Area of a Triangle - GED Math

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Question

Note: Figure NOT drawn to scale.

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

Answer

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:

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

$10 N = 20 \cdot 20$

$N = 400 \div 10 = 40$

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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Question

Note: Figure NOT drawn to scale.

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

Answer

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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Question

The above figure shows Square ; is the midpoint of $\overline{SU}$ ; is the midpoint of $\overline{RS}$ . What percent of the square is shaded?

Answer

The answer is independent of the sidelengths of the rectangle, so to ease calculations, we will arbitrarily assign to Square sidelength 4 inches.

The shaded region can be divided by a perpendicular segment from to $\overline{AR}$ , then another from to that segment, as follows:

By the fact that rectangles have opposite sides that are congruent, and by the fact that is the midpoint of $\overline{SU}$ and is the midpoint of $\overline{RS}$ , the shaded region is a composite of:

Square , which has sides of length 2 and therefore area $2^{2} = 4$ ;

Right triangle $\Delta QYE$ , which has two legs of length 2 and, thus, area $\frac{1}{2} \times 2 \times 2 = 2$ ; and,

Right triangle $\Delta QXA$ , which has legs of length 4 and 2 and, thus, area $\frac{1}{2} \times 4 \times 2 = 4$ .

The area of the shaded region is therefore .

Square has area $4^{2} = 16$ , so the shaded region is

$\frac{10}{16} \times 100 % = 62 \frac{1}{2} %$

of the square.

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Question

Give the area of the above triangle.

Answer

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:

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

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

Answer

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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Question

Find the area of a triangle with a base of and a height of .

Answer

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:

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Question

Find the area of a triangle with a base of 14in and a height of 9in.

Answer

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 14in and the height is 9in. So, we can substitute. We get

$A = \frac{1}{2} \cdot 14\text{in} \cdot 9\text{in}$

$A = 7\text{in} \cdot 9\text{in}$

$A = 63\text{in}^2$

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Question

Find the area of a triangle with a base of and a height of .

Answer

Write the formula for a triangle.

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

Substitute the dimensions and find the area.

$A = \frac{(12)(3)}{2} = 6(3) = 18$

The answer is:

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Question

Find the area of a triangle with a base length of 6 and a height of $\frac{1}{4}$ .

Answer

Write the area for a triangle.

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

Substitute the base and height into the equation.

$A = \frac{6(\frac{1}{4})}{2} = 3(\frac{1}{4}) = \frac{3}{4}$

The answer is: $\frac{3}{4}$

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Question

Determine the area of the triangle if the base is 12 and the height is 20.

Answer

Write the formula for the area of a triangle.

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

Substitute the base and height into the equation.

$A=\frac{(12)(20)}{2} = 6(20) =120$

The answer is:

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Question

Find the area of the triangle with a base of 6 and a height of 30.

Answer

Write the formula for the area of a triangle.

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

Substitute the base and height into the formula.

$A= \frac{6(30)}{2} = \frac{180}{2} = 90$

The answer is:

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Question

What is the area of a triangle with a length of and a height of ?

Answer

Write the formula for the area of a triangle.

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

Substitute the dimensions.

$A =\frac{(2x^2)(3x^2-4)}{2} = (x^2)(3x^2-4)$

Expand the terms.

The answer is:

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Question

Determine the area of a triangle with a base of 16 and a height of 12.

Answer

Write the formula for the area of a triangle.

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

Substitute the base and height into the equation.

$A= \frac{16(12)}{2} = 8(12) = 96$

The answer is:

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Question

Find the area of a triangle with a base of 7in and a height of 10in.

Answer

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

$A = \frac{1}{2} \cdot 7\text{in} \cdot 10\text{in}$

$A = \frac{1}{2} \cdot 10\text{in} \cdot 7\text{in}$

$A = 5\text{in} \cdot 7\text{in}$

$A = 35\text{in}^2$

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Question

Find the area of a triangle with a base of 14in and a height of 8in.

Answer

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 14in. We know the height of the triangle is 8in. So, we substitute. We get

$A = \frac{1}{2} \cdot 14\text{in} \cdot 8\text{in}$

$A = 7\text{in} \cdot 8\text{in}$

$A = 56\text{in}^2$

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