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Basic Geometry : Triangles

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Example Questions

Example Question #1442 : Basic Geometry

Find the area.

Possible Answers:

$64\sqrt3$

$48\sqrt3$

$60\sqrt3$

$32\sqrt3$

Correct answer:

$32\sqrt3$

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.

$\text{base}=\sqrt{16^2-8^2}$

Simplify.

$\text{base}=\sqrt{192}$

Reduce.

$\text{base}=8\sqrt3$

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

$\text{Area}=\frac{8 \times 8\sqrt3}{2}$

Solve.

$\text{Area}=32\sqrt3$

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Example Question #1443 : Basic Geometry

Find the area.

Possible Answers:

$\frac{81\sqrt3}{2}$

$99\sqrt3$

$\frac{117\sqrt3}{2}$

$81\sqrt3$

Correct answer:

$\frac{81\sqrt3}{2}$

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.

$\text{base}=\sqrt{18^2-9^2}$

Simplify.

$\text{base}=\sqrt{243}$

Reduce.

$\text{base}=9\sqrt3$

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

$\text{Area}=\frac{9 \times 9\sqrt3}{2}$

Solve.

$\text{Area}=\frac{81\sqrt3}{2}$

Report an Error

Example Question #1441 : Plane Geometry

Find the area.

Possible Answers:

$110\sqrt3$

$\frac{121\sqrt3}{2}$

$121\sqrt3$

$243\sqrt3$

Correct answer:

$\frac{121\sqrt3}{2}$

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.

$\text{base}=\sqrt{22^2-11^2}$

Simplify.

$\text{base}=\sqrt{363}$

Reduce.

$\text{base}=11\sqrt3$

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

$\text{Area}=\frac{11 \times 11\sqrt3}{2}$

Solve.

$\text{Area}=\frac{121\sqrt3}{2}$

Report an Error

Example Question #1445 : Basic Geometry

Find the area.

Possible Answers:

144

$72\sqrt3$

$144\sqrt3$

$36\sqrt3$

Correct answer:

$72\sqrt3$

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.

$\text{base}=\sqrt{24^2-12^2}$

Simplify.

$\text{base}=\sqrt{432}$

Reduce.

$\text{base}=12\sqrt3$

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

$\text{Area}=\frac{12 \times 12\sqrt3}{2}$

Solve.

$\text{Area}=72\sqrt3$

Report an Error

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.

$\text{Area}=18\sqrt3$

Report an Error

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.

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

Report an Error

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 #461 : Triangles

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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Basic Geometry : Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1442 : Basic Geometry

Example Question #1443 : Basic Geometry

Example Question #1441 : Plane Geometry

Example Question #1445 : Basic Geometry

Example Question #1442 : Plane Geometry

Example Question #1441 : Basic Geometry

Example Question #1448 : Basic Geometry

Example Question #1451 : Basic Geometry

Example Question #461 : Triangles

Example Question #1453 : Basic Geometry

All Basic Geometry Resources

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