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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 #175 : Triangles

If the hypotenuse of a right isosceles triangle is $4\sqrt{3}$ , what is the length of one side of the triangle?

Possible Answers:

$2\sqrt6$

$8\sqrt3$

$6\sqrt2$

$4\sqrt6$

Correct answer:

$2\sqrt6$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

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

Simplify.

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

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

Multiply the fraction by one in the form of $\frac{\sqrt{2}}{\sqrt{2}}$ .

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

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

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(4\sqrt3)(\sqrt2)}{2}$

Simplify.

$a=\frac{4\sqrt{6}}{2}$

Reduce.

$a=2\sqrt6$

Report an Error

Example Question #176 : Triangles

If the hypotenuse of a right isosceles triangle is $9\sqrt6$ , what is the length of one side of the triangle?

Possible Answers:

$9\sqrt2$

$9\sqrt3$

$\frac{9\sqrt2}{2}$

$\frac{9\sqrt3}{2}$

Correct answer:

$9\sqrt3$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

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

Simplify.

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

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

Multiply the fraction by one in the form of $\frac{\sqrt{2}}{\sqrt{2}}$ .

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

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

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{(9\sqrt6)(\sqrt2)}{2}$

Simplify.

$a=\frac{9\sqrt{12}}{2}$

$a=\frac{18\sqrt3}{2}$

Reduce.

$a=9\sqrt3$

Report an Error

Example Question #177 : Triangles

If the hypotenuse of a right isosceles triangle is 128 , what is the length of a side of the triangle?

Possible Answers:

$64\sqrt2$

128

$32\sqrt3$

Correct answer:

$64\sqrt2$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

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

Simplify.

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

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

Multiply the fraction by one in the form of $\frac{\sqrt{2}}{\sqrt{2}}$ .

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

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

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{128(\sqrt2)}{2}$

Simplify.

$a=64\sqrt2$

Report an Error

Example Question #181 : Triangles

If the hypotenuse of a right isosceles triangle is 1000 , what is the length of a side of the triangle?

Possible Answers:

$250\sqrt3$

$500\sqrt2$

250

$125\sqrt6$

Correct answer:

$500\sqrt2$

Explanation:

A right isosceles triangle is also a 45-45-90 triangle.

To find the length of a side, we will need to use the Pythagorean Theorem:

a^2+b^2=c^2

Since this is an isosceles triangle,

a=b

The Pythagorean Theorem can then be rewritten as the following:

a^2+a^2=c^2

2a^2=c^2

Since we are trying to find the length of a side of this triangle, solve for .

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

Simplify.

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

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

Multiply the fraction by one in the form of $\frac{\sqrt{2}}{\sqrt{2}}$ .

$a=\frac{c}{\sqrt2}\times\frac{\sqrt{2}}{\sqrt{2}}$

Solve.

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

Now, substitute in the length of the hypotenuse in for to solve for the side of the triangle in the question.

$a=\frac{1000(\sqrt2)}{2}$

Simplify.

$a=500\sqrt2$

Report an Error

Example Question #182 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

135

131

129

127

Correct answer:

135

Explanation:

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

First, recall how to find the area of a triangle.

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

In the figure, we are given the triangle’s hypotenuse, which is also the width of the rectangle. We can then use this information and the Pythagorean theorem to find the length of the sides of the triangle.

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{18\sqrt2}{2}$

$\text{base}=9\sqrt2$

Now, substitute this number in to find the area of the triangle.

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

$\text{Area of Triangle}=\frac{162}{2}$

$\text{Area of Triangle}=81$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=12\times18$

$\text{Area of Rectangle}=216$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=216-81$

Solve.

$\text{Area of Shaded Region}=135$

Report an Error

Example Question #183 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

245

214

221

291

Correct answer:

221

Explanation:

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

First, recall how to find the area of a triangle.

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

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{34\sqrt2}{2}$

$\text{base}=17\sqrt2$

Now, substitute this number in to find the area of the triangle.

