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SAT II Math I : Geometry

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

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6 Diagnostic Tests 113 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #2 : Diameter, Radius, And Circumference

Example circle

Find the diameter, circumference and area of the circle above.

Possible Answers:

Diameter= 6ft

Circumference=37.68 ft

Area= 28.27 ft²

Diameter= 3ft

Circumference=37.68 ft

Area= 28.7 ft²

Diameter= 9ft

Circumference=37.68 ft

Area= 28 ft²

Diameter= 6 ft

Circumference= 19 ft

Area= 30 ft²

Diameter= 6 ft

Circumference=18.84 ft

Area= 28.27 ft²

Correct answer:

Diameter= 6 ft

Circumference=18.84 ft

Area= 28.27 ft²

Explanation:

To find the diameter, you must know that radius is half the diameter (or the diameter is 2 times the radius.

$\small Diameter= 2(radius)$

$\small \small 2(3)= 6$

The diameter is 6ft.

To find the circumference, you must multiply the diameter (6ft) by pi.

$\small Circumference=\pi (diameter)$

$\small \small \pi(6)= 18.84$

The circumference is 18.84 ft.

To find the surface area, you must aquare the radius (3ft) and multiply by pi.

$\small Surface Area=\pi (radius^{2})$

$\small \pi(3{^2})=\pi(9)=28.27$

The surface area is 28.27 ft².

The diameter is 6ft, the circumference is 18.84 ft, and the surface area is 28.27 ft².

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Example Question #1 : Diameter, Radius, And Circumference

A circle has a diameter of 10cm. What is the circumference?

Possible Answers:

$25\pi cm$

$10\pi cm$

$5\pi cm$

$100\pi cm$

$20\pi cm$

Correct answer:

$10\pi cm$

Explanation:

The circumference of a circle is given by the equation:

$C=2\pi r$

The radius is half the diameter, in this case half of 10cm is 5cm

Plug in 5cm for r

$C=2\pi 5cm$

Simplify to get the final answer

$C=10\pi cm$

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Example Question #2 : Diameter, Radius, And Circumference

If the diameter of a circle is 10cm , what is the area of the circle?

Possible Answers:

$20\pi$

$25\pi$

$15\pi$

$10\pi$

Correct answer:

$25\pi$

Explanation:

Step 1: Recall the formula for an area of a circle...

$Area=\pi r^2$ .

Step 2: Given the diameter, find the radius..

We know that the diameter is twice the length of the radius...

D=2r

Plug in for :

10=2r

Divide by 2:

$\frac {10}{2}=5=r$

Step 3: Now that we know the radius, plug the radius into the area formula..

$Area=\pi (5)^2$

Simplify:

$Area=25(\pi)=25\pi$

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Example Question #3 : Diameter, Radius, And Circumference

Determine the circumference of the circle with an area of $5\pi$ .

Possible Answers:

$2\pi\sqrt{2}$

$\pi\sqrt{10}$

$2\pi\sqrt{5}$

$\frac{2\sqrt5}{\pi}$

$\frac{50}{\pi}$

Correct answer:

$2\pi\sqrt{5}$

Explanation:

Write the formula for the area of a circle.

$A=\pi r^2$

Substitute the area.

$5\pi=\pi r^2$

5=r^2

Square root both sides to solve for radius.

$\sqrt{5}=\sqrt{r^2}$

$r=\sqrt5$

Write the formula for circumference.

$C=2\pi r$

Substitute the radius.

The answer is: $C=2\pi\sqrt{5}$

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Example Question #3 : Diameter, Radius, And Circumference

Find the diameter of a circle if the circumference is 4x^2 .

Possible Answers:

$\frac{4x}{\pi}$

$\frac{2x}{\pi}$

$\frac{4x^2}{\pi}$

$\frac{8x}{\pi}$

$2\pi x^2$

Correct answer:

$\frac{4x^2}{\pi}$

Explanation:

Write the formula for the circumference of the circle.

