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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

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9 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

200

$400\sqrt{2}$

400

$100\sqrt{2}$

$200\sqrt{2}$

Correct answer:

400

Explanation:

The area of a triangle is:

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

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

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Example Question #4 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of the triangle below.

Possible Answers:

Correct answer:

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.

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

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Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of a 45/45/90 triangle that has a hypotenuse of $5\sqrt3$ .

Possible Answers:

$\sqrt{\frac{75}{2}}$

$\frac{75}{2}$

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

$\frac{75}{4}$

37.5

Correct answer:

$\frac{75}{4}$

Explanation:

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 a^2+b^2=c^2

2. Or using Find_the_leg_length_resolution

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 a=b=s .

That means the Pythagorean Theorem can be rewritten as:

s^2+s^2=c^2

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

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

$2s^2=(5\sqrt{3})^2$

$2s^2=25\sqrt{9}$

2s^2=75

$s^2=\frac{75}{2}$

$\sqrt{s^2}=\sqrt{\frac{75}{2}}$

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

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

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

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

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

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

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Example Question #5 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Find the area of a 45/45/90 triangle with a hypotenuse of $10\sqrt2$ .

Possible Answers:

200

$10\sqrt2$

$50\sqrt2$

100

Correct answer:

Explanation:

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.

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 a^2+b^2=c^2

2. Or using Find_the_leg_length_resolution

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 a=b=s .

That means the Pythagorean Theorem can be rewritten as:

s^2+s^2=c^2

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

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

$2s^2=(10\sqrt{2})^2$

$2s^2=100\sqrt{4}$

2s^2=200

$s^2=\frac{200}{2}$

$\sqrt{s^2}=\sqrt{100}$

s=10

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

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

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

Area = 50

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Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

144

Correct answer:

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

$\text{height}=\frac{14\sqrt2}{2}=7\sqrt2$

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.

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

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Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

300

100

400

200

Correct answer:

100

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

$\text{height}=\frac{20\sqrt2}{2}=10\sqrt2$

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.

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

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Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

Correct answer:

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

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

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.

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

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Example Question #81 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

Correct answer:

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

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

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.

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

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Example Question #12 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

$3\sqrt3$

Correct answer:

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

$\text{height}=\frac{6\sqrt2}{2}=3\sqrt2$

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.

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

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Example Question #1061 : Plane Geometry

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

Possible Answers:

Correct answer:

Explanation:

An isosceles right triangle is another way of saying that the triangle is a 45-45-90 triangle.

Now, recall the Pythagorean Theorem:

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

Because we are working with a 45-45-90 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.

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

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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #4 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #5 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #2 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #3 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #81 : 45/45/90 Right Isosceles Triangles

Example Question #12 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #1061 : Plane Geometry

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