$\text{Area of Triangle}=\frac{17\sqrt2 \times 17\sqrt2}{2}$

$\text{Area of Triangle}=\frac{578}{2}$

$\text{Area of Triangle}=289$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=15\times34$

$\text{Area of Rectangle}=510$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=510-289$

Solve.

$\text{Area of Shaded Region}=221$

Report an Error

Example Question #184 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

220

190

250

230

Correct answer:

220

Explanation:

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

First, recall how to find the area of a triangle.

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

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{20\sqrt2}{2}$

$\text{base}=10\sqrt2$

Now, substitute this number in to find the area of the triangle.

$\text{Area of Triangle}=\frac{10\sqrt2 \times 10\sqrt2}{2}$

$\text{Area of Triangle}=\frac{200}{2}$

$\text{Area of Triangle}=100$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=16\times 20$

$\text{Area of Rectangle}=320$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=320-100$

Solve.

$\text{Area of Shaded Region}=220$

Report an Error

Example Question #185 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

Correct answer:

Explanation:

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

First, recall how to find the area of a triangle.

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

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{14\sqrt2}{2}$

$\text{base}=7\sqrt2$

Now, substitute this number in to find the area of the triangle.

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

$\text{Area of Triangle}=\frac{98}{2}$

$\text{Area of Triangle}=49$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=8\times 14$

$\text{Area of Rectangle}=112$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=112-49$

Solve.

$\text{Area of Shaded Region}=63$

Report an Error

Example Question #186 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

143.75

156.25

145.50

159.75

Correct answer:

143.75

Explanation:

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

First, recall how to find the area of a triangle.

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

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{23\sqrt2}{2}$

Now, substitute this number in to find the area of the triangle.

$\text{Area of Triangle}=\frac{\frac{23\sqrt2}{2} \times \frac{23\sqrt2}{2}}{2}$

$\text{Area of Triangle}=\frac{529}{4}$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=12\times 23$

$\text{Area of Rectangle}=276$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=276-\frac{529}{4}$

Solve.

$\text{Area of Shaded Region}=143.75$

Report an Error

Example Question #187 : Triangles

In the figure, a right isosceles triangle is placed in a rectangle. What is the area of the shaded region?

Possible Answers:

120

Correct answer:

Explanation:

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

First, recall how to find the area of a triangle.

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

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

Now, because this is a right isosceles triangle,

$\text{base}=\text{height}$

We can then make the following substitution:

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

$\text{Hypotenuse}^2=2(\text{base})^2$

$\text{Hypotenuse}=\sqrt{2(\text{base})^2}$

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

Therefore:

$\text{base}=\frac{\text{Hypotenuse}}{\sqrt2}$

$\text{base}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the value of the hypotenuse to find the length of the base:

$\text{base}=\frac{28\sqrt2}{2}$

$\text{base}=14\sqrt2$

Now, substitute this number in to find the area of the triangle.

$\text{Area of Triangle}=\frac{14\sqrt2 \times 14\sqrt2}{2}$

$\text{Area of Triangle}=\frac{392}{2}$

$\text{Area of Triangle}=196$

Next, recall how to find the area of the rectangle:

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

Substitute in the given information from the question to find the area.

$\text{Area of Rectangle}=10\times 28$

$\text{Area of Rectangle}=280$

Now that we have the areas of both the rectangle and the triangle, we can find the area of the shaded region.

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

$\text{Area of Shaded Region}=280-196$

Solve.

$\text{Area of Shaded Region}=84$

Report an Error

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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #175 : Triangles

Example Question #176 : Triangles

Example Question #177 : Triangles

Example Question #181 : Triangles

Example Question #182 : Triangles

Example Question #183 : Triangles

Example Question #184 : Triangles

Example Question #185 : Triangles

Example Question #186 : Triangles

Example Question #187 : Triangles

All Basic Geometry Resources

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