How to find the area of an acute / obtuse triangle - Intermediate Geometry

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Question

In ΔABC: a = 8, b = 13, c = 9.

Find the area of ΔABC (to the nearest tenth).

Answer

In order to determine the area of a non-right triangle, we can use Heron's formula:

$Area =\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$

Using the information from the question, we obtain:
$s=\frac{8+13+9}{2}=15$
$Area =\sqrt{15(15-8)(15-13)(15-9)}=35.5$

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Question

In ΔABC: a = 16, b = 11, c = 19.

Find the area of ΔABC (to the nearest tenth).

Answer

In order to determine the area of a non-right triangle, we can use Heron's formula:

$Area =\sqrt{s(s-a)(s-b)(s-c)}$
$s=\frac{a+b+c}{2}$

Using the information from the question, we obtain:
$s=\frac{16+11+19}{2}=23$
$Area =\sqrt{23(23-16)(23-11)(23-19)}=87.9$

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Question

Find the height of a triangle if its base is long and its area is .

Answer

The formula to find the area of a triangle is

$\text{Area}=\frac{\text{base }\times\text{height}}{2}$

Substitute in the given values for area and base to solve for the height, :

$24=\frac{12h}{2}$

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Question

In terms of , what is the area of a triangle with a height of and a base of ?

Answer

The formula to find the area of a triangle is

$\text{Area}=\frac{base\times height}{2}$

Substitute in the given values for the base and the height to find the area.

$\text{Area}=\frac{5a\times 2a}{2}=\frac{10a^2}{2}=5a^2:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side, the triangle's height, and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(33^{\circ})=\frac{height}{11}$

Now, solve for the height.

$\text{Height}=11\cos(33^{\circ})=9.23$

Now you can find the area of the triangle:

$\text{Area}=\frac{45\times9.23}{2}=207.7:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side, the triangle's height, and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(24^{\circ})=\frac{height}{34}$

Now, solve for the height.

$\text{Height}=11\cos(24^{\circ})=10.05$

Now you can find the area.

$\text{Area}=\frac{45\times 10.05}{2}=226.1:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta =\frac{adjacent}{hypotenuse}$

$\cos(20^{\circ})=\frac{height}{5}$

Now, solve for the height.

$\text{Height}=5\cos(20^{\circ})=4.70$

Now you can find the area.

$\text{Area}=\frac{8\times 4.70}{2}=18.8:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(18^{\circ})=\frac{height}{3}$

Now, solve for the height.

$\text{Height}=3\cos(18^{\circ})=2.85$

Now you can find the area.

$\text{Area}=\frac{7\times 2.85}{2}=10.0:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{cosine}$ , because we are looking for the length of the adjacent side—the triangle's height—and we know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{cosine }\theta=\frac{adjacent}{hypotenuse}$

$\cos(24^{\circ})=\frac{height}{14}$

Now, solve for the height.

$\text{Height}=14\cos(24^{\circ})=12.79$

Now you can find the area.

$\text{Area}=\frac{24\times 12.79}{2}=153.5:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(55^{\circ})=\frac{height}{12}$

Now, solve for the height.

$\text{Height}=12\sin(55^{\circ})=9.83$

Now you can find the area.

$\text{Area}=\frac{14\times 9.83}{2}=68.8:units^2$

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Question

Find the area of the triangle. Round to the nearest tenths.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(41^{\circ})=\frac{height}{9}$

Now, solve for the height.

$\text{Height}=9\sin(41^{\circ})=5.90$

Now you can find the area.

$\text{Area}=\frac{12\times 5.90}{2}=35.4:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(36^{\circ})=\frac{height}{4}$

Now, solve for the height.

$\text{Height}=4\sin(36^{\circ})=2.35$

Now you can find the area.

$\text{Area}=\frac{14\times 2.35}{2}=16.5:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(42^{\circ})=\frac{height}{8}$

Now, solve for the height.

$\text{Height}=8\sin(42^{\circ})=5.35$

Now you can find the area.

$\text{Area}=\frac{30\times 5.35}{2}=80.3:units^2$

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Question

Find the area of the triangle below. Round to the nearest tenths place.

Answer

The formula used to find the area of the triangle is

$\text{Area}=\frac{base\times height}{2}$

Now, we will need to use a trigonometric ratio to find the length of the height. Because of the angle given, we will need to use $\text{sine}$ , because we are looking for the height of the triangle, which in this case is the side opposite to the known angle, and we also know the length of the hypotenuse of the smaller triangle formed by the height.

Remember that $\text{sin }\theta=\frac{opposite}{hypotenuse}$

$\sin(48^{\circ})=\frac{height}{12}$

Now, solve for the height.

$\text{Height}=12\sin(48^{\circ})=8.92$

Now you can find the area.

$\text{Area}=\frac{26\times 8.92}{2}=116.0:units^2$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=8 \times 12$

$\text{Area of Rectangle}= 96$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{8}{2}$

$\text{height}=4$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 4$

$\text{Area of Triangle}=24$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=96-24$

Solve.

$\text{Area of Shaded Region}=72$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=6 \times 12$

$\text{Area of Rectangle}= 72$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{6}{2}$

$\text{height}=3$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 12\times 3$

$\text{Area of Triangle}=18$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=72-18$

Solve.

$\text{Area of Shaded Region}=54$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=14 \times 16$

$\text{Area of Rectangle}= 224$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{14}{2}$

$\text{height}=7$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 16\times 7$

$\text{Area of Triangle}=56$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=224-56$

Solve.

$\text{Area of Shaded Region}=168$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=1 \times 4$

$\text{Area of Rectangle}= 4$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{1}{2}$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 4\times \frac{1}{2}$

$\text{Area of Triangle}=1$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=4-1$

Solve.

$\text{Area of Shaded Region}=3$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=2 \times 7$

$\text{Area of Rectangle}= 14$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{2}{2}$

$\text{height}=1$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 7\times 1$

$\text{Area of Triangle}=\frac{7}{2}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=14-\frac{7}{2}$

Solve.

$\text{Area of Shaded Region}=\frac{21}{2}$

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Question

In the figure, a triangle that shares its base with the width of the rectangle has a height that is half the length of the rectangle. Find the area of the shaded region.

Answer

In order to find the area of the shaded region, we must first find the areas of the triangle and the rectangle.

Recall how to find the area of a rectangle:

$\text{Area of Rectangle}=\text{length}\times\text{width}$

Substitute in the given length and width to find the area.

$\text{Area of Rectangle}=2 \times 5$

$\text{Area of Rectangle}= 10$

Next, recall how to find the area of a triangle:

$\text{Area of Triangle}=\frac{1}{2}\times\text{base}\times\text{height}$

From the question, we know that the height of the triangle is half the length of the rectangle. Use the length of the rectangle to find the height of the triangle:

$\text{height}=\frac{\text{length}}{2}$

$\text{height}=\frac{2}{2}$

$\text{height}=1$

Since the base of the triangle and the width of the rectangle are the same, we can find the area of the triangle:

$\text{Area of Triangle}=\frac{1}{2}\times 5\times 1$

$\text{Area of Triangle}=\frac{5}{2}$

To find the area of the shaded region, subtract the area of the triangle from the area of the rectangle.

$\text{Area of Shaded Region}=\text{Area of Rectangle}-\text{Area of Triangle}$

$\text{Area of Shaded Region}=10-\frac{5}{2}$

Solve.

$\text{Area of Shaded Region}=\frac{15}{2}$

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