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4 Diagnostic Tests 263 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Area Of A Triangle

Triangle

Give the area of the above triangle.

Possible Answers:

$36 \sqrt{2}$

$36 \sqrt{3}$

Correct answer:

$36 \sqrt{3}$

Explanation:

This triangle is isosceles by the converse of the Isosceles Triangle Theorem. The altitude of the triangle divides it into two triangles that are both right and that are congruent, as follows:

Triangle

Each triangle is a $30 ^{\circ } - 60 ^{\circ } -90^{\circ }$ triangle. The altitude is the short leg of each triangle, and, therefore, its length is half that of the common hypotenuse, or 6. The long leg of each right triangle has length $\sqrt{3}$ times that of the short leg, or $6 \sqrt{3}$ , and the base of the entire large triangle is twice this, or $12 \sqrt{3}$ .

The area of the triangle is half the product of the height and the base:

$A = \frac{1}{2} bh =\frac{1}{2} \cdot 6 \cdot 12 \sqrt{3} = 36 \sqrt{3}$

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Example Question #2 : Area Of A Triangle

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. In terms of area, $\Delta BDC$ is what percent of $\Delta ABC$ ?

Possible Answers:

$80 \%$

$64 \%$

$75 \%$

$87 \frac{1}{2} \%$

Correct answer:

$80 \%$

Explanation:

The area of a triangle is half the product of its baselength and its height.

To find the area of $\Delta BDC$ , we can use the lengths of the legs $\overline{CD}$ and $\overline{BD}$ :

$A = \frac{1}{2} (CD)(BD ) = \frac{1}{2} \cdot 24 \times 12 = 144$

To find the area of $\Delta ABC$ , we can use the hypotenuse $\overline{AC}$ , the length of which is 30, and the altitude $\overline{BD}$ perpendicular to it:

$A = \frac{1}{2} (AC)(BD ) = \frac{1}{2} \cdot 30 \times 12 = 180$

In terms of area, $\Delta BDC$ is

$\frac{144}{180} \times 100 \% = 80 \%$

of $\Delta ABC$ .

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Example Question #2 : Area Of A Triangle

Triangle

Note: Figure NOT drawn to scale.

Refer to the figure above. Give the area of the blue triangle.

Possible Answers:

400

600

1,200

800

Correct answer:

400

Explanation:

The inscribed rectangle is a 20 by 20 square. Since opposite sides of the square are parallel, the corresponding angles of the two smaller right triangles are congruent; therefore, the two triangles are similar and, by definition, their sides are in proportion.

The small top triangle has legs 10 and 20; the blue triangle has legs 20 and , where can be calculated with the following proportion:

$\frac{N}{20} = \frac{20}{10 }$

$10 N = 20 \cdot 20$

10 N = 400

$N = 400 \div 10 = 40$

The legs of the blue triangle are 20 and 40; half their product is the area:

$\frac{1}{2} \times 20 \times 40 = 400$

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Example Question #4 : Area Of A Triangle

Find the area of a triangle with a base of 7in and a height that is two times the base.

Possible Answers:

$49\text{in}^2$

$36\text{in}^2$

$28\text{in}^2$

$21\text{in}^2$

$98\text{in}^2$

Correct answer:

$49\text{in}^2$

Explanation:

To find the area of a triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

Now, we know the base of the triangle is 7in. We also know the height of the triangle is two times the base. Therefore, the height is 14in. So, we can substitute. We get

$A = \frac{1}{2} \cdot 7\text{in} \cdot 14\text{in}$

$A = \frac{1}{2} \cdot 14\text{in} \cdot 7\text{in}$

$A = 7\text{in} \cdot 7\text{in}$

$A = 49\text{in}^2$

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Example Question #31 : Triangles

Find the area of a triangle with a base of 8in and a height that is half the base.

