SAT Math - 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?

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

$9-9\pi$

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

$9-3\pi$

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

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

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?

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

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

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

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

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

Explanation

The sidelength of the garden is feet less than that of the entire lot - that is, . 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.

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

The triangle is right and scalene.

The triangle is right and isosceles, but not equilateral.

The triangle is acute and scalene.

The triangle is acute and equilateral.

The triangle is acute and isosceles, but not equilateral.

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.

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)

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:

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

You are given triangles $\Delta MNO$ and $\Delta PQR$ , with $\angle N \cong \angle Q$ . Which of these statements, along with what you are given, is enough to prove that $\Delta MNO \sim \Delta PQR$ ?

$\angle M \cong \angle P$

$\frac{MN}{PQ} = \frac{MO}{PR}$

$\Delta MNO$ and $\Delta PQR$ have the same perimeter.

None of the other responses is correct.

Explanation

gives us the congruence of two corresponding angles and one corresponding side; this is not enough to establish similarity.

The perimeters of the triangles are irrelevant to their similarity, so $\Delta MNO$ and $\Delta PQR$ having the same perimeter does not help to establish similarity, with or without what is given.

$\frac{MN}{PQ} = \frac{MP}{PR}$ establishes the proportionality of two nonincluded sides of the angles known to be congruent. However, there is no statement that establishes similarity as a result of this.

$\angle M \cong \angle P$ , along with $\angle N \cong \angle Q$ , sets up the conditions of the Angle-Angle Similarity Postulate, which states that if two triangles have two pairs of congruent angles between them, the triangles are similar. $\angle M \cong \angle P$ is the correct choice.

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

The triangle is right and scalene.

The triangle is right and isosceles.

The triangle is acute and scalene.

The triangle is acute and isosceles.

The triangle cannot exist.

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.

Thingy_5

Refer to the above diagram. Which of the following choices gives a set of collinear points?

Explanation

Collinear points are points that are contained in the same line. Of the four choices, only fit the description, since all are on Line .

Regular Octagon has perimeter 80. $\overline{AB }$ has as its midpoint; segment $\overline{ME}$ is drawn. To the nearest tenth, give the length of $\overline{ME}$ .

Explanation

Below is the regular Octagon , with the referenced midpoint and segment $\overline{ME}$ . Note that perpendiculars have also been constructed from and to meet $\overline{BE}$ at and , respectively.

Octagon 2

$\bigtriangleup MBE$ is a right triangle with legs $\overline{MB}$ and $\overline{ BE}$ and hypotenuse $\overline{ME}$ .

The perimeter of the regular octagon is 80, so the length of each side is one-eighth of 80, or 10. Consequently,

$MB = \frac{1}{2} AB = \frac{1}{2} \cdot 10 = 5$

To find the length of $\overline{ BE}$ , we can break it down as

Quadrilateral is a rectangle, so .

$\bigtriangleup BXC$ is a 45-45-90 triangle with leg $\overline{BX}$ and hypotenuse $\overline{BC}$ ; by the 45-45-90 Triangle Theorem,

$BX = \frac{BC }{\sqrt{2}} \approx \frac{10}{1.4142} \approx 7.0711$

For similar reasons, $EY \approx 7.0711$ .

Therefore,

$BE = BX + XY + YE \approx 7.0711 + 10 + 7.0711 \approx 24.1421$

can now be evaluated using the Pythagorean Theorem:

$ME = \sqrt{(MB)^{2}+ ( BE ) ^{2}}$

Substituting and evaluating:

$ME = \sqrt{5 ^{2}+24.1421^{2}}$

$\approx \sqrt{25 +582.8427}$

$\approx \sqrt{ 607.8427}$

$\approx 24.7$ ,

the correct choice.

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?

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

$12x\sqrt{2}$

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

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

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

Explanation

The sidelength of the garden is $2 \times 6 = 12$ less than that of the entire lot - that is, . 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.

Which of the following describes a triangle with sides of length 10 inches, 1 foot, and 2 feet?

This triangle cannot exist.

This is an acute triangle.

This is a right triangle.

This is an obtuse triangle.

More information is needed to answer this question.

Explanation

One foot is equal to 12 inches, so the triangle would have sides 10, 12, and 24 inches. Since

the triangle violates the Triangle Inequality, which states that the sum of the lengths of the two smaller sides must exceed the length of the third. The triangle cannot exist.

Other 2-Dimensional Geometry

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