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SAT II Math II : SAT Subject Test in Math II

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

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Example Questions

Example Question #1 : Geometry

You have a right triangle with a hypotenuse of 13 inches and a leg of 5 inches, what is the area of the triangle?

Possible Answers:

30 in^2

120in^2

12in^2

56in^2

Correct answer:

30 in^2

Explanation:

You have a right triangle with a hypotenuse of 13 inches and a leg of 5 inches, what is the area of the triangle?

So find the area of a triangle, we need the following formula:

$A_{tri}=\frac{1}{2}bh$

However, we only know one leg, so we only know b or h.

To find the other leg, we can either use Pythagorean Theorem, or recognize that this is a 5-12-13 triangle. Meaning, our final leg is 12 inches long.

To prove this:

(13in)^2=(5in)^2+(xin)^2

169in^2-25in^2=144in^2=(xin)^2

144in^2=x^2in

$x=\sqrt{144in^2}=12in$

Now, we know both legs, let's just plug in and solve for area:

$A_{tri}=\frac{1}{2}(5in)(12in)=6in*5in=30in^2$

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Example Question #1 : 2 Dimensional Geometry

You have a rectangular-shaped rug which you want to put in your living room. If the rug is 12.5 feet long and 18 inches wide, what is the area of the rug?

Possible Answers:

15ft^2

18.75ft

3.75ft^2

18.75ft^2

Correct answer:

18.75ft^2

Explanation:

You have a rectangular-shaped rug which you want to put in your living room. If the rug is 12.5 feet long and 18 inches wide, what is the area of the rug?

To begin, we need to realize two things.

1) Our given measurements are not in equivalent units, so we need to convert one of them before doing any solving.

2) The area of a rectangle is given by: A=l*w

Now, let's convert 18 inches to feet, because it seems easier than 12.5 feet to inches:

$18inches*\frac{1ft}{12in}=1.5ft$

Now, using what we know from 2) we can find our answer

A=12.5ft*1.5ft=18.75ft^2

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Example Question #1 : 2 Dimensional Geometry

Give the area of $\bigtriangleup ABC$ to the nearest whole square unit, where:

AB = 34

AC = 27

$m \angle A = 124^{\circ }$

Possible Answers:

761

257

513

459

381

Correct answer:

381

Explanation:

The area of a triangle with two sides of lengths and and included angle of measure $\gamma$ can be calculated using the formula

$A = \frac{1}{2} ab \sin \gamma$ .

Setting $a = AB = 34, b= AC = 27 , \gamma = m \angle A = 124^{\circ }$ and evaluating :

$A = \frac{1}{2} (34 )(27)\sin 124 ^{\circ }$

$\approx 459 ( 0.8290 )$

$\approx 381$

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Example Question #311 : Sat Subject Test In Math Ii

Give the area of $\bigtriangleup ABC$ to the nearest whole square unit, where:

AB = 20

BC = 26

AC = 32

Possible Answers:

260

129

320

416

Correct answer:

260

Explanation:

The area of a triangle, given its three sidelengths, can be calculated using Heron's formula:

$A = \sqrt{s(s-a)(s-b)(s-c)}$ ,

where , , and are the lengths of the sides, and $s = \frac{a+b+ c}{2}$ .

Setting a = AB = 20 , b = BC = 26 , and c = AC = 32 , evaluate :

$s = \frac{20+26+32}{2} = \frac{78}{2} = 39$

and, substituting in Heron's formula:

$A = \sqrt{39(39-20)(39-26)(39-32)}$

$= \sqrt{39(19)(13)(7)}$

$= \sqrt{67,431 }$

$\approx 259.7$

To the nearest whole, this is 260.

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Example Question #11 : Area

Find the area of a triangle with a base length of x^2 and a height of $\frac{1}{2}x$ .

Possible Answers:

x^3

4x^2

$\frac{1}{2}x^3$

$\frac{1}{4}x^3$

x^2

Correct answer:

$\frac{1}{4}x^3$

Explanation:

Write the formula for the area of a triangle.

$A=\frac{BH}{2}$

Substitute the dimensions.

$A = \frac{x^2\cdot \frac{1}{2}x}{2} = \frac{1}{2}(x^2\cdot \frac{1}{2}x)$

The answer is: $\frac{1}{4}x^3$

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Example Question #11 : Geometry

Find the area of a circle with a radius of $\frac{3}{2}$ .

