45/45/90 Right Isosceles Triangles - Math

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$What;is;the;area;of;\Delta ABC?$

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$Area;\Delta=\frac{1}{2}$(base)(height)$

$$\frac{1}{2}$(3)(3)=\frac{9}{2}$=4.5$

If the length of one leg of a right isosceles triangle is 16cm, what is the length of the other leg?

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First we need to know that a right isosceles triangle has two legs with the same length.

In this case we do not need to worry about the hypotenuse.

Since one leg measures 16 cm, the other leg is the same length and so our answer is 16cm.

Consider an isosceles triangle with a height of 24 and a base of 12. What is the area of this triangle?

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The formula for the area of a trianlge is A = base * height * (1/2).

We're lucky here, because the question gives us all of the values we need. We simply need to plug them in:

A = base * height * (1/2) = 12 * 24 * (1/2) = 12 * 12 = 144

The side length of the 45-45-90 right triangle is $20$\sqrt{2}$$ , find the area of the right triangle.

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The area of a triangle is:

$A=\frac{bh}{2}$$ where b=base, h=height, and A=area

$A=\frac{20\sqrt{2}$20$\sqrt{2}$}{2}=\frac{400\sqrt{4}$}{2}=200$\sqrt{4}$=200*2=400$

Calculate the area of an isosceles right triangle who's hypotenuse is $4$\sqrt{2}$$ inches.

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The formula for the area of a triangle, right or not, is one half the base times height.

In this case, they are both $4 $\text{ in.}$$ Therefore, the respective values are entered, yielding:

$\frac{1}{2}$(4$\text{ in.}$)(4$\text{ in.}$)=8 $\text{ $in.}$^2$

Find the area of the triangle below.

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The key to finding the area of our triangle is to reaize that it is isosceles and therefore is a 45-45-90 triangle; therefore, we know the legs of our triangle are congruent and that each can be found by dividing the length of the hypotenuse by $\sqrt{2}$ .

$$\frac{8}{\sqrt{2}$}=\frac{8}{\sqrt{2}$}\cdot $\frac{\sqrt{2}$}{$\sqrt{2}$}=\frac{8\sqrt{2}$}{2}$

Rationalizing the denominator simplifies our result; however, we are interested in the area, not just the length of a leg; we remember that the formula for the area of a triangle is

$A = $\frac{1}{2}$bh$

where is the base and is the height; however, in our right triangle, the base and height are simply the two legs; therefore, we can calculate the area by substituting.

$A=\frac{1}{2}$($\frac{8\sqrt2}{2}$)($\frac{8\sqrt2}{2}$)=\frac{128}{8}$=16$

Find the area of a triangle that has a hypotenuse of $5\sqrt3$ .

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To find the area of a triangle, we must use $Area=\frac{1}{2}$b\cdot h$ where b=base and h=height.

In the problem, the only information given is what type the triangle is and what its hypotenuse is.

Given the area equation, the problem hasn't given any numbers that can be substituted into the equation to solve for an area. This means that the hypotenuse value must be used to determine the height and the base.

Because this is a 45/45/90 triangle, this means that it is also isosceles. Therefore, we can logic out that the base and the height must be the same.

The missing sides can be calulated in one of two ways:

1. Using the Pythagorean Theorem

2. Or using

If we were to use the Pythagorean Theorem, since we've already determined that b=h, that mean a=b in the equation. Let's say that .

That means the Pythagorean Theorem can be rewritten as:

${\color{Blue} $2s^2$$=c^2$}$

Now to substitute in the value of c to solve for the height and base.

$s=\frac{5\sqrt{6}$}{2}$

Now that we have the base and the height, we can substitute the values into the area equation and get the triangle's area.

$Area=\frac{1}{2}$\cdot$\frac{5\sqrt{6}$}{2}\cdot$\frac{5\sqrt{6}$}{2}$

$Area=\frac{1}{2}$\cdot$\frac{(25\cdot 6)}{4}$$

$Area = $\frac{1}{2}$\cdot$\frac{150}{4}$$

$Area = $\frac{1}{2}$\cdot $\frac{75}{2}$ = {\color{Blue} $\frac{75}{4}$}$

Find the area of a triangle with a hypotenuse of $10\sqrt2$ .

