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Basic Geometry : Basic Geometry

Study concepts, example questions & explanations for Basic Geometry

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All Basic Geometry Resources

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

Example Questions

Example Question #91 : Squares

Find the area of a square if its diagonal is $7\sqrt{10}$ .

Possible Answers:

245

$490\sqrt2$

$\frac{245}{2}$

$\frac{245\sqrt3}{2}$

Correct answer:

245

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(7\sqrt{10})^2}{2}=\frac{490}{2}=245$

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Example Question #341 : Quadrilaterals

Find the area of a square if its diagonal is $6\sqrt{10}$ .

Possible Answers:

$360\sqrt3$

$180\sqrt2$

180

$90\sqrt3$

Correct answer:

180

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(6\sqrt{10})^2}{2}=\frac{360}{2}=180$

Report an Error

Example Question #345 : Quadrilaterals

Find the area of a square if its diagonal is $8\sqrt{10}$ .

Possible Answers:

$320\sqrt2$

$320\sqrt3$

320

$160\sqrt3$

Correct answer:

320

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(8\sqrt{10})^2}{2}=\frac{640}{2}=320$

Report an Error

Example Question #346 : Quadrilaterals

Find the area of a square if its diagonal is $11\sqrt2$ .

Possible Answers:

121

$\frac{121}{2}$

$121\sqrt3$

$\frac{121\sqrt2}{2}$

Correct answer:

121

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(11\sqrt{2})^2}{2}=\frac{242}{2}=121$

Report an Error

Example Question #341 : Quadrilaterals

Find the area of a square if its diagonal is $15\sqrt{10}$ .

Possible Answers:

2250

1125

875

775

Correct answer:

1125

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(15\sqrt{10})^2}{2}=\frac{2250}{2}=1125$

Report an Error

Example Question #348 : Quadrilaterals

Find the area of a square if its diagonal is $12\sqrt{14}$ .

Possible Answers:

$504\sqrt{3}$

$1008\sqrt5$

1008

$1008\sqrt2$

Correct answer:

1008

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(12\sqrt{14})^2}{2}=\frac{2016}{2}=1008$

Report an Error

Example Question #342 : Quadrilaterals

Find the area of a square if its diagonal is $\sqrt{30}$ .

Possible Answers:

$30\sqrt2$

$15\sqrt2$

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(\sqrt{30})^2}{2}=\frac{30}{2}=15$

Report an Error

Example Question #350 : Quadrilaterals

Find the area of a square if its diagonal is $2\sqrt{15}$ .

Possible Answers:

$30\sqrt2$

$30\sqrt3$

Correct answer:

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Plug in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(2\sqrt{15})^2}{2}=\frac{60}{2}=30$

Report an Error

Example Question #351 : Quadrilaterals

Find the area of a square if it has a diagonal of $\sqrt5$ .

Possible Answers:

$\frac{5}{2}$

$5\sqrt2$

$\frac{15}{2}$

Correct answer:

$\frac{5}{2}$

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Substitute in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(\sqrt5)^2}{2}$

Simplify.

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

Report an Error

Example Question #751 : Plane Geometry

Find the area of a square if it has a diagonal of $\sqrt3$ .

Possible Answers:

$\sqrt3$

$\frac{3}{2}$

$3\sqrt2$

$\frac{3\sqrt2}{2}$

Correct answer:

$\frac{3}{2}$

Explanation:

The diagonal of a square is also the hypotenuse of a 45-45-90 triangle.

Recall how to find the area of a square:

$\text{Area}=\text{side}^2$

Now, use the Pythagorean theorem to find the area of the square.

$\text{side}^2+\text{side}^2=\text{Diagonal}^2$

$2(\text{side})^2=\text{Diagonal}^2$

$\text{side}^2=\frac{\text{Diagonal}^2}{2}$

$\text{Area}=\text{side}^2=\frac{\text{Diagonal}^2}{2}$

Substitute in the length of the diagonal to find the area of the square.

$\text{Area}=\frac{(\sqrt3)^2}{2}$

Simplify.

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

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 #91 : Squares

Example Question #341 : Quadrilaterals

Example Question #345 : Quadrilaterals

Example Question #346 : Quadrilaterals

Example Question #341 : Quadrilaterals

Example Question #348 : Quadrilaterals

Example Question #342 : Quadrilaterals

Example Question #350 : Quadrilaterals

Example Question #351 : Quadrilaterals

Example Question #751 : Plane Geometry

All Basic Geometry Resources

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