Possible Answers:

$9\pi$

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

$3\pi$

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

Correct answer:

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

Explanation:

Write the formula for the area of a circle.

$A=\pi r^2$

Substitute the radius into the equation.

$A=\pi (\frac{3}{2})^2 = \frac{9}{4} \pi$

The area is: $\frac{9}{4}\pi$

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Example Question #12 : Geometry

Determine the area of a rectangle if the length is x+3 and the height is $\frac{1}{4}(x+3)$ .

Possible Answers:

$\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

$\frac{1}{4}x^2 + 3x + \frac{9}{4}$

$\frac{1}{2}x^2 +\frac{1}{2}x+ \frac{3}{2}$

$\frac{1}{2}x^2 + \frac{3}{2}$

$\frac{1}{4}x^2 + 6x + \frac{9}{4}$

Correct answer:

$\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

Explanation:

The area of a rectangle is: A =LH

Substitute the length and height into the formula.

$A = (x+3)\cdot \frac{1}{4} (x+3)$

We will move the constant to the front and apply the FOIL method to simplify the binomials.

$A = \frac{1}{4} (x+3)(x+3) = \frac{1}{4} (x^2+6x+9)$

Distribute the fraction through all the terms of the trinomial.

The answer is: $\frac{1}{4}x^2 + \frac{3}{2}x + \frac{9}{4}$

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Example Question #315 : Sat Subject Test In Math Ii

What's the area of triangle with a side of $\frac{1}{2}x+\frac{1}{4}$ and a height of 2x+2 ?

Possible Answers:

$\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

x^2+3x+1

$x+\frac{1}{2}$

$x^2+\frac{3}{2}x+\frac{1}{2}$

$\frac{1}{2}x+\frac{1}{4}$

Correct answer:

$\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

Explanation:

Write the formula for the area of a triangle.

$A= \frac{BH}{2}$

Substitute the dimensions.

$A= \frac{(\frac{1}{2}x+\frac{1}{4})(2x+2)}{2}=\frac{1}{2}(\frac{1}{2}x+\frac{1}{4})(2x+2)$

$A= (\frac{1}{2}x+\frac{1}{4})(x+1)$

Use the FOIL method to expand this.

$A = (\frac{1}{2}x)(x)+ (\frac{1}{2}x)(1)+ (\frac{1}{4})(x)+ (\frac{1}{4})(1)$

Simplify the terms.

$A = \frac{1}{2}x^2+\frac{1}{2}x+\frac{1}{4}x+\frac{1}{4}$

Combine like-terms.

The answer is: $\frac{1}{2}x^2+\frac{3}{4}x+\frac{1}{4}$

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Example Question #311 : Sat Subject Test In Math Ii

Determine the side of a square with an area of 100x .

Possible Answers:

10000x^2

$10\sqrt{x}$

10x

10x^2

$\sqrt{10x}$

Correct answer:

$10\sqrt{x}$

Explanation:

Write the formula for the area of a square.

A = s^2

Substitute the area into the equation.

100x= s^2

Square root both sides.

$\sqrt{100x}= \sqrt{s^2}$

$s = 10\sqrt{x}$

The answer is: $10\sqrt{x}$

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Example Question #312 : Sat Subject Test In Math Ii

Find the area of a circle with a radius of 3x^3 .

Possible Answers:

$9\pi x^6$

$6\pi x^6$

$6\pi^2 x^6$

$9\pi x^9$

$9\pi^2 x^6$

Correct answer:

$9\pi x^6$

Explanation:

The area of a circle is $A=\pi r^2$ .

Substitute the radius and solve for the area.

$A=\pi (3x^3)^2 = \pi (3x^3)(3x^3) = 9\pi x^6$

The answer is: $9\pi x^6$

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All SAT II Math II Resources

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SAT II Math II : SAT Subject Test in Math II

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Geometry

Example Question #1 : 2 Dimensional Geometry

Example Question #1 : 2 Dimensional Geometry

Example Question #311 : Sat Subject Test In Math Ii

Example Question #11 : Area

Example Question #11 : Geometry

Example Question #12 : Geometry

Example Question #315 : Sat Subject Test In Math Ii

Example Question #311 : Sat Subject Test In Math Ii

Example Question #312 : Sat Subject Test In Math Ii

All SAT II Math II Resources

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