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High School Math : Geometry

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Example Questions

Example Question #1 : Clock Math

What is the measure of the smaller angle formed by the hands of an analog watch if the hour hand is on the 10 and the minute hand is on the 2?

Possible Answers:

120°

45°

56°

30°

90°

Correct answer:

120°

Explanation:

A analog clock is divided up into 12 sectors, based on the numbers 1–12. One sector represents 30 degrees (360/12 = 30). If the hour hand is directly on the 10, and the minute hand is on the 2, that means there are 4 sectors of 30 degrees between then, thus they are 120 degrees apart (30 * 4 = 120).

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Example Question #1 : How To Find The Angle Of Clock Hands

What is the measure, in degrees, of the acute angle formed by the hands of a 12-hour clock that reads exactly 3:10?

Possible Answers:

65°

60°

55°

72°

35°

Correct answer:

35°

Explanation:

The entire clock measures 360°. As the clock is divided into 12 sections, the distance between each number is equivalent to 30° (360/12). The distance between the 2 and the 3 on the clock is 30°. One has to account, however, for the 10 minutes that have passed. 10 minutes is 1/6 of an hour so the hour hand has also moved 1/6 of the distance between the 3 and the 4, which adds 5° (1/6 of 30°). The total measure of the angle, therefore, is 35°.

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Example Question #1 : How To Find The Area Of A Kite

What is the area of a kite with diagonals of 5 and 7?

Possible Answers:

17.5

Correct answer:

17.5

Explanation:

To find the area of a kite using diagonals you use the following equation $\frac{a*b}{2}=Area\: of a\: kite$

That diagonals ( and )are the lines created by connecting the two sides opposite of each other.

Plug in the diagonals for and to get $\frac{5*7}{2}=Area$

Then multiply and divide to get the area. $\frac{35}{2}=17.5$

The answer is 17.5

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Example Question #2 : How To Find The Area Of A Kite

Find the area of the following kite:

Kite

Possible Answers:

27m^2

39m^2

54m^2

26m^2

78m^2

Correct answer:

39m^2

Explanation:

The formula for the area of a kite is:

$A = \frac{1}{2}(d_{1}\cdot d_{2})$

Where $d_{1}$ is the length of one diagonal and $d_{2}$ is the length of the other diagonal

Plugging in our values, we get:

$A = \frac{1}{2}(d_{1}\cdot d_{2})$

$A = \frac{1}{2}(6m\cdot 13m) = 39m^2$

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Example Question #2 : How To Find The Area Of A Kite

Find the area of the following kite:

Screen_shot_2014-03-01_at_9.16.34_pm

Possible Answers:

$18\sqrt{3}m^2$

$18\sqrt{3}-18m^2$

$18\sqrt{3}+18m^2$

18m^2

20m^2

Correct answer:

$18\sqrt{3}+18m^2$

Explanation:

The formula for the area of a kite is:

$A = \frac{1}{2} (d_1)(d_2)$

where d_1 is the length of one diagonal and d_2 is the length of another diagonal.

Use the formulas for a 45-45-90 triangle and a 30-60-90 triangle to find the lengths of the diagonals. The formula for a triangle is $a-a-a \sqrt{2}$ and the formula for a triangle is $a-a \sqrt{3}-2a$ .

Our 45-45-90 triangle is: $3 \sqrt{2}m-3 \sqrt{2}m-6m$

Our 30-60-90 triangle is: $3 \sqrt{2}m-3 \sqrt{6}m-6\sqrt{2}m$

Plugging in our values, we get:

$A = \frac{1}{2} (d_1)(d_2)$

$A = \frac{1}{2} (6\sqrt{2}m) (3\sqrt{6}m + 3\sqrt{2}m)$

$A = \frac{1}{2} (18\sqrt{12}+18\sqrt{4}) = 18\sqrt{3}+18m^2$

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Example Question #1 : Kites

Find the perimeter of the following kite:

Kite

Possible Answers:

$18+\sqrt{10} m$

$16+\sqrt{10} m$

$8+6\sqrt{10} m$

$10+6\sqrt{10} m$

$10+9\sqrt{10} m$

Correct answer:

$10+6\sqrt{10} m$

Explanation:

In order to find the length of the two shorter edges, use a Pythagorean triple:

3-4-5

3m-4m-5m

In order to find the length of the two longer edges, use the Pythagorean theorem:

