45/45/90 Right Isosceles Triangles

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Math › 45/45/90 Right Isosceles Triangles

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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?

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.

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

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

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.

$\text{Hypotenuse}=\sqrt{9^2+9^2}$

$\text{Hypotenuse}=9\sqrt2$

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

$\text{Width}=\frac{9\sqrt2}{2}$

Now, find the area of the rectangle.

$\text{Area of Rectangle}=5\times\frac{9\sqrt2}{2}$

$\text{Area of Rectangle}=\frac{45\sqrt2}{2}$

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.

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

$\text{Area of Triangle}=\frac{81}{2}$

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

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

$\text{Area of Shaded Region}=\frac{81}{2}-\frac{45\sqrt2}{2}$

Solve.

$\text{Area of Shaded Region}=8.68$

What is the area of a square that has a diagonal whose endpoints in the coordinate plane are located at (-8, 6) and (2, -4)?

100

100√2

50√2

200√2

Explanation

Square_part1

Square_part2

Square_part3

A right isosceles triangle has leg lengths of $\frac{1}{8}$ . What is the length of the hypotenuse?

$\frac{\sqrt2}{8}$

$8\sqrt2$

$4\sqrt2$

$12\sqrt2$

Explanation

Recall the Pythagorean Theorem:

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

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

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.

$c=\frac{\sqrt2}{8}$

What is the area of a square that has a diagonal whose endpoints in the coordinate plane are located at (-8, 6) and (2, -4)?

100

100√2

50√2

200√2

Explanation

Square_part1

Square_part2

Square_part3

Find the length of the hypotenuse.

$13\sqrt2$

$13\sqrt3$

$\sqrt{26}$

Explanation

To find the hypotenuse of a triangle, you can use the Pythagorean Theorem. For any right triangle with leg lengths of and and a hypotenuse of ,

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

Now, plug in the values of and from the question.

$\text{Hypotenuse}=\sqrt{13^2+13^2}$

$\text{Hypotenuse}=\sqrt{169+169}$

Solve.

$\text{Hypotenuse}=\sqrt{338}$

Simplify.

$\text{Hypotenuse}=13\sqrt2$

Find the length of the hypotenuse.

$12\sqrt2$

$2\sqrt6$

$4\sqrt6$

$12\sqrt3$

Explanation

To find the hypotenuse of a triangle, you can use the Pythagorean Theorem. For any right triangle with leg lengths of and and a hypotenuse of ,

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

Now, plug in the values of and from the question.

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

$\text{Hypotenuse}=\sqrt{144+144}$

Solve.

$\text{Hypotenuse}{}=\sqrt{288}$

Simplify.

$\text{Hypotenuse}=12\sqrt2$

An isosceles triangle has a base of 6 and a height of 4. What is the perimeter of the triangle?

None of these

Explanation

An isosceles triangle is basically two right triangles stuck together. The isosceles triangle has a base of 6, which means that from the midpoint of the base to one of the angles, the length is 3. Now, you have a right triangle with a base of 3 and a height of 4. The hypotenuse of this right triangle, which is one of the two congruent sides of the isosceles triangle, is 5 units long (according to the Pythagorean Theorem).

The total perimeter will be the length of the base (6) plus the length of the hypotenuse of each right triangle (5).

5 + 5 + 6 = 16

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

Explanation

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a 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.

$\text{height}=\frac{8\sqrt2}{2}=4\sqrt2$

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.

$\text{Area}=\frac{1}{2}(4\sqrt2 \times 4\sqrt2)=16$

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

$\frac{1}{2}$

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:

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