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Basic Geometry : Plane 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 #11 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $\displaystyle 24$, what is the area of the triangle?

Possible Answers:

144 $\displaystyle 144$

100 $\displaystyle 100$

120 $\displaystyle 120$

$\displaystyle 96$

Correct answer:

144 $\displaystyle 144$

Explanation:

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a $\displaystyle 45-45-90$ triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$ $\displaystyle 2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$ $\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}$

$\text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}=\frac{\text{hypotenuse}}{\sqrt2}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$ $\displaystyle \text{height}=\sqrt{\frac{\text{hypotenuse}^2}{2}}=\frac{\text{hypotenuse}}{\sqrt2}=\frac{\text{hypotenuse}(\sqrt2)}{2}{}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{24\sqrt2}{2}=12\sqrt2$ $\displaystyle \text{height}=\frac{24\sqrt2}{2}=12\sqrt2$

Now, recall how to find the area of a triangle:

$\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(12\sqrt2 \times 12\sqrt2)=144$ $\displaystyle \text{Area}=\frac{1}{2}(12\sqrt2 \times 12\sqrt2)=144$

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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 $\displaystyle 18$, what is the area of the triangle?

Possible Answers:

$\displaystyle 64$

$\displaystyle 81$

$\displaystyle 90$

100 $\displaystyle 100$

Correct answer:

$\displaystyle 81$

Explanation:

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a $\displaystyle 45-45-90$ triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$ $\displaystyle 2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$ $\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{18\sqrt2}{2}=9\sqrt2$ $\displaystyle \text{height}=\frac{18\sqrt2}{2}=9\sqrt2$

Now, recall how to find the area of a triangle:

$\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(9\sqrt2 \times 9\sqrt2)=81$ $\displaystyle \text{Area}=\frac{1}{2}(9\sqrt2 \times 9\sqrt2)=81$

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Example Question #16 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $\displaystyle 22$, what is the area of the triangle?

Possible Answers:

$22\sqrt3$ $\displaystyle 22\sqrt3$

$11\sqrt2$ $\displaystyle 11\sqrt2$

121 $\displaystyle 121$

144 $\displaystyle 144$

Correct answer:

121 $\displaystyle 121$

Explanation:

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a $\displaystyle 45-45-90$ triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$ $\displaystyle 2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$ $\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{22\sqrt2}{2}=11\sqrt2$ $\displaystyle \text{height}=\frac{22\sqrt2}{2}=11\sqrt2$

Now, recall how to find the area of a triangle:

$\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(11\sqrt2 \times 11\sqrt2)=121$ $\displaystyle \text{Area}=\frac{1}{2}(11\sqrt2 \times 11\sqrt2)=121$

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Example Question #17 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

If the hypotenuse of a right isosceles triangle is $\displaystyle 40$, what is the area of the triangle?

Possible Answers:

369 $\displaystyle 369$

400 $\displaystyle 400$

100 $\displaystyle 100$

200 $\displaystyle 200$

Correct answer:

400 $\displaystyle 400$

Explanation:

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a $\displaystyle 45-45-90$ triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$ $\displaystyle 2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$ $\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{40\sqrt2}{2}=20\sqrt2$ $\displaystyle \text{height}=\frac{40\sqrt2}{2}=20\sqrt2$

Now, recall how to find the area of a triangle:

$\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(20\sqrt2 \times 20\sqrt2)=400$ $\displaystyle \text{Area}=\frac{1}{2}(20\sqrt2 \times 20\sqrt2)=400$

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Example Question #91 : 45/45/90 Right Isosceles Triangles

If the hypotenuse of a right isosceles triangle is $\displaystyle 26$, what is the area of the triangle?

Possible Answers:

225 $\displaystyle 225$

144 $\displaystyle 144$

169 $\displaystyle 169$

158 $\displaystyle 158$

Correct answer:

169 $\displaystyle 169$

Explanation:

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

Now, recall the Pythagorean Theorem:

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

Because we are working with a $\displaystyle 45-45-90$ triangle, the base and the height have the same length. We can rewrite the above equation as the following:

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

$2(\text{height})^2=\text{hypotenuse}^2$ $\displaystyle 2(\text{height})^2=\text{hypotenuse}^2$

$\text{height}^2=\frac{\text{hypotenuse}^2}{2}$ $\displaystyle \text{height}^2=\frac{\text{hypotenuse}^2}{2}$

Now, plug in the value of the hypotenuse to find the height for the given triangle.

$\text{height}=\frac{26\sqrt2}{2}=13\sqrt2$ $\displaystyle \text{height}=\frac{26\sqrt2}{2}=13\sqrt2$

Now, recall how to find the area of a triangle:

$\text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$ $\displaystyle \text{Area}=\frac{1}{2}\times\text{base}\times\text{height}$

Since the base and the height are the same length, we can then find the area of the given triangle.

$\text{Area}=\frac{1}{2}(13\sqrt2 \times 13\sqrt2)=169$ $\displaystyle \text{Area}=\frac{1}{2}(13\sqrt2 \times 13\sqrt2)=169$

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Example Question #1071 : Plane Geometry

Find the area of the triangle if the diameter of the circle is $2\sqrt2$ $\displaystyle 2\sqrt2$.