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

Substitute the circumference into the equation.

$4x^2 = \pi D$

Divide by pi on both sides to get the diameter.

$\frac{4x^2 }{ \pi }=\frac{ \pi D}{ \pi }$

The answer is: $\frac{4x^2}{\pi}$

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Example Question #3 : Diameter, Radius, And Circumference

Find the area of a circle if the circumference is $5\pi$ .

Possible Answers:

$\frac{25}{2}\pi$

$\frac{5}{2}\pi$

$\frac{5}{2}\pi^2$

$\frac{5}{4}\pi^2$

$\frac{25}{4}\pi$

Correct answer:

$\frac{25}{4}\pi$

Explanation:

Write the formula for the circumference of a circle.

$C=2\pi r$

Substitute the circumference.

$5\pi=2\pi r$

Divide by $2\pi$ to isolate the .

$\frac{5\pi}{2\pi}=\frac{2\pi r}{2\pi}$

The radius is: $r= \frac{5}{2}$

Write the formula for the area of the circle.

$A=\pi r^2$

Substitute the radius.

$A=\pi (\frac{5}{2})^2 = \frac{25}{4}\pi$

The answer is: $\frac{25}{4}\pi$

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Example Question #1 : Finding Sides

Triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate the length of the hypotenuse of the blue triangle.

Possible Answers:

$10\sqrt{2}$

$20 \sqrt{5}$

$20\sqrt{2}$

$10 \sqrt{5}$

$40\sqrt{2}$

Correct answer:

$20 \sqrt{5}$

Explanation:

The inscribed rectangle is a 20 by 20 square. Since opposite sides of the square are parallel, the corresponding angles of the two smaller right triangles are congruent; therefore, the two triangles are similar and, by definition, their sides are in proportion.

The small top triangle has legs 10 and 20. Therefore, the length of its hypotenuse can be determined using the Pythagorean Theorem:

$c = \sqrt{10^{2}+ 20^{2}} = \sqrt{100 + 400} = \sqrt{500} = \sqrt{100} \cdot \sqrt{5} = 10 \sqrt{5}$

The small top triangle has short leg 10 and hypotenuse $10 \sqrt{5}$ . The blue triangle has short leg 20 and unknown hypotenuse , where can be calculated with the proportion statement

$\frac{N}{10\sqrt{5}} = \frac{20}{10}$

$\frac{N}{10\sqrt{5}} \cdot 10\sqrt{5} = \frac{20}{10} \cdot 10\sqrt{5}$

$N = 20 \sqrt{5}$

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Example Question #31 : Geometry

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

AC = 25, a= 60, b = 45 .

Evaluate .

Possible Answers:

$BC = 25\sqrt{2}$

$BC = \frac{25\sqrt{2}}{ 2 }$

$B C= 33\frac{1}{3}$

$BC = \frac{50\sqrt{6}}{ 3 }$

$BC = \frac{25\sqrt{6}}{ 2 }$

Correct answer:

$BC = \frac{25\sqrt{6}}{ 2 }$

Explanation:

By the Law of Sines,

$\frac{BC}{\sin a^{\circ } } = \frac{AC}{\sin b^{\circ }}$

Substitute AC = 25, a= 60, b = 45 and solve for :

$\frac{BC}{\sin 60^{\circ } } = \frac{25}{\sin 45^{\circ }}$

$\frac{BC}{\sin 60^{\circ } } \cdot \sin 60^{\circ } = \frac{25}{\sin 45^{\circ }} \cdot \sin 60^{\circ }$

$BC = \frac{25\sin 60^{\circ }}{\sin 45^{\circ }}$

$BC= \frac{25 \cdot \frac{\sqrt{3}}{2} }{ \frac{\sqrt{2}}{2}}$

$BC= \frac{ \frac{25 \sqrt{3}}{2} }{ \frac{\sqrt{2}}{2}}$

$BC= \frac{25 \sqrt{3}}{2} \cdot \frac{2}{\sqrt{2}}$

$BC= \frac{25\sqrt{3}}{ \sqrt{2} }$

$BC = \frac{25\sqrt{3}}{ \sqrt{2} } \cdot \frac{\sqrt{2}}{\sqrt{2}}$

$BC = \frac{25\sqrt{6}}{ 2 }$

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Example Question #1 : Finding Sides

Triangle

Note: figure NOT drawn to scale.

