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Basic Geometry : Basic Geometry

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

Example Questions

Example Question #181 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

399

423

403

410

Correct answer:

403

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:

$\text{base}=\frac{26\sqrt2}{2}$

$\text{base}=13\sqrt2$

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

$\text{Area of Triangle}=\frac{13\sqrt2 \times 13\sqrt2}{2}$

$\text{Area of Triangle}=\frac{338}{2}$

$\text{Area of Triangle}=169$

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.

$\text{Area of Rectangle}=22\times 26$

$\text{Area of Rectangle}=572$

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

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

$\text{Area of Shaded Region}=572-169$

Solve.

$\text{Area of Shaded Region}=403$

Report an Error

Example Question #189 : 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:

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

$\text{base}=10\sqrt2$

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

$\text{Area of Triangle}=\frac{16\sqrt2 \times 16\sqrt2}{2}$

$\text{Area of Triangle}=\frac{512}{2}$

$\text{Area of Triangle}=256$

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.

$\text{Area of Rectangle}=12\times 16$

$\text{Area of Rectangle}=192$

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

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

$\text{Area of Shaded Region}=256-192$

Solve.

$\text{Area of Shaded Region}=64$

Report an Error

Example Question #21 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

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

Possible Answers:

142

144

148

146

Correct answer:

144

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:

$\text{base}=\frac{24\sqrt2}{2}$

$\text{base}=12\sqrt2$

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

$\text{Area of Triangle}=\frac{12\sqrt2 \times 12\sqrt2}{2}$

$\text{Area of Triangle}=\frac{288}{2}$

$\text{Area of Triangle}=144$

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.

$\text{Area of Rectangle}=12\times 24$

$\text{Area of Rectangle}=288$

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

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

$\text{Area of Shaded Region}=288-144$

Solve.

$\text{Area of Shaded Region}=144$

Report an Error

Example Question #191 : Triangles

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

Possible Answers:

195

225

215

205

Correct answer:

195

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:

$\text{base}=\frac{26\sqrt2}{2}$

$\text{base}=13\sqrt2$

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

$\text{Area of Triangle}=\frac{13\sqrt2 \times 13\sqrt2}{2}$

$\text{Area of Triangle}=\frac{338}{2}$

$\text{Area of Triangle}=169$

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.

$\text{Area of Rectangle}=14\times 26$

$\text{Area of Rectangle}=364$

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

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

$\text{Area of Shaded Region}=364-169$

Solve.

$\text{Area of Shaded Region}=195$

Report an Error

Example Question #192 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

135

145

150

165

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.

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

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

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:

$\text{base}=\frac{30\sqrt2}{2}$

$\text{base}=15\sqrt2$

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

$\text{Area of Triangle}=\frac{15\sqrt2 \times 15\sqrt2}{2}$

$\text{Area of Triangle}=\frac{450}{2}$

$\text{Area of Triangle}=225$

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.

$\text{Area of Rectangle}=12\times 30$

$\text{Area of Rectangle}=360$

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

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

$\text{Area of Shaded Region}=360-225$

Solve.

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

Report an Error

Example Question #193 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

120

180

150

Correct answer:

120

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:

$\text{base}=\frac{24\sqrt2}{2}$

$\text{base}=12\sqrt2$

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

$\text{Area of Triangle}=\frac{12\sqrt2 \times 12\sqrt2}{2}$

$\text{Area of Triangle}=\frac{288}{2}$

$\text{Area of Triangle}=144$

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.

$\text{Area of Rectangle}=11\times 24$

$\text{Area of Rectangle}=264$

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

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

$\text{Area of Shaded Region}=264-144$

Solve.

$\text{Area of Shaded Region}=120$

Report an Error

Example Question #191 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

$6\text{ cm}$

$16\text{ cm}$

$8\text{ cm}$

$32\text{ cm}$

$45\text{ cm}$

Correct answer:

$16\text{ cm}$

Explanation:

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.

Report an Error

Example Question #1 : How To Find If Right Triangles Are Similar

$\bigtriangleup ABC \sim \bigtriangleup DEF$ ; $\angle F$ is a right angle; $\overline{AB}=20$ ; $\overline{AC}=10$ ; $\overline{DE}=30$

Find $\overline{EF}$ .

Possible Answers:

$10 \sqrt{3}$

$15 \sqrt{3}$

$30\sqrt{3}$

Correct answer:

$15 \sqrt{3}$

Explanation:

Since $\bigtriangleup ABC \sim \bigtriangleup DEF$ and $\angle F$ is a right angle, $\angle C$ is also a right angle.

$\overline{AB}$ is the hypotenuse of the first triangle; since one of its legs $\overline{AC}$ is half the length of that hypotenuse, $\bigtriangleup ABC$ is 30-60-90 with $\overline{AC}$ the shorter leg and $\overline{BC}$ the longer.

Because the two are similar triangles, $\overline{DE}$ is the hypotenuse of the second triangle, and $\overline{EF}$ is its longer leg.

Therefore, $EF = \frac{\sqrt{3}}{2} \cdot DE = \frac{\sqrt{3}}{2} \cdot 30 = 15 \sqrt{3}$ .

Report an Error

Example Question #2 : How To Find If Right Triangles Are Similar

Which of the following is sufficient to say that two right triangles are similar?

Possible Answers:

Two of the sides are the same.

Two sides and one angle are congruent.

All the angles are congruent.

Two angles and one side are congruent.

Correct answer:

All the angles are congruent.

Explanation:

If all three angles of a triangle are congruent but the sides are not, then one of the triangles is a scaled up version of the other. When this happens the proportions between the sides still remains unchanged which is the criteria for similarity.

Report an Error

Example Question #2 : How To Find If Right Triangles Are Similar

Which of the following statements is true regarding the two triangles?

Possible Answers:

We do not have enough information to show the triangles are similar.

The triangles are not similar because they are not the same size.

The triangles are similar by Hypotenuse-Leg

The triangles are similar because they are both right triangles.

The triangles are similar by Side-Angle-Side

Correct answer:

The triangles are similar by Side-Angle-Side

Explanation:

Though we must do a little work, we can show these triangles are similar. First, right triangles are not necessarily always similar. They must meet the necessary criteria like any other triangles; furthermore, there is no Hypotenuse-Leg Theorem for similarity, only for congruence; therefore, we can eliminate two answer choices.

However, we can use the Pythagorean Theorem with the smaller triangle to find the missing leg. Doing so gives us a length of 48. Comparing the ratio of the shorter legs in each trangle to the ratio of the longer legs we get

$\frac{20}{10}=\frac{2}{1}$

$\frac{48}{24}=\frac{2}{1}$

In both cases, the leg of the larger triangle is twice as long as the corresponding leg in the smaller triangle. Given that the angle between the two legs is a right angle in each triangle, these angles are congruent. We now have enough evidence to conclude similarity by Side-Angle-Side.

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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #181 : 45/45/90 Right Isosceles Triangles

Example Question #189 : Triangles

Example Question #21 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #191 : Triangles

Example Question #192 : 45/45/90 Right Isosceles Triangles

Example Question #193 : 45/45/90 Right Isosceles Triangles

Example Question #191 : 45/45/90 Right Isosceles Triangles

Example Question #1 : How To Find If Right Triangles Are Similar

Example Question #2 : How To Find If Right Triangles Are Similar

Example Question #2 : How To Find If Right Triangles Are Similar

All Basic Geometry Resources

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