A^2+B^2=C^2

(9)^2+(3)^2=C^2

C^2=90

$C=3\sqrt{10}m$

The formula of the perimeter of a kite is:

$P = 2(side_{short})+2(side_{long})$

Plugging in our values, we get:

$P = 2(5m)+2(3\sqrt{10}m) = 10+6\sqrt{10}m$

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Example Question #1 : How To Find The Perimeter Of Kite

Find the perimeter of the following kite:

Screen_shot_2014-03-01_at_9.16.34_pm

Possible Answers:

$12\sqrt{2}m + 12m$

$12\sqrt{2}m - 12m$

24m

$12\sqrt{3}m - 12m$

$12\sqrt{3}m + 12m$

Correct answer:

$12\sqrt{2}m + 12m$

Explanation:

The formula for the perimeter of a kite is:

$P = 2(s_{long}) + 2(s_{short})$

Where $s_{long}$ is the length of the longer side and $s_{short}$ is the length of the shorter side

Use the formulas for a 45-45-90 triangle and a 30-60-90 triangle to find the lengths of the longer sides. The formula for a triangle is $a-a-a \sqrt{2}$ and the formula for a triangle is $a-a \sqrt{3}-2a$ .

Our 45-45-90 triangle is: $3 \sqrt{2}m-3 \sqrt{2}m-6m$

Our 30-60-90 triangle is: $3 \sqrt{2}m-3 \sqrt{6}m-6\sqrt{2}m$

Plugging in our values, we get:

$P = 2(6\sqrt{2}m) + 2(6m)$

$P = 12\sqrt{2}m + 12m$

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Example Question #1 : Trapezoids

The following quadrilaterals are similar. Solve for .

(Figure not drawn to scale).

Possible Answers:

3.5

Correct answer:

3.5

Explanation:

When polygons are similar, the sides will have the same ratio to one another. Set up the appropriate proportions.

$\small \frac{2}{8}=\frac{x}{14}$

Cross multiply.

$\small 28=8x$

$\small x=3.5$

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

Find the area of the following trapezoid:

Trapezoid

Possible Answers:

126m^2

252m^2

202m^2

260m^2

180m^2

Correct answer:

252m^2

Explanation:

The formula for the area of a trapezoid is:

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

Where b_1 is the length of one base, b_2 is the length of the other base, and is the height.

To find the height of the trapezoid, use a Pythagorean triple:

5-12-13

5m-12m-13m

Plugging in our values, we get:

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

$A = \frac{1}{2}(16m+26m)(12m)=252m^2$

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Example Question #1 : Trapezoids

Find the area of the following trapezoid:

Trapezoid_angles

Possible Answers:

$48+24\sqrt{3}m^2$

$96+12\sqrt{3}m^2$

$54+24\sqrt{6}m^2$

$96+24\sqrt{3}m^2$

$60+24\sqrt{3}m^2$

Correct answer:

$96+24\sqrt{3}m^2$

Explanation:

Use the formula for 30-60-90 triangles in order to find the length of the bottom base and the height.

The formula is:

$a-a\sqrt{3}-2a$

Where is the length of the side opposite the $\measuredangle 30$ .

Beginning with the 12m side, if we were to create a 30-60-90 triangle, the length of the base is $6\sqrt{3}m$ , and the height is .

Creating another 30-60-90 triangle on the left, we find the height is , the length of the base is $2\sqrt{3}m$ , and the side is $4\sqrt{3}m$ .

The formula for the area of a trapezoid is:

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

Where b_1 is the length of one base, b_2 is the length of the other base, and is the height.

Plugging in our values, we get:

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

$A = \frac{1}{2}(16m+16m+6\sqrt{3}m+2\sqrt{3}m)(6m)$

$A = \frac{1}{2}(32m+8\sqrt{3}m)(6m)$

$A = (32m+8\sqrt{3}m)(3m)=96+24\sqrt{3}m^2$

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High School Math : Geometry

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Clock Math

Example Question #1 : How To Find The Angle Of Clock Hands

Example Question #1 : How To Find The Area Of A Kite

Example Question #2 : How To Find The Area Of A Kite

Example Question #2 : How To Find The Area Of A Kite

Example Question #1 : Kites

Example Question #1 : How To Find The Perimeter Of Kite

Example Question #1 : Trapezoids

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

Example Question #1 : Trapezoids

All High School Math Resources

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