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High School Math : Quadrilaterals

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Example Question #1 : Quadrilaterals

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 : Quadrilaterals

Find the area of the following kite:

Kite

Possible Answers:

39m^2

26m^2

78m^2

54m^2

27m^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 #3 : Quadrilaterals

Find the area of the following kite:

Screen_shot_2014-03-01_at_9.16.34_pm

Possible Answers:

$18\sqrt{3}m^2$

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

20m^2

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

18m^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 : How To Find The Perimeter Of Kite

Find the perimeter of the following kite:

Kite

Possible Answers:

$8+6\sqrt{10} m$

$16+\sqrt{10} m$

$10+6\sqrt{10} m$

$18+\sqrt{10} m$

$10+9\sqrt{10} m$

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 #5 : Quadrilaterals

Find the perimeter of the following kite:

Screen_shot_2014-03-01_at_9.16.34_pm

Possible Answers:

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

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

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

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

24m

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 #7 : Quadrilaterals

Find the area of the following trapezoid:

Trapezoid

Possible Answers:

260m^2

202m^2

180m^2

252m^2

126m^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 #8 : Quadrilaterals

Find the area of the following trapezoid:

Trapezoid_angles

Possible Answers:

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

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

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

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

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

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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Example Question #9 : Quadrilaterals

Determine the area of the following trapezoid:

Screen_shot_2014-02-27_at_6.39.24_pm

Possible Answers:

25m^2

24m^2

28m^2

20m^2

35m^2

Correct answer:

24m^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 another base, and is the length of the height.

Plugging in our values, we get:

$A=\frac{1}{2}(7m+5m)(4m)=24m^2$

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Example Question #10 : Quadrilaterals

Find the area of the following trapezoid:

Screen_shot_2014-02-27_at_6.47.22_pm

Possible Answers:

$52\sqrt{101}m^2$

$48\sqrt{81}m^2$

$48\sqrt{101}m^2$

$48\sqrt{91}m^2$

$52\sqrt{91}m^2$

Correct answer:

$48\sqrt{91}m^2$

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 another base, and is the length of the height.

Use the Pythagorean Theorem to find the height of the trapezoid:

A^2+B^2=C^2

6m^2+B^2=20m^2

$B=2\sqrt{91}m$

Plugging in our values, we get:

$A=\frac{1}{2}(18m+30m)(2\sqrt{91}m)=48\sqrt{91}m^2$

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High School Math : Quadrilaterals

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Quadrilaterals

Example Question #2 : Quadrilaterals

Example Question #3 : Quadrilaterals

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

Example Question #5 : Quadrilaterals

Example Question #1 : Trapezoids

Example Question #7 : Quadrilaterals

Example Question #8 : Quadrilaterals

Example Question #9 : Quadrilaterals

Example Question #10 : Quadrilaterals

All High School Math Resources

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