Refer to the triangle in the above diagram.

AB = 15, BC = 20, b=45

Evaluate . Round to the nearest tenth, if applicable.

Possible Answers:

$AC \approx 14.2$

$AC \approx 10.3$

$AC \approx 32.4$

$AC \approx 30.4$

$AC \approx 18.0$

Correct answer:

$AC \approx 14.2$

Explanation:

By the Law of Cosines,

$\left (AC \right )^{2} =\left (AB \right )^{2}+ \left (BC \right )^{2} - 2 (AB)(BC) \cos b^{\circ }$

Substitute AB = 15, BC = 20, b=45 :

$\left (AC \right )^{2} =15^{2}+20^{2} - 2 \cdot15 \cdot 20 \cos 45^{\circ }$

$\left (AC \right )^{2} =225+400 - 600 \cdot \frac{\sqrt{2}}{2}$

$\left (AC \right )^{2} \approx 625 -424.26$

$\left (AC \right )^{2} \approx 200.74$

$AC \approx \sqrt{200.74} \approx 14.2$

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Example Question #4 : Finding Sides

Pentagon

The above figure is a regular pentagon. Evaluate to the nearest tenth.

Possible Answers:

$x \approx 5.4$

$x \approx 6.9$

$x \approx 6.5$

$x \approx 6.1$

$x \approx 5.7$

Correct answer:

$x \approx 6.5$

Explanation:

Two sides of the triangle formed measure 4 each; the included angle is one angle of the regular pentagon, which measures

$m = \left [\frac{180(5-2)}{5} \right ] ^{\circ }= 108^{\circ }$

The length of the third side can be found by applying the Law of Cosines:

$x^{2} = a^{2} + b^{2} - 2ab \cos C^{\circ }$

where a = b = 4, C = 108 :

$x^{2} = 4^{2} + 4^{2} - 2 \cdot 4 \cdot 4 \cos 108^{\circ }$

$x^{2} = 16 +16 -32 \cos 108^{\circ }$

$x^{2} \approx 32 -32 \left ( -0.3090\right )$

$x^{2} \approx 41.88$

$x \approx \sqrt{41.88} \approx 6.5$

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SAT II Math I : Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #2 : Diameter, Radius, And Circumference

Find the diameter, circumference and area of the circle above.

The diameter is 6ft.

The circumference is 18.84 ft.

The surface area is 28.27 ft².

The diameter is 6ft, the circumference is 18.84 ft, and the surface area is 28.27 ft².

Example Question #1 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #1 : Finding Sides

Example Question #31 : Geometry

Example Question #1 : Finding Sides

Example Question #4 : Finding Sides

All SAT II Math I Resources

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SAT II Math I : Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #2 : Diameter, Radius, And Circumference

Find the diameter, circumference and area of the circle above.

The diameter is 6ft.

The circumference is 18.84 ft.

The surface area is 28.27 ft2.

The diameter is 6ft, the circumference is 18.84 ft, and the surface area is 28.27 ft2.

Example Question #1 : Diameter, Radius, And Circumference

Example Question #2 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #3 : Diameter, Radius, And Circumference

Example Question #1 : Finding Sides

Example Question #31 : Geometry

Example Question #1 : Finding Sides

Example Question #4 : Finding Sides

All SAT II Math I Resources

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The surface area is 28.27 ft².

The diameter is 6ft, the circumference is 18.84 ft, and the surface area is 28.27 ft².