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Basic Geometry : How to find the area of a 45/45/90 right isosceles triangle

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Example Questions

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:

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.

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

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.

$\text{Area}=8$

Report an Error

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.

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

Report an Error

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.

$\text{Area}=18$

Report an Error

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.

$\text{Area}=200$

Report an Error

Example Question #101 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is $22\sqrt2$ .

Possible Answers:

144

$3\sqrt{111}$

484

121

242

Correct answer:

242

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.

$\text{side}=\frac{22\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=22$

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.

$\text{Area}=\frac{22\times 22}{2}$

Solve.

$\text{Area}=242$

Report an Error

Example Question #102 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is .

Possible Answers:

$4\sqrt{3}$

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.

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

Simplify.

$\text{side}=4\sqrt2$

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.

$\text{Area}=\frac{4\sqrt2\times 4\sqrt2}{2}$

Solve.

$\text{Area}=16$

Report an Error

Example Question #103 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the diameter of the circle is $10\sqrt2$ .

Possible Answers:

110

125

$2\sqrt{45}$

100

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.

$\text{side}=\frac{10\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=10$

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.

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

Solve.

$\text{Area}=50$

Report an Error

Example Question #104 : 45/45/90 Right Isosceles Triangles

Find the area of the triangle if the radius of the circle is .

Possible Answers:

$3\sqrt3$

Correct answer:

Explanation:

The two tick marks on the image indicate that those sides are congruent; therefore, this is an isosceles right triangle.

Notice that the hypotenuse of the triangle is also the diameter of the circle. Recall the relationship between the diameter of a circle and its radius:

$\text{Hypotenuse}=\text{Diameter}=2\times\text{radius}$

Substitute in the value of the radius to find the length of the diameter.

$\text{Hypotenuse}=2\times 3$

Simplify.

$\text{Hypotenuse}=6$

Now, use the Pythagorean theorem to find the lengths of the missing sides of the triangle.

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

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

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

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

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

In this case, because we have an isosceles right triangle,

$\text{base}=\text{height}=\text{side}$

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

Take the equation derived from the Pythagorean theorem and plug it in to the equation above.

$\text{Area}=\frac{\text{Hypotenuse}^2}{2}\times \frac{1}{2}$

Simplify.

$\text{Area}=\frac{\text{Hypotenuse}^2}{4}$

Now, substitute in the value of the hypotenuse to find the area of the triangle.

$\text{Area}=\frac{6^2}{4}$

Solve.

$\text{Area}=18$

Report an Error

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Basic Geometry : How to find the area of a 45/45/90 right isosceles triangle

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

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

Example Question #101 : 45/45/90 Right Isosceles Triangles

Example Question #102 : 45/45/90 Right Isosceles Triangles

Example Question #103 : 45/45/90 Right Isosceles Triangles

Example Question #104 : 45/45/90 Right Isosceles Triangles

All Basic Geometry Resources

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