Area of Triangle}=\frac{1}{2}(\text{base})(\text{height})$

$\text{Area of Triangle}=\frac{1}{2}(6)(12)=36$

To find the area of the figure, add the two areas together.

$\text{Area of Figure}=\text{Area of Trapezoid}+\text{Area of Triangle}$

$\text{Area of Figure}=20+2\sqrt{180}+36=82.83$

Make sure to round to places after the decimal.

Screen_shot_2015-03-06_at_2.30.09_pm

What is the area of the trapezoid pictured above in square units?

Explanation

The formula for the area of a trapezoid is the average of the bases times the height,

$A=\frac{b_{1}+b_{2}}{2}h$ .

Looking at this problem and when the appropriate values are plugged in, the formula yields:

$A=(\frac{10+24}{2})7$

$A=\frac{34}{2}\cdot 7$

$A=\frac{238}{2}=119$

Screen_shot_2015-03-06_at_2.30.09_pm

What is the area of the trapezoid pictured above in square units?

Explanation

The formula for the area of a trapezoid is the average of the bases times the height,

$A=\frac{b_{1}+b_{2}}{2}h$ .

Looking at this problem and when the appropriate values are plugged in, the formula yields:

$A=(\frac{10+24}{2})7$

$A=\frac{34}{2}\cdot 7$

$A=\frac{238}{2}=119$

A right triangle and rectangle are placed adjacent to one another such that the composite figure formed by the triangle and rectangle is a trapezoid.

Find the area of the trapezoid given that the base of the triangle is 7 ft and the hypotenuse of the triangle is 25ft. The base of the rectangle is 9 feet.

$300ft^{2}$

$2880 ft^{2}$

$312.5 ft^{2}$

$144ft^{2}$

Explanation

$First\ we\ can\ find\ the\ height\ of\ the\ right\ triangle\ and\ assume\ that\ this\ is\ also\ the\ height\ of\ the\ rectangle\ and\ the\ trapezoid.$

$The\ height\ of\ the\ right\ triangle\ is\ obtained\ by\ using\ the\ Pythagorean\ theorem.$

$7^{2}+h^{2}=25^{2}$

$49+h^{2}=625$

$h^{2}=576$

$We\ know\ the\ top\ of\ the\ trapezoid\ is\ the\ same\ as\ the\ base\ of\ the\ rectangle\ which\ is\ 9\ cm.$

$We\ know\ that\ the\ base\ of\ the\ trapezoid\ is\ the\ sum\ of\ the\ base\ of\ the\ triangle\ and\ the\ base\ of\ the\ rectangle: 9+7=16$

$Now\ we\ have\ all\ parts\ needed\ to\ plug\ into\ the\ formula\ for\ the\ area\ of\ a\ trapezoid: A=\frac{b_{1}+b^{_{2}}}{2}*h$

$A=\frac{9+16}{2}*24=300cm^{2}$

A right triangle and rectangle are placed adjacent to one another such that the composite figure formed by the triangle and rectangle is a trapezoid.

Find the area of the trapezoid given that the base of the triangle is 7 ft and the hypotenuse of the triangle is 25ft. The base of the rectangle is 9 feet.

$300ft^{2}$

$2880 ft^{2}$

$312.5 ft^{2}$

$144ft^{2}$

Explanation

$First\ we\ can\ find\ the\ height\ of\ the\ right\ triangle\ and\ assume\ that\ this\ is\ also\ the\ height\ of\ the\ rectangle\ and\ the\ trapezoid.$

$The\ height\ of\ the\ right\ triangle\ is\ obtained\ by\ using\ the\ Pythagorean\ theorem.$

$7^{2}+h^{2}=25^{2}$

$49+h^{2}=625$

$h^{2}=576$

$We\ know\ the\ top\ of\ the\ trapezoid\ is\ the\ same\ as\ the\ base\ of\ the\ rectangle\ which\ is\ 9\ cm.$

$We\ know\ that\ the\ base\ of\ the\ trapezoid\ is\ the\ sum\ of\ the\ base\ of\ the\ triangle\ and\ the\ base\ of\ the\ rectangle: 9+7=16$

$Now\ we\ have\ all\ parts\ needed\ to\ plug\ into\ the\ formula\ for\ the\ area\ of\ a\ trapezoid: A=\frac{b_{1}+b^{_{2}}}{2}*h$

$A=\frac{9+16}{2}*24=300cm^{2}$

What is the area of the following trapezoid?

Trapezoid