Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

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Example Questions

Example Question #881 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 2\sqrt2\).

Possible Answers:

\(\displaystyle 8\sqrt2\)

\(\displaystyle 8\)

\(\displaystyle 12\)

\(\displaystyle 16\)

Correct answer:

\(\displaystyle 8\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

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

Simplify.

\(\displaystyle \text{side}=2\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(2)\)

Solve.

\(\displaystyle \text{Perimeter}=8\)

Example Question #37 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 3\sqrt2\).

Possible Answers:

\(\displaystyle 12\)

\(\displaystyle 12\sqrt2\)

\(\displaystyle 24\)

\(\displaystyle 16\)

Correct answer:

\(\displaystyle 12\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{3\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=3\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(3)\)

Solve.

\(\displaystyle \text{Perimeter}=12\)

Example Question #38 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 4\sqrt2\).

Possible Answers:

\(\displaystyle 16\)

\(\displaystyle 12\)

\(\displaystyle 16\sqrt2\)

\(\displaystyle 24\)

Correct answer:

\(\displaystyle 16\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

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

Simplify.

\(\displaystyle \text{side}=4\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(4)\)

Solve.

\(\displaystyle \text{Perimeter}=16\)

Example Question #39 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 5\sqrt2\).

Possible Answers:

\(\displaystyle 20\)

\(\displaystyle 20\sqrt2\)

\(\displaystyle 50\)

\(\displaystyle 25\sqrt2\)

Correct answer:

\(\displaystyle 20\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

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

Simplify.

\(\displaystyle \text{side}=5\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(5)\)

Solve.

\(\displaystyle \text{Perimeter}=20\)

Example Question #40 : How To Find The Perimeter Of A Square

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 7\sqrt2\).

Possible Answers:

\(\displaystyle 28\sqrt2\)

\(\displaystyle 98\)

\(\displaystyle 56\)

\(\displaystyle 28\)

Correct answer:

\(\displaystyle 28\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{7\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=7\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(7)\)

Solve.

\(\displaystyle \text{Perimeter}=28\)

Example Question #881 : Basic Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 9\sqrt2\).

Possible Answers:

\(\displaystyle 18\sqrt2\)

\(\displaystyle 72\)

\(\displaystyle 36\sqrt2\)

\(\displaystyle 36\)

Correct answer:

\(\displaystyle 36\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{9\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=9\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(9)\)

Solve.

\(\displaystyle \text{Perimeter}=36\)

Example Question #882 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 11\sqrt2\).

Possible Answers:

\(\displaystyle 22\)

\(\displaystyle 44\)

\(\displaystyle 121\sqrt2\)

\(\displaystyle 22\sqrt2\)

Correct answer:

\(\displaystyle 44\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{11\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=11\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(11)\)

Solve.

\(\displaystyle \text{Perimeter}=44\)

Example Question #883 : Plane Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 13\sqrt2\).

Possible Answers:

\(\displaystyle 52\sqrt3\)

\(\displaystyle 52\)

\(\displaystyle 104\)

\(\displaystyle 52\sqrt2\)

Correct answer:

\(\displaystyle 52\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{13\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=13\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(13)\)

Solve.

\(\displaystyle \text{Perimeter}=52\)

Example Question #231 : Squares

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 15\sqrt2\).

Possible Answers:

\(\displaystyle 180\)

\(\displaystyle 90\)

\(\displaystyle 60\)

\(\displaystyle 120\)

Correct answer:

\(\displaystyle 60\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

\(\displaystyle \text{side}=\frac{15\sqrt2(\sqrt2)}{2}\)

Simplify.

\(\displaystyle \text{side}=15\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(15)\)

Solve.

\(\displaystyle \text{Perimeter}=60\)

Example Question #881 : Basic Geometry

Find the perimeter of a square inscribed in a circle that has a diameter of \(\displaystyle 12\sqrt2\).

Possible Answers:

\(\displaystyle 60\)

\(\displaystyle 48\)

\(\displaystyle 36\)

\(\displaystyle 24\)

Correct answer:

\(\displaystyle 48\)

Explanation:

1

Notice that the diameter of the circle is also the diagonal of the square. The diagonal of the square is also the hypotenuse of a right isosceles triangle that has the sides of the square as its legs.

\(\displaystyle \text{diameter}=\text{diagonal}\)

Now, use the Pythagorean theorem to find the length of the sides of the square.

\(\displaystyle \text{Diagonal}^2=\text{side}^2+\text{side}^2\)

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

\(\displaystyle \text{side}^2=\frac{\text{Diagonal}^2}{2}\) 

\(\displaystyle \text{side}=\sqrt{\frac{\text{Diagonal}^2}{2}}=\frac{\text{Diagonal}\sqrt2}{2}\)

Now, substitute in the value of the diagonal to find the length of a side of the square.

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

Simplify.

\(\displaystyle \text{side}=12\)

Now, recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side})\)

Substitute in the value of the side to find the perimeter of the square.

\(\displaystyle \text{Perimeter}=4(12)\)

Solve.

\(\displaystyle \text{Perimeter}=48\)

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