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Advanced Geometry : Quadrilaterals

Study concepts, example questions & explanations for Advanced Geometry

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Example Questions

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Example Question #21 : Trapezoids

Find the area of the figure below.

Possible Answers:

233.97

296.67

280.79

281.05

Correct answer:

280.79

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.

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

$\text{Hypotenuse}=\sqrt{7^2+16^2}=\sqrt{305}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(20+\sqrt{305})(12)=120+6\sqrt{305}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(7)(16)=56$

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

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

$\text{Area of Figure}=120+6\sqrt{305}+56=280.79$

Make sure to round to places after the decimal.

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Example Question #21 : How To Find The Area Of A Trapezoid

Find the area of the figure.

Possible Answers:

774.26

749.06

703.31

746.17

Correct answer:

746.17

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.

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

$\text{Hypotenuse}=\sqrt{20^2+24^2}=\sqrt{976}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(25+\sqrt{976})(18)=225+9\sqrt{976}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(20)(24)=240$

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

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

$\text{Area of Figure}=225+9\sqrt{976}+240=746.17$

Make sure to round to places after the decimal.

Report an Error

Example Question #71 : Plane Geometry

Find the area of the figure.

Possible Answers:

354.01

305.22

396.84

332.22

Correct answer:

332.22

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.

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

$\text{Hypotenuse}=\sqrt{6^2+20^2}=\sqrt{436}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(21+\sqrt{436})(13)=\frac{273+13\sqrt{436}}{2}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(6)(20)=60$

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

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

$\text{Area of Figure}=\frac{273+13\sqrt{436}}{2}+60=332.22$

Make sure to round to places after the decimal.

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Example Question #31 : How To Find The Area Of A Trapezoid

Find the area of the figure below.

Possible Answers:

161.57

150.36

148.78

132.67

Correct answer:

161.57

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.

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

$\text{Hypotenuse}=\sqrt{7^2+12^2}=\sqrt{193}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(16+\sqrt{193})(8)=64+4\sqrt{193}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(7)(12)=42$

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

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

$\text{Area of Figure}=64+4\sqrt{193}+42=161.57$

Make sure to round to places after the decimal.

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Example Question #31 : Trapezoids

Find the area of the figure below.

Possible Answers:

59.38

55.30

50.67

62.12

Correct answer:

62.12

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.

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

$\text{Hypotenuse}=\sqrt{5^2+11^2}=\sqrt{146}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(11+\sqrt{146})(3)=\frac{33+3\sqrt{146}}{2}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(5)(11)=\frac{55}{2}$

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

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

$\text{Area of Figure}=\frac{33+3\sqrt{146}}{2}+\frac{55}{2}=62.12$

Make sure to round to places after the decimal.

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Example Question #71 : Quadrilaterals

Find the area of the figure below.

Possible Answers:

906.39

877.23

803.92

898.53

Correct answer:

898.53

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.

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

$\text{Hypotenuse}=\sqrt{22^2+26^2}=\sqrt{1160}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(34+\sqrt{1160})(18)=306+9\sqrt{1160}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(22)(26)=286$

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

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

$\text{Area of Figure}=306+9\sqrt{1160}+286=898.53$

Make sure to round to places after the decimal.

Report an Error

Example Question #31 : Trapezoids

Find the area of the figure below.

Possible Answers:

140.58

136.01

150.81

131.95

Correct answer:

150.81

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.

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

$\text{Hypotenuse}=\sqrt{9^2+10^2}=\sqrt{181}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(13+\sqrt{181})(8)=52+4\sqrt{181}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(9)(10)=45$

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

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

$\text{Area of Figure}=52+4\sqrt{181}+45=150.81$

Make sure to round to places after the decimal.

Report an Error

Example Question #31 : How To Find The Area Of A Trapezoid

Find the area of the figure.

Possible Answers:

111.73

94.37

91.85

103.37

Correct answer:

111.73

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.

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

$\text{Hypotenuse}=\sqrt{7^2+12^2}=\sqrt{193}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(14+\sqrt{193})(5)=35+\frac{5\sqrt{193}}{2}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(7)(12)=42$

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

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

$\text{Area of Figure}=35+\frac{5\sqrt{117}}{2}+42=111.73$

Make sure to round to places after the decimal.

Report an Error

Example Question #71 : Plane Geometry

Trapezoid

Figure NOT drawn to scale.

Examine the above trapezoid.

MN = 20 , TM = MP = 5 , RM = MA

True, false, or inconclusive: the area of Trapezoid TRAP is 200.

Possible Answers:

True

False

Correct answer:

True

Explanation:

The area of a trapezoid is equal to one half the product of half the height of the trapezoid and the sum of the lengths of the bases. This is

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

$A = \frac{1}{2} \cdot TP \cdot (TR + PA)$

or, equivalently,

$A = TP \cdot \frac{TR + PA}{2}$

The height of the trapezoid is

TP = TM + MP = 5 + 5 = 10 .

The lengths of bases $\overline{TR}$ and $\overline{PA}$ are not given, so it might appear that determining the area of the trapezoid is impossible.

However, it is given that TM = MP and RM = MA - that is, the segment $\overline{MN}$ bisects both legs of the trapezoid. This makes $\overline{MN}$ the midsegment of the trapezoid, the length of which is the arithmetic mean of those of the bases:

$MN = \frac{TR+ PA}{2}$ .

Therefore, the formula for the area of the trapezoid can be rewritten as

$A = TP \cdot MN$ ,

the product of the height and the length of the midsegment.

TP = 10 and MN = 20 , so

$A =10 \cdot 20 = 200$ ,

making the statement true.

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Example Question #1 : How To Find The Length Of The Diagonal Of A Trapezoid

What is the length of the diagonals of trapezoid ABCD ? Assume the figure is an isoceles trapezoid.

Trapezoid

Possible Answers:

$\sqrt{95}m$

$\sqrt{98}m$

$\sqrt{99}m$

$\sqrt{97}m$

$\sqrt{96}m$

Correct answer:

$\sqrt{97}m$

Explanation:

To find the length of the diagonal, we need to use the Pythagorean Theorem. Therefore, we need to sketch the following triangle within trapezoid ABCD :

Trapezoid

We know that the base of the triangle has length $9\: m$ . By subtracting the top of the trapezoid from the bottom of the trapezoid, we get:

$12\: m - 6\: m = 6\: m$

Dividing by two, we have the length of each additional side on the bottom of the trapezoid:

$\frac{6\: m}{2} = 3\: m$

Adding these two values together, we get $9\: m$ .

The formula for the length of diagonal uses the Pythagoreon Theorem:

AC^2 = AE^2 + EC^2 , where is the point between and representing the base of the triangle.

Plugging in our values, we get:

$AC^2 = (9\: m)^2 + (4\: m)^2$

$AC^2 = 81\: m^2 + 16\: m^2$

$AC^2 = 97\: m^2$

$AC = \sqrt{97}\: m$

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Advanced Geometry : Quadrilaterals

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #21 : Trapezoids

Example Question #21 : How To Find The Area Of A Trapezoid

Example Question #71 : Plane Geometry

Example Question #31 : How To Find The Area Of A Trapezoid

Example Question #31 : Trapezoids

Example Question #71 : Quadrilaterals

Example Question #31 : Trapezoids

Example Question #31 : How To Find The Area Of A Trapezoid

Example Question #71 : Plane Geometry

Example Question #1 : How To Find The Length Of The Diagonal Of A Trapezoid

All Advanced Geometry Resources

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