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To find the area of a triangle, we must use $Area=\frac{1}{2}$b\cdot h$ where b=base and h=height.

In the problem, the only information given is what type the triangle is and what its hypotenuse is.

Given the area equation, the problem hasn't given any numbers that can be substituted into the equation to solve for an area. This means that the hypotenuse value must be used to determine the height and the base.

Because this is a 45/45/90 triangle, this means that it is also isosceles. Therefore, we can logic out that the base and the height must be the same.

The missing sides can be calulated in one of two ways:

1. Using the Pythagorean Theorem

2. Or using

If we were to use the Pythagorean Theorem, since we've already determined that b=h, that means a=b in the equation. Let's say that .

That means the Pythagorean Theorem can be rewritten as:

${\color{Blue} $2s^2$$=c^2$}$

Now to substitute in the value of c to solve for the height and base.

Now that we have the base and the height, we can substitute the values into the area equation and get the triangle's area.

$Area=\frac{1}{2}$\cdot 10 \cdot 10$

$Area=\frac{1}{2}$\cdot 100$

If the hypotenuse of an isosceles right triangle is cm, what is the area of the triangle in square centimeters?

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Isosceles right triangles are special triangles because they possess angles of the following measures: $45^$\text{o}$$ , $45^$\text{o}$$ , and $90^$\text{o}$$ . These triangles are known as 45-45-90 triangles and have special characteristics. Recall the Pythagorean theorem:

In this equation, is the length of the triangle's base, is equal to its height, and is equal to the length of its hypotenuse. In an isosceles right triangle, the base and the height have the same length; therefore, is equal to , and you can rewrite the Pythagorean theorem like this:

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

Rearrange the equation so that is isolated on one side of the equals sign. First, simplify by dividing both sides of the equation by 2.

Next, take the square root of both sides.

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

$a=\frac{10\sqrt2}{2}$$

$a=5\sqrt2$

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 find the area of the given triangle.

$$\text{Area}$=\frac{1}{2}$(5\sqrt2 \times 5\sqrt2)$

$$\text{Area}$=\frac{1}{2}$(25\times 2)$

Solve.

$\text{Area}$=25 $\text{ $cm}$^2$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{14\sqrt2}{2}$=7\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}$=\frac{1}{2}$(7\sqrt2 \times 7\sqrt2)=49$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{20\sqrt2}{2}$=10\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}$=\frac{1}{2}$(10\sqrt2 \times 10\sqrt2)=100$

Find the length of the hypotenuse.

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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 ,

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

$\text{Hypotenuse}$=$\sqrt{25+25}$

Solve.

$\text{Hypotenuse}$=$\sqrt{50}$

Simplify.

$$\text{Hypotenuse}$=5\sqrt2$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{2\sqrt2}{2}$=\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}$=\frac{1}{2}$(\sqrt2 \times \sqrt2)=1$

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

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

Now, recall the Pythagorean Theorem:

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:

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

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

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{6\sqrt2}{2}$=3\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}$=\frac{1}{2}$(3\sqrt2 \times 3\sqrt2)=9$

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

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

Now, recall the Pythagorean Theorem:

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:

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 length of the hypotenuse.

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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 ,

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

$\text{Hypotenuse}$=$\sqrt{36+36}$

Solve.

$\text{Hypotenuse}$=$\sqrt{72}$

Simplify.

$$\text{Hypotenuse}$=6\sqrt2$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{24\sqrt2}{2}$=12\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}$=\frac{1}{2}$(12\sqrt2 \times 12\sqrt2)=144$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{18\sqrt2}{2}$=9\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}$=\frac{1}{2}$(9\sqrt2 \times 9\sqrt2)=81$

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

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

Now, recall the Pythagorean Theorem:

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:

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

$$\text{height}$=\frac{22\sqrt2}{2}$=11\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}$=\frac{1}{2}$(11\sqrt2 \times 11\sqrt2)=121$