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SAT Math : Plane Geometry

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Example Questions

Example Question #281 : Plane Geometry

Inscribed quad

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Possible Answers:

The question cannot be answered from the information given.

Correct answer:

The question cannot be answered from the information given.

Explanation:

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (180-t)^{\circ }$ , and solving for :

t + (180-t)= 180

180+t-t= 180

180= 180 ,

The statement turns out to be true regardless of the value of . Therefore, without further information, the value of cannot be determined.

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Example Question #281 : Sat Mathematics

Inscribed quad

Figure NOT drawn to scale.

The above figure shows a quadrilateral inscribed in a circle. Evaluate .

Possible Answers:

t= 78

t= 84

t = 80

t = 82

t = 86

Correct answer:

t= 84

Explanation:

If a quadrilateral is inscribed in a circle, then each pair of its opposite angles are supplementary - that is, their degree measures total $180^{\circ }$ .

$\angle A$ and $\angle C$ are two such angles, so

$m \angle A + m \angle C= 180 ^{\circ }$

Setting $m \angle A = t^{\circ }$ and $m \angle C = (t+12)^{\circ }$ , and solving for :

t + (t+12 )= 180

2t +12 = 180

2t +12 - 12 = 180 - 12

2t = 168

$2t \div 2 = 168 \div 2$

t = 84 ,

the correct response.

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Example Question #282 : Geometry

Secant 2

Figure NOT drawn to scale.

Refer to the above diagram. $\overline{AB}$ is a diameter. Evaluate $m \angle CBA$

Possible Answers:

$28^{\circ }$

$31^{\circ }$

$45^{\circ }$

$62^{\circ }$

$56^{\circ }$

Correct answer:

$31^{\circ }$

Explanation:

$\overline{AB}$ is a diameter, so $\overarc{ACB}$ is a semicircle - therefore, $m \overarc{ACB} = 180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc{AC}+ m \overarc{CB} = 180 ^{\circ }$

If we let $t ^{\circ }= m \overarc{AC}$ , then

$t ^{\circ }+ m \overarc{CB} = 180 ^{\circ }$ ,

and

$m \overarc{CB} = (180 -t )^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NC}$ and $\overline{NB}$ are such segments intercepting $\overarc{AC}$ and $\overarc{CB}$ , it holds that

$\frac{1}{2}\left ( m \overarc{CB}- m \overarc{AC} \right ) = m \angle CNB$

Setting $m \overarc{AC} = t ^{\circ }$ , $m \overarc{CB} = (180 -t )^{\circ }$ , and $m \angle CNB = 28 ^{\circ }$ :

$\frac{1}{2} [ \left ( 180-t \right ) - t ] = 28$

$\frac{1}{2} \left ( 180-2 t \right ) = 28$

90-t = 28

90-t + t - 28 = 28 + t - 28

62 = t

$m \overarc{AC} =62^{\circ }$

The inscribed angle that intercepts this arc, $\angle CBA$ , has half this measure:

$m \angle CBA = \frac{1}{2} \cdot m \overarc{AC} = \frac{1}{2} \cdot 62^{\circ } = 31^{\circ }$ .

This is the correct response.

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Example Question #33 : Circles

Secant 3 Figure NOT drawn to scale.

In the above figure, $\overline{AB}$ is a diameter. Also, the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5. Give the measure of $\angle TNA$ .

Possible Answers:

$15^{\circ }$

$18^{\circ }$

$21 ^{\circ }$

$24 ^{\circ }$

The measure of $\angle TNA$ cannot be determine from the information given.

Correct answer:

$15^{\circ }$

Explanation:

$\overline{AB}$ is a diameter, so $\overarc {ATB}$ is a semicircle, which has measure $180 ^{\circ }$ . By the Arc Addition Principle,

$m \overarc {AT } + m \overarc {BT} = m \overarc {ATB}$

If we let $t ^{\circ }= m \overarc {AT }$ , then, substituting:

$t ^{\circ }+ m \overarc {BT} = 180 ^{\circ }$ ,

and

$m \overarc {BT} =( 180 - t ) ^{\circ }$

the ratio of the length of $\overarc{BT}$ to that of $\overarc{AT}$ is 7 to 5; this is also the ratio of their degree measures; that is,

$\frac{m \overarc{BT}}{m \overarc{AT}} = \frac{5}{3}$

Setting $m \overarc {AT } = t ^{\circ }$ and $m \overarc {BT} =( 180 - t ) ^{\circ }$ :

