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Advanced Geometry : Plane Geometry

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Example Questions

Example Question #61 : Plane Geometry

$\small \small Find\ the\ height\ of\ a\ trapezoid\ with\ bases\ of\ 8\ and\ 12\ and\ an\ area\ of\ 80m^{2}.$

Possible Answers:

$\small 12\ m$

$\small 60\ m$

$\small \small \small 10\ m$

$\small \small 8\ m$

Correct answer:

$\small \small 8\ m$

Explanation:

$\small Use\ the\ area\ formula\ for\ a\ trapezoid.$

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

$\small Substitute\ each\ known\ value\ then\ use\ algebra\ to\ solve\ for\ the\ height.$

$\small \small 80=\frac{8+12}{2}h$

$\small 80=10h$

$\small 8=h$

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Example Question #352 : Advanced Geometry

Find the area of the trapezoid given below:

Trap1

Possible Answers:

$300\ cm^2$

$325\ cm^2$

$240\ cm^2$

$260\ cm^2$

Correct answer:

$300\ cm^2$

Explanation:

In order to find the area of a trapezoid, we must use the formula below:

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

$In\ this\ trapezoid\ b_{1}=30\ and\ b_{2}=20.$

$We\ can\ also\ see\ that\ h=12.$

$When\ we\ plug\ in\ all\ of\ the\ above\ values\ into\ the\ formula\ we\ get:$

$A=\frac{20+30}{2}(12)$

A=300cm^2

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Example Question #16 : Trapezoids

Show algebraically how to develop the trapezoid area formula.

Varsity7

Possible Answers:

$\frac{1}{2}b_{1}h+\frac{1}{2}b_{2}h$

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

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

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

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

$h(b_{1}\cdot b_{2})$

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

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

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

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

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

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

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

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

Correct answer:

$\frac{1}{2}b_{1}h+\frac{1}{2}b_{2}h$

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

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

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

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

Explanation:

1) The area $A_{1}$ of the triangle that has base $b_{1}$ is $A_{1} = \frac{1}{2}(b_{1})h$ .

2) The area $A_{2}$ of the triangle that has base $b_{2}$ is $A_{2} = \frac{1}{2}(b_{2})h$ .

3) The sum of the two triangel areas is

$A_{1} + A_{2}$

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

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

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

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

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

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

Find the area of the trapezoid.

Varsity2

Possible Answers:

For area ,

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

$= \frac{2a}{2}(a + 4a)$

= a(5a)

$= 5a^{2}$

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

$= \frac{1}{2}(3a)$

$A = a\cdot2a\cdot4a$

$= 8a^{3}$

$A = 4\cdot \frac{1}{2}(4a-a)2a$

$= 4\cdot \frac{1}{2}(3a)2a$

$= 2\cdot3a\cdot2a$

$= 12a^{2}$

$A = \frac{1}{2}\cdot4a\cdot2a + a$

$= 4a^{2} + a$

$= 5a^{2}$

Correct answer:

For area ,

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

$= \frac{2a}{2}(a + 4a)$

= a(5a)

$= 5a^{2}$

Explanation:

1) Using the trapezoid area formula, $A = \frac{h}{2}(b_{1} + b_{2})$ ,

height h = 2a , base $b_{1} = a$ , base $b_{2} = 4a$ .

2) After substition, the resulting expression is $= \frac{2a}{2}(a + 4a)$ .

3) The resulting expression is then simplified,

= a(5a)

$= 5a^{2}$

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Example Question #22 : Trapezoids

The rectangle is circumscribed about the trapezoid EBCF . Find the area $A_{T}$ of the trapezoid.

Varsity6

Possible Answers:

49.56

37.5

Correct answer:

49.56

Explanation:

1) The measure of the trapezoid top base $\overline{BC}$ has to be found.

2) To find $m(\overline{BC})$ , and have to be found, then the sum x + y is subtracted from the length of the top side of the rectangle, resulting in the difference 10 - (x + y) .

3) $m(\overline{BC}) = 10 - (x + y)$ .

