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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 #108 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

100

125

Correct answer:

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.

$\text{Area}=25$

Report an Error

Example Question #109 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

144

Correct answer:

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.

$\text{Hypotenuse}=2 \times6$

Simplify.

$\text{Hypotenuse}=12$

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.

$\text{Area}=\frac{12^2}{4}$

Solve.

$\text{Area}=36$

Report an Error

Example Question #110 : 45/45/90 Right Isosceles Triangles

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

Possible Answers:

Correct answer:

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.

$\text{Hypotenuse}=2\times 7$

Simplify.

$\text{Hypotenuse}=14$

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.

$\text{Area}=\frac{14^2}{4}$

Solve.

$\text{Area}=49$

Report an Error

Example Question #1091 : Basic Geometry

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

Possible Answers:

$64\pi$

128

256

Correct answer:

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.

$\text{Hypotenuse}=2\times 8$

Simplify.

$\text{Hypotenuse}=16$

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.

$\text{Area}=\frac{16^2}{4}$

Solve.

$\text{Area}=64$

Report an Error

Example Question #1092 : Basic Geometry

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

Possible Answers:

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{16}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{4}$

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.

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

Report an Error

Example Question #1093 : Basic Geometry

Find the area of the triangle if the radius of the circle is $\frac{3}{2}$ .

Possible Answers:

$\frac{9}{4}$

$\frac{9}{8}$

$\frac{9}{2}$

Correct answer:

$\frac{9}{4}$

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.

$\text{Hypotenuse}=2\times \frac{3}{2}$

Simplify.

$\text{Hypotenuse}=3$

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.

$\text{Area}=\frac{3^2}{4}$

Solve.

$\text{Area}=\frac{9}{4}$

Report an Error

Example Question #1094 : Basic Geometry

Find the area of the triangle if the radius of the circle is $\frac{9}{2}$ .

Possible Answers:

$\frac{81}{2}$

$\frac{81}{4}$

Correct answer:

$\frac{81}{4}$

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.

$\text{Hypotenuse}=2\times \frac{9}{2}$

Simplify.

$\text{Hypotenuse}=9$

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.

$\text{Area}=\frac{9^2}{4}$

Solve.

$\text{Area}=\frac{81}{4}$

Report an Error

Example Question #1095 : Basic Geometry

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

Possible Answers:

$\frac{11}{4}$

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

$\frac{121}{4}$

$\frac{121}{2}$

Correct answer:

$\frac{121}{4}$

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.

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

Report an Error

Example Question #1091 : Plane Geometry

Find the area of this triangle:

Isosceles right

Possible Answers:

7.9

7.16

6.24

5.12

Correct answer:

7.16

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:

(1.5)^2 + x^2 = 5^ 2

2.25 + x^2 = 25

x^2 = 22.75

$x \approx 4.77$

Now we can find the area using the formula

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

In this case,

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

Report an Error

Example Question #1097 : Basic Geometry

Isoscelesrighttriangle 1000

In the isosceles right triangle shown here, its hypotenuse is $3\sqrt{2}$ units long. What is its area?

Possible Answers:

4.5

7.5

Correct answer:

4.5

Explanation:

By the Pythagorean Theorem, we can deduce the lengths of both of the sides of the given isosceles right triangle, since they will be congruent according to the definition of an isosceles triangle:

a^2+b^2=c^2

$a^2+b^2=(3\sqrt{2})^2$

$2a^2=(3\sqrt{2})^2$

because a=b .

2a^2=18

a^2=9

$a=\sqrt{9}=3$ .

Now we can calculate this triangle's area according to the formula for the area of a right triangle:

$A=\frac{bh}{2}=\frac{(3)(3)}{2}=\frac{9}{2}=4.5$ .

Hence, the area of this isosceles right triangle is 4.5 square units.

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All Basic Geometry Resources

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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #108 : 45/45/90 Right Isosceles Triangles

Example Question #109 : 45/45/90 Right Isosceles Triangles

Example Question #110 : 45/45/90 Right Isosceles Triangles

Example Question #1091 : Basic Geometry

Example Question #1092 : Basic Geometry

Example Question #1093 : Basic Geometry

Example Question #1094 : Basic Geometry

Example Question #1095 : Basic Geometry

Example Question #1091 : Plane Geometry

Example Question #1097 : Basic Geometry

All Basic Geometry Resources

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