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Basic Geometry : 45/45/90 Right Isosceles Triangles

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Example Questions

Example Question #31 : Triangles

A right isosceles triangle has leg lengths of $\sqrt5$ . What is the length of the hypotenuse?

Possible Answers:

$\sqrt7$

$\sqrt3$

$5\sqrt2$

$\sqrt{10}$

Correct answer:

$\sqrt{10}$

Explanation:

Recall the Pythagorean Theorem:

a^2+b^2=c^2

Since we know that this is an isosceles right triangle, we know the following:

a=b

The Pythagorean Theorem can then be simplifed to the following equation:

a^2+a^2=2a^2=c^2

Now, solve for since the question asks for the length of the hypotenuse.

$c=\sqrt{2a^2}$

$c=a\sqrt2$

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

$c=(\sqrt5)(\sqrt2)$

Simplify.

$c=\sqrt{10}$

Report an Error

Example Question #32 : Triangles

A right isosceles triangle has leg lengths of $2\sqrt{2}$ . What is the length of the hypotenuse?

Possible Answers:

$4\sqrt2$

Correct answer:

Explanation:

Recall the Pythagorean Theorem:

a^2+b^2=c^2

Since we know that this is an isosceles right triangle, we know the following:

a=b

The Pythagorean Theorem can then be simplifed to the following equation:

a^2+a^2=2a^2=c^2

Now, solve for since the question asks for the length of the hypotenuse.

$c=\sqrt{2a^2}$

$c=a\sqrt2$

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

$c=(2\sqrt2)(\sqrt2)$

Simplify.

c=4

Report an Error

Example Question #33 : Triangles

A right isosceles triangle has leg lengths of $3\sqrt2$ . What is the length of the hypotenuse?

Possible Answers:

$2\sqrt6$

$6\sqrt2$

Correct answer:

Explanation:

Recall the Pythagorean Theorem:

a^2+b^2=c^2

Since we know that this is an isosceles right triangle, we know the following:

a=b

The Pythagorean Theorem can then be simplifed to the following equation:

a^2+a^2=2a^2=c^2

Now, solve for since the question asks for the length of the hypotenuse.

$c=\sqrt{2a^2}$

$c=a\sqrt2$

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

$c=(3\sqrt2)(\sqrt2)$

Simplify.

c=6

Report an Error

Example Question #31 : 45/45/90 Right Isosceles Triangles

If the width of the rectangle is half of the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

37.06

39.07

38.06

40.09

Correct answer:

38.06

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{12^2+12^2}$

$\text{Hypotenuse}=12\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{12\sqrt2}{2}$

$\text{Width}=6\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=4\times6\sqrt2$

$\text{Area of Rectangle}=24\sqrt2$

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

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

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

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

$\text{Area of Triangle}=72$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=72-24\sqrt2$

Solve.

$\text{Area of Shaded Region}=38.06$

Report an Error

Example Question #1012 : Basic Geometry

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

0.44

0.36

0.59

0.91

Correct answer:

0.59

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

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

$\text{Hypotenuse}=2\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{2\sqrt2}{2}$

$\text{Width}=\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=1\times\sqrt2$

$\text{Area of Rectangle}=\sqrt2$

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

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

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

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

$\text{Area of Triangle}=2$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=2-\sqrt2$

Solve.

$\text{Area of Shaded Region}=0.59$

Report an Error

Example Question #36 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

2.48

2.71

2.34

2.57

Correct answer:

2.34

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{4^2+4^2}$

$\text{Hypotenuse}=4\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{4\sqrt2}{2}$

$\text{Width}=2\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=2\times2\sqrt2$

$\text{Area of Rectangle}=4\sqrt2$

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

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

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

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

$\text{Area of Triangle}=8$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=8-4\sqrt2$

Solve.

$\text{Area of Shaded Region}=2.34$

Report an Error

Example Question #37 : Triangles

If the wiidth of the rectangle is half the length of the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

2.69

2.38

2.10

2.49

Correct answer:

2.38

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{3^2+3^2}$

$\text{Hypotenuse}=3\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{3\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=1\times \frac{3\sqrt2}{2}$

$\text{Area of Rectangle}=\frac{3\sqrt2}{2}$

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

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

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

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

$\text{Area of Triangle}=\frac{9}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{9}{2}-\frac{3\sqrt2}{2}$

Solve.

$\text{Area of Shaded Region}=2.38$

Report an Error

Example Question #38 : Triangles

If the width of the rectangle is half the length of the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

5.29

5.85

5.43

5.10

Correct answer:

5.43

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{5^2+5^2}$

$\text{Hypotenuse}=5\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{5\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=2\times\frac{5\sqrt2}{2}$

$\text{Area of Rectangle}=5\sqrt2$

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

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

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

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

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

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{25}{2}-5\sqrt2$

Solve.

$\text{Area of Shaded Region}=5.43$

Report an Error

Example Question #39 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

9.57

9.51

9.92

9.22

Correct answer:

9.51

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{6^2+6^2}=6\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{6\sqrt2}{2}$

$\text{Width}=3\sqrt2$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=2\times3\sqrt2$

$\text{Area of Rectangle}=6\sqrt2$

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

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

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

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

$\text{Area of Triangle}=18$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=18-6\sqrt2$

Solve.

$\text{Area of Shaded Region}=9.51$

Report an Error

Example Question #40 : Triangles

If the width of the rectangle is half the hypotenuse of the triangle, then what is the area of the shaded region?

Possible Answers:

9.29

9.34

9.18

9.65

Correct answer:

9.65

Explanation:

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

First, let's recall how to find the area of a rectangle.

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

Now, the question tells us the following relationship between the width of the rectangle and the hypotenuse of the triangle:

$\text{Width}=\frac{\text{Hypotenuse}}{2}$

Now, let's use the Pythagorean theorem to find the length of the hypotenuse.

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

$\text{Hypotenuse}=\sqrt{\text{base}^2+\text{height}^2}$

Substitute in the values of the base and of the height to find the hypotenuse.

$\text{Hypotenuse}=\sqrt{7^2+7^2}$

$\text{Hypotenuse}=7\sqrt2$

Now, substitute this value in to find the width of the rectangle.

$\text{Width}=\frac{7\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=3\times\frac{7\sqrt2}{2}$

$\text{Area of Rectangle}=\frac{21\sqrt2}{2}$

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

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

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

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

$\text{Area of Triangle}=\frac{49}{2}$

Finally, we are ready to find the area of the shaded region.

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

$\text{Area of Shaded Region}=\frac{49}{2}-\frac{21\sqrt2}{2}$

Solve.

$\text{Area of Shaded Region}=9.65$

Report an Error

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Basic Geometry : 45/45/90 Right Isosceles Triangles

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #31 : Triangles

Example Question #32 : Triangles

Example Question #33 : Triangles

Example Question #31 : 45/45/90 Right Isosceles Triangles

Example Question #1012 : Basic Geometry

Example Question #36 : Triangles

Example Question #37 : Triangles

Example Question #38 : Triangles

Example Question #39 : Triangles

Example Question #40 : Triangles

All Basic Geometry Resources

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