3) By using the Pythagorean Theorem on $\bigtriangleup EAB$ and $\bigtriangleup FCD$ , and can be found.

4) Using $\bigtriangleup EAB$ for ,

$(7.5)^{2} + x^{2} = 8^{2}$

$x^{2} = 64 - 56.25$

$x^{2} = 7.75$

$x = \sqrt{7.75}$

$x \approx 2.784$

5) Using $\bigtriangleup FCD$ for ,

$(7.5)^{2} + y^{2} = (8.5)^{2}$

$y^{2} = 72.25 - 56.25$

$y^{2} = 16$

y = 4

6) $BC = 10 - (\sqrt{7.75} + 4) \approx 3.216$

7) $A_{T} = \frac{7.5}{2} \cdot ( 3.216 + 10)$

= (3.75)(13.216)

= 49.56

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Example Question #22 : How To Find The Area Of A Trapezoid

Find the area of the shaded region.

Varsity3

Possible Answers:

105

322

22.5

200

Correct answer:

22.5

Explanation:

The shaded region is between the outer and inner trapezoid. To find the area of the shaded region, subtract the area of the inner trapezoid from the area of the outer trapezoid.

1) Area of the shaded region = $A_{S}$ , area of the outer trapezoid = $A_{O}$ , area of the inner trapezoid = $A_{I}$ .

2) $A_{O} = \frac{5}{2}(5 + 10)$ , $A_{I} = \frac{3}{2}(4 + 6)$

3) $A_{S} = A_{O} - A_{I}$

$= \frac{5}{2}(5 + 10) - \frac{3}{2}(4 + 6)$

$= \frac{5}{2}(15) - \frac{3}{2}(10)$

$= \frac{75}{2} - \frac{30}{2}$

$= \frac{45}{2}$

= 22.5

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Example Question #23 : How To Find The Area Of A Trapezoid

Find the area of the figure.

Possible Answers:

82.83

79.06

81.27

85.33

Correct answer:

82.83

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.

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Example Question #25 : Trapezoids

Find the area of the figure.

Possible Answers:

Correct answer:

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.

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

$\text{Hypotenuse}=\sqrt{5^2+12^2}=\sqrt{169}=13$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(11+13)(3)=36$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(5)(12)=30$

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

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

$\text{Area of Figure}=30+36=66$

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Example Question #26 : Trapezoids

Find the area of the figure below.

Possible Answers:

136.28

150.09

164.67

114.00

Correct answer:

164.67

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.

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

$\text{Hypotenuse}=\sqrt{10^2+16^2}=\sqrt{356}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(15+\sqrt{356})(5)=\frac{75+5\sqrt{356}}{2}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(10)(16)=80$

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

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

$\text{Area of Figure}=\frac{75+5\sqrt{356}}{2}+80=164.67$

Make sure to round to places after the decimal.

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Example Question #27 : Trapezoids

Find the area of the figure.

Possible Answers:

75.64

68.09

64.63

71.22

Correct answer:

64.63

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.

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

$\text{Hypotenuse}=\sqrt{6^2+9^2}=\sqrt{117}$

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.

$\text{Area of Trapezoid}=\frac{1}{2}(8+\sqrt{117})(4)=16+2\sqrt{117}$

Next, find the area of the triangle.

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

$\text{Area of Triangle}=\frac{1}{2}(6)(9)=27$

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

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

$\text{Area of Figure}=16+2\sqrt{117}+27=64.63$

Make sure to round to places after the decimal.

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Advanced Geometry : Plane Geometry

Study concepts, example questions & explanations for Advanced Geometry

All Advanced Geometry Resources

Example Questions

Example Question #61 : Plane Geometry

Example Question #352 : Advanced Geometry

Example Question #16 : Trapezoids

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

Example Question #22 : Trapezoids

Example Question #22 : How To Find The Area Of A Trapezoid

Example Question #23 : How To Find The Area Of A Trapezoid

Example Question #25 : Trapezoids

Example Question #26 : Trapezoids

Example Question #27 : Trapezoids

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