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Basic Geometry : Right Triangles

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Example Questions

Example Question #1442 : Plane Geometry

Find the area.

Possible Answers:

$27\sqrt3$

$40\sqrt3$

$36\sqrt3$

$18\sqrt3$

Correct answer:

$18\sqrt3$

Explanation:

Recall how to find the area of a triangle:

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

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{12^2-6^2}$

Simplify.

$\text{base}=\sqrt{108}$

Reduce.

$\text{base}=6\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{6 \times 6\sqrt3}{2}$

Solve.

$\text{Area}=18\sqrt3$

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Example Question #1441 : Basic Geometry

Find the area.

Possible Answers:

$14\sqrt3$

$49\sqrt3$

$56\sqrt3$

$\frac{49\sqrt3}{2}$

Correct answer:

$\frac{49\sqrt3}{2}$

Explanation:

Recall how to find the area of a triangle:

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

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{14^2-7^2}$

Simplify.

$\text{base}=\sqrt{147}$

Reduce.

$\text{base}=7\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{7 \times 7\sqrt3}{2}$

Solve.

$\text{Area}=\frac{49\sqrt3}{2}$

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Example Question #1448 : Basic Geometry

Find the area.

Possible Answers:

$25\sqrt3$

$\frac{75\sqrt3}{2}$

$50\sqrt3$

$100\sqrt3$

Correct answer:

$50\sqrt3$

Explanation:

Recall how to find the area of a triangle:

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

Now, we have the height and the hypotenuse from the question. Use the Pythagorean Theorem to find the length of the base.

$\text{hypotenuse}^2=\text{base}^2+\text{height}^2$

$\text{base}^2=\text{hypotenuse}^2-\text{height}^2$

$\text{base}=\sqrt{\text{hypotenuse}^2-\text{height}^2}$

Substitute in the values of the height and hypotenuse.

$\text{base}=\sqrt{20^2-10^2}$

Simplify.

$\text{base}=\sqrt{300}$

Reduce.

$\text{base}=10\sqrt3$

Now, substitute in the values of the base and the height to find the area.

$\text{Area}=\frac{10 \times 10\sqrt3}{2}$

Solve.

$\text{Area}=50\sqrt3$

Report an Error

Example Question #1451 : Basic Geometry

Given that the two legas of a right triangle have lengths of 9ft and 13ft , find the area.

Possible Answers:

52.8 ft

65.7 ft

50.3 ft

58.5 ft

60.1 ft

Correct answer:

58.5 ft

Explanation:

To find the area of a triangle, the formula is Area = 1/2 (Base) (Height) . By plugging in the information given, we get:

$Area = \frac{1}{2} (9) (13)$

Area = 58.5 ft

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Example Question #61 : How To Find The Area Of A Right Triangle

Find the area of a right triangle whose side lengths are 5 and 4.

Possible Answers:

Correct answer:

Explanation:

To solve, simply use the formula for the area of a triangle. Thus,

$A=\frac{1}{2}Bh=\frac{1}{2}*5*4=10$

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Example Question #1453 : Basic Geometry

What is the area of a right triangle that has two side lengths of 7 and 8 inches respectively and the third side is the hypotenuse.

Possible Answers:

56in^2

34in^2

28in^2

72in^2

15in^2

Correct answer:

28in^2

Explanation:

The formula for the area of a triangle is, $\frac{1}{2}*base * height$ .

In the case of a right triangle you can use the two side lengths other than the hypotenuse to solve the equation.

$\frac{1}{2}* 8 * 7 =$

4 * 7 =

28in^2

Area is in units squared.

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Example Question #271 : Right Triangles

Find the area of a right triangle with base 4 and height 5.

Possible Answers:

Correct answer:

Explanation:

To solve, simply use the formula for the area of a triangle. Thus,

$A=\frac{1}{2}Bh=\frac{1}{2}*4*5=10$

If the formula escapes you, simply remember that two equivalent triangles put together equal a rectangle. So, the area of a triangle must be half the area of a rectangle.

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Example Question #472 : Triangles

The diameter of the circle is , find the area of the shaded region.

Possible Answers:

19.21

21.28

20.09

20.86

Correct answer:

20.86

Explanation:

To find the area of the shaded region, we will first need to find the area of the right triangle and the area of the circle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Now, recall how to find the length of the radius from the length of the diameter.

$\text{Radius}=\frac{\text{Diameter}}{2}$

Substitute in the given diameter to find the radius.

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

$\text{Radius}=1$

Now, substitute in the radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 1^2$

$\text{Area of Circle}=\pi$

Next, recall how to find the area of a right triangle.

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

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

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

$\text{Area of Triangle}=24$

We can now find the area of the shaded region:

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

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

Solve and round to two decimal places.

$\text{Area of Shaded Region}=20.86$

Report an Error

Example Question #473 : Triangles

The diameter of the circle is , find the area of the shaded region.

Possible Answers:

13.04

11.82

12.48

12.93

Correct answer:

12.93

Explanation:

To find the area of the shaded region, we will first need to find the area of the right triangle and the area of the circle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Now, recall how to find the length of the radius from the length of the diameter.

$\text{Radius}=\frac{\text{Diameter}}{2}$

Substitute in the given diameter to find the radius.

$\text{Radius}=\frac{3}{2}$

Now, substitute in the radius to find the area of the circle.

$\text{Area of Circle}=\pi\times\left( \frac{3}{2}\right)^2$

$\text{Area of Circle}=\frac{9}{4}\pi$

Next, recall how to find the area of a right triangle.

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

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

$\text{Area of Triangle}=\frac{4 \times 10}{2}$

$\text{Area of Triangle}=20$

We can now find the area of the shaded region:

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

$\text{Area of Shaded Region}=20-\frac{9}{4}\pi$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=12.93$

Report an Error

Example Question #474 : Triangles

The diameter of the circle is 1.2 , find the area of the shaded region.

Possible Answers:

14.90

12.13

13.87

10.71

Correct answer:

13.87

Explanation:

To find the area of the shaded region, we will first need to find the area of the right triangle and the area of the circle.

Recall how to find the area of a circle:

$\text{Area of Circle}=\pi\times\text{radius}^2$

Now, recall how to find the length of the radius from the length of the diameter.

$\text{Radius}=\frac{\text{Diameter}}{2}$

Substitute in the given diameter to find the radius.

$\text{Radius}=\frac{1.2}{2}$

$\text{Radius}=0.6$

Now, substitute in the radius to find the area of the circle.

$\text{Area of Circle}=\pi\times 0.6^2$

$\text{Area of Circle}=0.36\pi$

Next, recall how to find the area of a right triangle.

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

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

$\text{Area of Triangle}=\frac{2 \times 15}{2}$

$\text{Area of Triangle}=15$

We can now find the area of the shaded region:

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

$\text{Area of Shaded Region}=15-0.36\pi$

Solve and round to two decimal places.

$\text{Area of Shaded Region}=13.87$

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Basic Geometry : Right Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1442 : Plane Geometry

Example Question #1441 : Basic Geometry

Example Question #1448 : Basic Geometry

Example Question #1451 : Basic Geometry

Example Question #61 : How To Find The Area Of A Right Triangle

Example Question #1453 : Basic Geometry

Example Question #271 : Right Triangles

Example Question #472 : Triangles

Example Question #473 : Triangles

Example Question #474 : Triangles

All Basic Geometry Resources

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