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Basic Geometry : Triangles

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9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #17 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is , what is the area of the triangle?

Possible Answers:

369

400

100

200

Correct answer:

400

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$

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

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}=\frac{\text{hypotenuse}}{\sqrt2}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{40\sqrt2}{2}=20\sqrt2$

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

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

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(20\sqrt2 \times 20\sqrt2)=400$

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Example Question #18 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is , what is the area of the triangle?

Possible Answers:

158

144

169

225

Correct answer:

169

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$

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

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}=\frac{\text{hypotenuse}}{\sqrt2}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{26\sqrt2}{2}=13\sqrt2$

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

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

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(13\sqrt2 \times 13\sqrt2)=169$

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Example Question #91 : Triangles

Find the area of the triangle if the diameter of the circle is $2\sqrt2$ .

Possible Answers:

$\sqrt{3}$

$\sqrt2$

Correct answer:

Explanation:

Notice that the given triangle is a right isosceles triangle. The hypotenuse of the triangle is the same as the diameter of the circle; therefore, we can use the Pythagorean theorem to find the length of the legs of this triangle.

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{2\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=2$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

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

Solve.

$\text{Area}=2$

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Example Question #91 : Triangles

Find the area of the triangle if the diameter of the circle is $12\sqrt2$ .

Possible Answers:

$\sqrt{71}$

144

$\sqrt{111}$

Correct answer:

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{12\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=12$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

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

Solve.

$\text{Area}=72$

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Example Question #91 : Triangles

Find the area of the triangle if the diameter of the circle is $4\sqrt2$ .

Possible Answers:

$\sqrt{7}$

Correct answer:

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{4\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=4$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{4\times 4}{2}$

Solve.

$\text{Area}=8$

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Example Question #21 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of the triangle if the diameter of the circle is $50\sqrt2$ .

Possible Answers:

1420

1265

1150

1250

1520

Correct answer:

1250

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{50\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=50$

Now, recall how to find the area of a triangle.

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

Since we have a right isocseles triangle, the base and the height are the same length.

$\text{Area}=\frac{50\times 50}{2}$

Solve.

$\text{Area}=1250$

Report an Error

Example Question #93 : Triangles

Find the area of the triangle if the diameter of the circle is $5\sqrt2$ .

Possible Answers:

$5\sqrt{3}$

$5\sqrt2$

$\frac{25}{2}$

Correct answer:

$\frac{25}{2}$

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{5\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=5$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{5\times 5}{2}$

Solve.

$\text{Area}=\frac{25}{2}$

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Example Question #94 : Triangles

Find the area of the triangle if the diameter of the circle is $6\sqrt2$ .

Possible Answers:

$9\sqrt{3}$

Correct answer:

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{6\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=6$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

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

Solve.

$\text{Area}=18$

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Example Question #95 : Triangles

Find the area of the triangle if the diameter of the circle is $8\sqrt2$ .

Possible Answers:

$8\sqrt{3}$

128

Correct answer:

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{8\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=8$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

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

Solve.

$\text{Area}=32$

Report an Error

Example Question #96 : Triangles

Find the area of the triangle if the diameter of the circle is $20\sqrt2$ .

Possible Answers:

$2\sqrt{211}$

800

200

100

400

Correct answer:

200

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{20\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=20$

Now, recall how to find the area of a triangle.

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{20\times 20}{2}$

Solve.

$\text{Area}=200$

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All Basic Geometry Resources

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Basic Geometry : Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #17 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #18 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #91 : Triangles

Example Question #91 : Triangles

Example Question #91 : Triangles

Example Question #21 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #93 : Triangles

Example Question #94 : Triangles

Example Question #95 : Triangles

Example Question #96 : Triangles

All Basic Geometry Resources

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