$\frac{180-t }{t} = \frac{7}{5}$

Cross-multiply, then solve for :

7t = 5 (180-t)

7t = 900 - 5t

7t + 5t = 900 - 5t + 5t

12t = 900

$\frac{12t }{12}= \frac{900}{12}$

t = 75

$m \overarc {AT } =75^{\circ }$ , and $m \overarc {BT} =( 180 -75 ) ^{\circ } = 105 ^{\circ }$

If a secant and a tangent are drawn from a point to a circle, the measure of the angle they form is half the difference of the measures of the intercepted arcs. Since $\overline{NB}$ and $\overline{NT}$ are such segments whose angle $\angle TNA$ intercepts $\overarc{AT}$ and $\overarc{BT}$ , it holds that:

$m \angle TNA = \frac{1}{2}\left ( m \overarc{BT}- m \overarc{AT} \right )$

$m \angle TNA = \frac{1}{2}\left ( 105 ^{\circ } - 75^{\circ } \right )$

$m \angle TNA = \frac{1}{2}\left ( 30 ^{\circ } \right )$

$m \angle TNA = 15^{\circ }$

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Example Question #1 : Diameter

If the area of a circle is four times larger than the circumference of that same circle, what is the diameter of the circle?

Possible Answers:

Correct answer:

Explanation:

Set the area of the circle equal to four times the circumference πr² = 4(2πr).

Cross out both π symbols and one r on each side leaves you with r = 4(2) so r = 8 and therefore d = 16.

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Example Question #282 : Plane Geometry

The perimeter of a circle is 36 π. What is the diameter of the circle?

Possible Answers:

Correct answer:

Explanation:

The perimeter of a circle = 2 πr = πd

Therefore d = 36

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Example Question #283 : Plane Geometry

Sat_math_picture

If the area of the circle touching the square in the picture above is $81\pi$ , what is the closest value to the area of the square?

Possible Answers:

100

162

144

211

Correct answer:

162

Explanation:

Obtain the radius of the circle from the area.

$A=\pi r^2=81\pi$

r^2=81

r=9

Split the square up into 4 triangles by connecting opposite corners. These triangles will have a right angle at the center of the square, formed by two radii of the circle, and two 45-degree angles at the square's corners. Because you have a 45-45-90 triangle, you can calculate the sides of the triangles to be , , and $x\sqrt{2}$ . The radii of the circle (from the center to the corners of the square) will be 9. The hypotenuse (side of the square) must be $9\sqrt{2}$ .

The area of the square is then $(9\sqrt{2})^2=162$ .

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Example Question #284 : Plane Geometry

Two legs of a right triangle measure 3 and 4, respectively. What is the area of the circle that circumscribes the triangle?

Possible Answers:

$25\pi$

$5\pi$

$12.5\pi$

$6\pi$

$6.25\pi$

Correct answer:

$6.25\pi$

Explanation:

For the circle to contain all 3 vertices, the hypotenuse must be the diameter of the circle. The hypotenuse, and therefore the diameter, is 5, since this must be a 3-4-5 right triangle.

The equation for the area of a circle is A = πr².

$A=\pi (5/2)^2=6.25\pi$

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Example Question #285 : Plane Geometry

The circumference of the circle is $100\pi$ . What is the diameter?

Possible Answers:

100

Correct answer:

100

Explanation:

Write the formula for the circumference.

$C=2\pi r= \pi D$

Substitute the circumference.

$100\pi=\pi D$

D=100

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Example Question #284 : Geometry

Find the diameter of a circle whose area is $6\pi$ .

Possible Answers:

$\sqrt6$

$2\sqrt6$

Correct answer:

$2\sqrt6$

Explanation:

To solve, simply use the formula for the area of a circle to find the radius, and then multiply it by 2 to find the diameter. Thus,

$A=\pi{r^2}\rightarrowr=\sqrt{\frac{A}{\pi}}$

$r=\sqrt{\frac{6\pi}{\pi}}=\sqrt{6}$

$d=2*\sqrt{6}=2\sqrt6$

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SAT Math : Plane Geometry

Study concepts, example questions & explanations for SAT Math

All SAT Math Resources

Example Questions

Example Question #281 : Plane Geometry

Example Question #281 : Sat Mathematics

Example Question #282 : Geometry

Example Question #33 : Circles

Example Question #1 : Diameter

Example Question #282 : Plane Geometry

Example Question #283 : Plane Geometry

Example Question #284 : Plane Geometry

Example Question #285 : Plane Geometry

Example Question #284 : Geometry

All SAT Math Resources

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