High School Math : Geometry

Study concepts, example questions & explanations for High School Math

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Example Questions

Example Question #151 : Geometry

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:

30°

56°

45°

90°

120°

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).

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:

72°

60°

55°

35°

65°

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°.

 

 

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:

\(\displaystyle 24\)

\(\displaystyle 35\)

\(\displaystyle 17.5\)

\(\displaystyle 25\)

Correct answer:

\(\displaystyle 17.5\)

Explanation:

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

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

Plug in the diagonals for \(\displaystyle a\) and \(\displaystyle b\) to get \(\displaystyle \frac{5*7}{2}=Area\)

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

The answer is \(\displaystyle 17.5\)

Example Question #1 : Quadrilaterals

Find the area of the following kite:

Kite

Possible Answers:

\(\displaystyle 26m^2\)

\(\displaystyle 39m^2\)

\(\displaystyle 78m^2\)

\(\displaystyle 54m^2\)

\(\displaystyle 27m^2\)

Correct answer:

\(\displaystyle 39m^2\)

Explanation:

The formula for the area of a kite is:

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

Where \(\displaystyle d_{1}\) is the length of one diagonal and \(\displaystyle d_{2}\) is the length of the other diagonal

Plugging in our values, we get:

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

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

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:

\(\displaystyle 18\sqrt{3}m^2\)

\(\displaystyle 18\sqrt{3}-18m^2\)

\(\displaystyle 18\sqrt{3}+18m^2\)

\(\displaystyle 18m^2\)

\(\displaystyle 20m^2\)

Correct answer:

\(\displaystyle 18\sqrt{3}+18m^2\)

Explanation:

The formula for the area of a kite is:

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

where \(\displaystyle d_1\) is the length of one diagonal and \(\displaystyle d_2\) is the length of another diagonal.

 

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

Our \(\displaystyle 45-45-90\) triangle is: \(\displaystyle 3 \sqrt{2}m-3 \sqrt{2}m-6m\)

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

 

Plugging in our values, we get:

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

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

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

Example Question #1 : Quadrilaterals

Find the perimeter of the following kite:

Kite

Possible Answers:

\(\displaystyle 18+\sqrt{10} m\)

\(\displaystyle 10+6\sqrt{10} m\)

\(\displaystyle 16+\sqrt{10} m\)

\(\displaystyle 8+6\sqrt{10} m\)

\(\displaystyle 10+9\sqrt{10} m\)

Correct answer:

\(\displaystyle 10+6\sqrt{10} m\)

Explanation:

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

\(\displaystyle 3-4-5\)

\(\displaystyle 3m-4m-5m\)

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

\(\displaystyle A^2+B^2=C^2\)

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

\(\displaystyle C^2=90\)

\(\displaystyle C=3\sqrt{10}m\)

The formula of the perimeter of a kite is:

\(\displaystyle P = 2(side_{short})+2(side_{long})\)

Plugging in our values, we get:

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

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:

\(\displaystyle 24m\)

\(\displaystyle 12\sqrt{2}m + 12m\)

\(\displaystyle 12\sqrt{2}m - 12m\)

\(\displaystyle 12\sqrt{3}m - 12m\)

\(\displaystyle 12\sqrt{3}m + 12m\)

Correct answer:

\(\displaystyle 12\sqrt{2}m + 12m\)

Explanation:

The formula for the perimeter of a kite is:

\(\displaystyle P = 2(s_{long}) + 2(s_{short})\)

Where \(\displaystyle s_{long}\) is the length of the longer side and \(\displaystyle s_{short}\) is the length of the shorter side

 

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

 

Our \(\displaystyle 45-45-90\) triangle is: \(\displaystyle 3 \sqrt{2}m-3 \sqrt{2}m-6m\)

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

 

Plugging in our values, we get:

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

\(\displaystyle P = 12\sqrt{2}m + 12m\)

Example Question #1 : Trapezoids

The following quadrilaterals are similar. Solve for \(\displaystyle x\).

Question_9

(Figure not drawn to scale).

Possible Answers:

\(\displaystyle 7\)

\(\displaystyle 4\)

\(\displaystyle 3\)

\(\displaystyle 3.5\)

Correct answer:

\(\displaystyle 3.5\)

Explanation:

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

\(\displaystyle \small \frac{2}{8}=\frac{x}{14}\)

Cross multiply.

\(\displaystyle \small 28=8x\)

\(\displaystyle \small x=3.5\)

Example Question #1 : Trapezoids

Find the area of the following trapezoid:

Trapezoid

Possible Answers:

\(\displaystyle 202m^2\)

\(\displaystyle 260m^2\)

\(\displaystyle 252m^2\)

\(\displaystyle 126m^2\)

\(\displaystyle 180m^2\)

Correct answer:

\(\displaystyle 252m^2\)

Explanation:

The formula for the area of a trapezoid is:

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

Where \(\displaystyle b_1\) is the length of one base, \(\displaystyle b_2\) is the length of the other base, and \(\displaystyle h\) is the height.

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

\(\displaystyle 5-12-13\)

\(\displaystyle 5m-12m-13m\)

Plugging in our values, we get:

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

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

Example Question #1 : Trapezoids

Find the area of the following trapezoid:

Trapezoid_angles

Possible Answers:

\(\displaystyle 48+24\sqrt{3}m^2\)

\(\displaystyle 96+12\sqrt{3}m^2\)

\(\displaystyle 54+24\sqrt{6}m^2\)

\(\displaystyle 96+24\sqrt{3}m^2\)

\(\displaystyle 60+24\sqrt{3}m^2\)

Correct answer:

\(\displaystyle 96+24\sqrt{3}m^2\)

Explanation:

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

The formula is:

\(\displaystyle a-a\sqrt{3}-2a\)

Where \(\displaystyle a\) is the length of the side opposite the \(\displaystyle \measuredangle 30\).

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

Creating another \(\displaystyle 30-60-90\) triangle on the left, we find the height is \(\displaystyle 6m\), the length of the base is \(\displaystyle 2\sqrt{3}m\), and the side is \(\displaystyle 4\sqrt{3}m\).

 

The formula for the area of a trapezoid is:

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

Where \(\displaystyle b_1\) is the length of one base, \(\displaystyle b_2\) is the length of the other base, and \(\displaystyle h\) is the height.

Plugging in our values, we get:

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

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

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

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

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