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SAT II Math II : Other 2-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math II

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Example Questions

Example Question #1 : Other 2 Dimensional Geometry

Which of the following describes a triangle with sides of length 9 feet, 4 yards, and 180 inches?

Possible Answers:

The triangle is right and scalene.

The triangle is acute and scalene.

The triangle is acute and isosceles, but not equilateral.

The triangle is acute and equilateral.

The triangle is right and isosceles, but not equilateral.

Correct answer:

The triangle is right and scalene.

Explanation:

3 feet make a yard, so 9 feet is equal to 3 yards. 36 inches make a yard, so 180 inches is equal to $180 \div 36 = 5$ yards. That makes this a 3-4-5 triangle. 3-4-5 is a well-known Pythagorean triple; that is, they have the relationship

$3^{2}+ 4^{2}=5^{2}$

and any triangle with these three sidelengths is a right triangle. Also, since the three sides are of different lengths, the triangle is scalene.

The correct response is that the triangle is right and scalene.

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Example Question #1 : Other 2 Dimensional Geometry

Which of the following describes a triangle with sides of length two yards, eight feet, and ten feet?

Possible Answers:

The triangle cannot exist.

The triangle is right and scalene.

The triangle is right and isosceles.

The triangle is acute and scalene.

The triangle is acute and isosceles.

Correct answer:

The triangle is right and scalene.

Explanation:

Two yards is equal to six feet. The sidelengths are 6, 8, and 10, which form a well-known Pythagorean triple with the relationship

$6^{2}+ 8^{2} = 10^{2}$

The triangle is therefore right. Since no two sides have the same length, it is also scalene.

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Example Question #74 : Geometry

Garden

The above figure shows a square garden (in green) surrounded by a dirt path feet wide throughout. Which of the following expressions gives the distance, in feet, from one corner of the garden to the opposite corner?

Possible Answers:

$\left (40 - x \right )\sqrt{2}$

$\left (80 - x \right )\sqrt{2}$

$\left (40 + x \right )\sqrt{2}$

$\left (160- 2x \right )\sqrt{2}$

$\left (80 - 2x \right )\sqrt{2}$

Correct answer:

$\left (80 - 2x \right )\sqrt{2}$

Explanation:

The sidelength of the garden is feet less than that of the entire lot - that is, 80 - 2x . Since the garden is square, the path from one corner to the other is a diagonal of a square, which has length $\sqrt{2}$ times the sidelength. This is

$d = \left (80 - 2x \right )\sqrt{2}$ feet.

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Example Question #4 : Other 2 Dimensional Geometry

Garden

The above figure shows a square garden (in green) surrounded by a dirt path six feet wide throughout. Which of the following expressions gives the distance, in feet, from one corner of the garden to the opposite corner?

Possible Answers:

$\left (2 x - 24\right )\sqrt{2}$

$12x\sqrt{2}$

$\left ( x - 12\right )\sqrt{2}$

$\left ( x - 6\right )\sqrt{2}$

$\left (2 x - 12\right )\sqrt{2}$

Correct answer:

$\left ( x - 12\right )\sqrt{2}$

Explanation:

The sidelength of the garden is $2 \times 6 = 12$ less than that of the entire lot - that is, x - 12 . Since the garden is square, the path from one corner to the other is a diagonal of a square, which has length $\sqrt{2}$ times the sidelength. This is

$d = (x - 12)\sqrt{2}$ feet.

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Example Question #5 : Other 2 Dimensional Geometry

A circle is inscribed inside a square that touches all edges of the square. The square has a length of 3. What is the area of the region inside the square and outside the edge of the circle?

Possible Answers:

$9-3\pi$

$\textup{The answer is not given.}$

$9-9\pi$

$9-\frac{9}{2}\pi$

$9-\frac{9}{4}\pi$

Correct answer:

$9-\frac{9}{4}\pi$

Explanation:

Solve for the area of the square.

$A_{square}=s^2 = 3^2 = 9$

Solve for the area of the circle. Given the information that the circle touches all sides of the square, the diameter is equal to the side length of the square.

This means that the radius is half the length of the square: $r=\frac{3}{2}$

Substitute the radius.

$A_{circle}=\pi r^2 = \pi (\frac{3}{2})^2= \pi (\frac{3}{2}) (\frac{3}{2}) = \frac{9}{4}\pi$

Subtract the area of the square and the circle to determine the area desired.

The answer is: $9-\frac{9}{4}\pi$

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Example Question #71 : Geometry

Inscribed

Figure is not drawn to scale

$\overline{AC}$ is a diameter of the circle; its length is ten; furthermore we know the following:

$m \overarc {AB} = 126^{\circ }$

Give the length of $\overline{AB}$ (nearest tenth)

Possible Answers:

7.0

9.5

5.8

7.6

8.9

Correct answer:

8.9

Explanation:

Locate , the center of the circle, which is the midpoint of $\overline{AC}$ ; draw radius $\overline{OB}$ . $\bigtriangleup AOB$ is formed. The central angle that intercepts $\overarc {AB}$ is $\angle AOB$ , so $m \angle AOB = m \overarc {AB} = 126^{\circ }$ . $\overline{OA}$ and $\overline{OB}$ , being radii of the circle, have length half the diameter of ten, or five. The diagram is below.

Inscribed

By the Law of Cosines, given two sides of a triangle of length and , and their included angle of measure $\gamma$ , the length of the third side can be calculated using the formula

$c ^{2} = a^{2} + b^{2} - 2ab \cos \gamma$

Setting $a = OA = 5, b = OB = 5, \gamma = m \angle AOB = 126 ^{\circ } , c = AB$ , solve for :

$(AB) ^{2} = 5^{2} +5 ^{2} - 2 \cdot 5 \cdot 5 \cdot \cos 126^{\circ }$

$(AB) ^{2} =25 + 25 - 50 \cos 126^{\circ }$

$(AB) ^{2} \approx 50 - 50 (-0.5878)$

$(AB) ^{2} \approx 50 + 29.39$

$(AB) ^{2} \approx 79.39$

Taking the square root of both sides:

$AB \approx\sqrt{ 79.39} \approx 8.9$

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SAT II Math II : Other 2-Dimensional Geometry

Study concepts, example questions & explanations for SAT II Math II

All SAT II Math II Resources

Example Questions

Example Question #1 : Other 2 Dimensional Geometry

Example Question #1 : Other 2 Dimensional Geometry

Example Question #74 : Geometry

Example Question #4 : Other 2 Dimensional Geometry

Example Question #5 : Other 2 Dimensional Geometry

Example Question #71 : Geometry

All SAT II Math II Resources

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