Possible Answers:

$\displaystyle 2$

$\displaystyle 4$

$\sqrt{3}$ $\displaystyle \sqrt{3}$

$\sqrt2$ $\displaystyle \sqrt2$

$\displaystyle 3$

Correct answer:

$\displaystyle 2$

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.

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$ $\displaystyle \text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$ $\displaystyle 2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$ $\displaystyle \text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$ $\displaystyle \text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{2\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{2\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=2$ $\displaystyle \text{side}=2$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{2 \times 2}{2}$ $\displaystyle \text{Area}=\frac{2 \times 2}{2}$

Solve.

$\text{Area}=2$ $\displaystyle \text{Area}=2$

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Example Question #1072 : Basic Geometry

Find the area of the triangle if the diameter of the circle is $12\sqrt2$ $\displaystyle 12\sqrt2$.

Possible Answers:

$\displaystyle 72$

144 $\displaystyle 144$

$\displaystyle 71$

$\sqrt{71}$ $\displaystyle \sqrt{71}$

$\sqrt{111}$ $\displaystyle \sqrt{111}$

Correct answer:

$\displaystyle 72$

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$ $\displaystyle \text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$ $\displaystyle 2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$ $\displaystyle \text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$ $\displaystyle \text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{12\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{12\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=12$ $\displaystyle \text{side}=12$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{12 \times 12}{2}$ $\displaystyle \text{Area}=\frac{12 \times 12}{2}$

Solve.

$\text{Area}=72$ $\displaystyle \text{Area}=72$

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Example Question #91 : Triangles

Find the area of the triangle if the diameter of the circle is $4\sqrt2$ $\displaystyle 4\sqrt2$.

Possible Answers:

$\displaystyle 18$

$\displaystyle 8$

$\displaystyle 24$

$\displaystyle 16$

$\sqrt{7}$ $\displaystyle \sqrt{7}$

Correct answer:

$\displaystyle 8$

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$ $\displaystyle \text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$ $\displaystyle 2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$ $\displaystyle \text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$ $\displaystyle \text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{4\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{4\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=4$ $\displaystyle \text{side}=4$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{4\times 4}{2}$ $\displaystyle \text{Area}=\frac{4\times 4}{2}$

Solve.

$\text{Area}=8$ $\displaystyle \text{Area}=8$

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Example Question #1072 : Plane Geometry

Find the area of the triangle if the diameter of the circle is $50\sqrt2$ $\displaystyle 50\sqrt2$.

Possible Answers:

1150 $\displaystyle 1150$

1250 $\displaystyle 1250$

1265 $\displaystyle 1265$

1420 $\displaystyle 1420$

1520 $\displaystyle 1520$

Correct answer:

1250 $\displaystyle 1250$

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$ $\displaystyle \text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$ $\displaystyle 2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$ $\displaystyle \text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$ $\displaystyle \text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{50\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{50\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=50$ $\displaystyle \text{side}=50$

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

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

Since we have a right isocseles triangle, the base and the height are the same length.

$\text{Area}=\frac{50\times 50}{2}$ $\displaystyle \text{Area}=\frac{50\times 50}{2}$

Solve.

$\text{Area}=1250$ $\displaystyle \text{Area}=1250$

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Example Question #93 : Triangles

Find the area of the triangle if the diameter of the circle is $5\sqrt2$ $\displaystyle 5\sqrt2$.

Possible Answers:

$5\sqrt{3}$ $\displaystyle 5\sqrt{3}$

$5\sqrt2$ $\displaystyle 5\sqrt2$

$\displaystyle 50$

$\frac{25}{2}$ $\displaystyle \frac{25}{2}$

$\displaystyle 10$

Correct answer:

$\frac{25}{2}$ $\displaystyle \frac{25}{2}$

Explanation:

$\text{Hypotenuse}^2=\text{side}^2+\text{side}^2$ $\displaystyle \text{Hypotenuse}^2=\text{side}^2+\text{side}^2$

$2(\text{side}^2)=\text{Hypotenuse}^2$ $\displaystyle 2(\text{side}^2)=\text{Hypotenuse}^2$

$\text{side}^2=\frac{\text{Hypotenuse}^2}{2}$ $\displaystyle \text{side}^2=\frac{\text{Hypotenuse}^2}{2}$

$\text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$ $\displaystyle \text{side}=\sqrt{\frac{\text{Hypotenuse}^2}{2}}=\frac{\text{Hypotenuse}\sqrt2}{2}$

Substitute in the given hypotenuse to find the length of the leg of a triangle.

$\text{side}=\frac{5\sqrt2(\sqrt2)}{2}$ $\displaystyle \text{side}=\frac{5\sqrt2(\sqrt2)}{2}$

Simplify.

$\text{side}=5$ $\displaystyle \text{side}=5$

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

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

Since we have a right isosceles triangle, the base and the height are the same length.

$\text{Area}=\frac{5\times 5}{2}$ $\displaystyle \text{Area}=\frac{5\times 5}{2}$

Solve.

$\text{Area}=\frac{25}{2}$ $\displaystyle \text{Area}=\frac{25}{2}$

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Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #11 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

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

Example Question #16 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #17 : How To Find The Area Of A 45/45/90 Right Isosceles Triangle

Example Question #91 : 45/45/90 Right Isosceles Triangles

Example Question #1071 : Plane Geometry

Example Question #1072 : Basic Geometry

Example Question #91 : Triangles

Example Question #1072 : Plane Geometry

Example Question #93 : Triangles

All Basic Geometry Resources

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