Geometry - 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 $4\sqrt2$ .

$\sqrt{7}$

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.

$\text{Area}=8$

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

$8\sqrt{3}$

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.

$\text{Area}=32$

Find the area of the triangle below.

Explanation

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.

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

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

Isosceles right

Explanation

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:

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

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.

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

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

Next, take the square root of both sides.

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

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

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.

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

Find the area of the triangle if the radius of the circle is $\frac{11}{2}$ .

$\frac{121}{4}$

$\frac{11}{4}$

$\frac{11\sqrt2}{4}$

$\frac{121}{2}$

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:

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

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

$\text{Hypotenuse}=2\times \frac{11}{2}$

Simplify.

$\text{Hypotenuse}=11$

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.

$\text{Area}=\frac{11^2}{4}$

Solve.

$\text{Area}=\frac{121}{4}$

Find the area of this triangle:

Isosceles right

Explanation

To find the height, use the Pythagorean Theorem. One of the legs is the missing side, and the other is 1.5, half of 3. The hypotenuse is 5:

$x \approx 4.77$

Now we can find the area using the formula

$area = \frac{1}{2} (base)(height)$

In this case,

$\frac{1}{2}(3)(4.77) \approx 7.16$

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

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.

$\text{Hypotenuse}=2\times 4$

Simplify.

$\text{Hypotenuse}=8$

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.

$\text{Area}=\frac{8^2}{4}$

Solve.

$\text{Area}=16$

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:

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

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

$\text{Hypotenuse}=2\times 1$

Simplify.

$\text{Hypotenuse}=2$

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.

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

Solve.

$\text{Area}=1$

Find the area of the triangle if the radius of the circle is $\frac{1}{2}$ .

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{2}$

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:

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

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

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

Simplify.

$\text{Hypotenuse}=1$

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.

$\text{Area}=\frac{1^2}{4}$

Solve.

$\text{Area}=\frac{1}{4}$

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

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.

$\text{Hypotenuse}=2\times 5$

Simplify.

$\text{Hypotenuse}=10$

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.

$\text{Area}=\frac{10^2}{4}$

Solve.

$\text{Area}=25$

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

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