Possible Answers:

$36\text{in}^2$

$16\text{in}^2$

$32\text{in}^2$

$24\text{in}^2$

$12\text{in}^2$

Correct answer:

$16\text{in}^2$

Explanation:

To find the area of a triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

Now, we know the base of the triangle is 8in. We also know the height of the triangle is half the base. Therefore, the height is 4in. So, we can substitute. We get

$A = \frac{1}{2} \cdot 8\text{in} \cdot 4\text{in}$

$A = 4\text{in} \cdot4\text{in}$

$A = 16\text{in}^2$

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Example Question #3 : Area Of A Triangle

What is the area of a triangle with a base of $\frac{1}{2}$ and a height of $\frac{2}{3}$ ?

Possible Answers:

$\frac{1}{6}$

$\frac{1}{2}$

$\frac{1}{12}$

$\frac{1}{9}$

$\frac{1}{3}$

Correct answer:

$\frac{1}{6}$

Explanation:

Write the formula for the area of a triangle.

$A = \frac{1}{2}bh$

Substitute the base and height.

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

Simplify the fractions.

The answer is: $\frac{1}{6}$

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Example Question #5 : Area Of A Triangle

Find the area of a triangle with a base of 6cm and a height that is three times the base.

Possible Answers:

$36\text{cm}^2$

$108\text{cm}^2$

$54\text{cm}^2$

$48\text{cm}^2$

$18\text{cm}^2$

Correct answer:

$54\text{cm}^2$

Explanation:

To find the area of a triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

Now, we know the base of the triangle is 6cm. We also know the height is three times the base. Therefore, the height is 18cm. So, we substitute. We get

$A = \frac{1}{2} \cdot 6\text{cm} \cdot 18\text{cm}$

$A = 3\text{cm} \cdot 18\text{cm}$

$A = 54\text{cm}^2$

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Example Question #3 : Area Of A Triangle

Find the area of a triangle with a base of 10in and a height of 9in.

Possible Answers:

$28\text{in}^2$

$40\text{in}^2$

$90\text{in}^2$

$35\text{in}^2$

$45\text{in}^2$

Correct answer:

$45\text{in}^2$

Explanation:

To find the area of a triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

Now, we know the base of the triangle is 10in. We know the height of the triangle is 9in. So, we can substitute. We get

$A = \frac{1}{2} \cdot 10\text{in} \cdot 9\text{in}$

$A = 5\text{in} \cdot 9\text{in}$

$A = 45\text{in}^2$

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Example Question #7 : Area Of A Triangle

Find the area of a triangle with a base of 8cm and a height of 12cm.

Possible Answers:

$96\text{cm}^2$

$32\text{cm}^2$

$24\text{cm}^2$

$20\text{cm}^2$

$48\text{cm}^2$

Correct answer:

$48\text{cm}^2$

Explanation:

To find the area of a triangle, we will use the following formula:

$A = \frac{1}{2} \cdot b \cdot h$

where b is the base and h is the height of the triangle.

We know the base of the triangle is 8cm.

We know the height of the triangle is 12cm.

Now, we can substitute. We get

$A = \frac{1}{2} \cdot 8\text{cm} \cdot 12\text{cm}$

$A = 4\text{cm} \cdot 12\text{cm}$

$A = 48\text{cm}^2$

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Example Question #1172 : Ged Math

Find the area of a triangle with a base of 16x^2 and a height of .

Possible Answers:

64x^3

16x^2+4x

64x^5

32x^3

8x^2+2x

Correct answer:

32x^3

Explanation:

Write the area for a triangle.

$A= \frac{BH}{2}$

Substitute the base and height.

$A= \frac{(16x^2)(4x)}{2} = \frac{64x^3}{2} = 32x^3$

The answer is: 32x^3

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GED Math : GED Math

Study concepts, example questions & explanations for GED Math

All GED Math Resources

Example Questions

Example Question #1 : Area Of A Triangle

Example Question #2 : Area Of A Triangle

Example Question #2 : Area Of A Triangle

Example Question #4 : Area Of A Triangle

Example Question #31 : Triangles

Example Question #3 : Area Of A Triangle

Example Question #5 : Area Of A Triangle

Example Question #3 : Area Of A Triangle

Example Question #7 : Area Of A Triangle

Example Question #1172 : Ged Math

All GED Math Resources

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