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SAT II Math I : SAT Subject Test in Math I

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

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Example Questions

Example Question #11 : Geometry

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

AB = 19

AC = 25

$m \angle A = 57^{\circ }$

Possible Answers:

238

259

129

398

199

Correct answer:

199

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 = 19 , b = AC = 25 , and $\gamma = m \angle A = 57^{\circ }$ , then evaluating :

$A = \frac{1}{2} (19 )(25)\sin 57^{\circ }$

$A = \frac{1}{2} (19 )(25) (0.8387)$

$A \approx 199$ .

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Example Question #11 : 2 Dimensional Geometry

Find the area of a triangle with a height of $\frac{1}{2}$ and a base of $\frac{1}{4}$ .

Possible Answers:

$\frac{3}{4}$

$\frac{1}{4}$

$\frac{3}{8}$

$\frac{1}{2}$

$\frac{1}{16}$

Correct answer:

$\frac{1}{16}$

Explanation:

Write the formula for the area of a triangle.

$A=\frac{1}{2}bh$

Substitute the base and height into the formula.

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

Simplify the fractions.

The answer is: $\frac{1}{16}$

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Example Question #11 : Area

Find the area of a circle if the circumference is 3.

Possible Answers:

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

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

$\frac{9}{4\pi^2}$

$\frac{3}{2\pi^2}$

$\frac{3}{2\pi}$

Correct answer:

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

Explanation:

Write the formula for the circumference of a circle.

$C=2\pi r$

Substitute the circumference.

$3=2\pi r$

Divide by $2\pi$ on both sides.

$\frac{3}{2\pi}= r$

We will need the formula for the area of the circle.

$A=\pi r^2$

Substitute the radius to find the area.

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

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

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Example Question #11 : Area

What is the area of a square if the length of the side is $\pi\sqrt2$ ?

Possible Answers:

$\sqrt{2\pi}$

$4\pi$

$4\pi ^2$

$2\pi$

$2\pi ^2$

Correct answer:

$2\pi ^2$

Explanation:

Write the formula for the area of a square.

A=s^2

Substitute the side into the formula.

$A=(\pi\sqrt2)^2 = \pi^2 (2) = 2\pi ^2$

The answer is: $2\pi ^2$

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Example Question #451 : Sat Subject Test In Math I

On the XY plane, line segment AB has endpoints (0, a) and (b, 0). If a square is drawn with segment AB as a side, in terms of a and b what is the area of the square?

Possible Answers:

A=(a + b) (a - b)

A=2(ab)^2

$A=\frac{1}{2} (a + b)$

A=b^2+a^2

Cannot be determined

Correct answer:

A=b^2+a^2

Explanation:

Since the question is asking for area of the square with side length AB, recall the formula for the area of a square.

$A=\text{side}^2$

Now, use the distance formula to calculate the length of AB.

$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

let

$\\(x_1,y_1)=(0,a) \\(x_2,y_2)=(b,0)$

Now substitute the values into the distance formula.

$\\d=\sqrt{(b-0)^2+(0-2)^2} \\d=\sqrt{b^2+a^2}$

From here substitute the side length value into the area formula.

$\\A=\text{side}^2 \\A=\left(\sqrt{b^2+a^2}\right)^2 \\A=b^2+a^2$

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Example Question #22 : Area

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

AB = 13

BC = 25

AC = 17

Possible Answers:

Cannot be determined

102

Correct answer:

102

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 = 13 , b = AC = 17 , and c= BC = 25 ,

$s = \frac{13+17+25}{2} = \frac{55}{2} = 27.5$

and, substituting in Heron's formula:

$A = \sqrt{27.5(27.5-13)(27.5-17)(27.5-25 )}$

$= \sqrt{27.5(14.5)(10.5)(2.5 )}$

$= \sqrt{10,467.1875}$

$\approx 102$

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Example Question #1 : Perimeter

Triangle

Note: Figure NOT drawn to scale.

Refer to the figure above, which shows a square inscribed inside a large triangle. What is the difference between the perimeter of the white trapezoid and the blue triangle?

Possible Answers:

$-10 +10 \sqrt{3}$

$-10 +10 \sqrt{5}$

$-10 +10 \sqrt{2}$

Correct answer:

$-10 +10 \sqrt{5}$

Explanation:

One side of the white trapezoid is the hypotenuse of the small top triangle, which has legs 10 and 20. Therefore, the length of this side 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 trapezoid has perimeter

$30 + 20 + 20 + 10 \sqrt{5} = 70 + 10 \sqrt{5}$ .

The small top triangle has legs 10 and 20 and hypotenuse $10 \sqrt{5}$ , and, therefore, perimeter

$10 + 20 + 10 \sqrt{5} = 30 + 10 \sqrt{5}$ . The blue triangle, which is similar as a result of the parallelism of the opposite sides of the square, has short leg 20. Since the perimeter of two similar triangles is directly proportional to a side, we can set up and solve the proportion statement to find the perimeter of the blue triangle:

$\frac{P}{30 + 10 \sqrt{5}} = \frac{20}{10}$

$\frac{P}{30 + 10 \sqrt{5}}\left ( 30 + 10 \sqrt{5} \right ) = \frac{20}{10}\left ( 30 + 10 \sqrt{5} \right )$

$P =60 + 20 \sqrt{5}$

The difference between the perimeters is

$\left (60 + 20 \sqrt{5} \right )- \left (70 + 10 \sqrt{5} \right )=-10 +10 \sqrt{5}$

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Example Question #1 : Perimeter

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Give the ratio of the perimeter of $\Delta BDC$ to that of $\Delta ADB$ .

Possible Answers:

3:1

3:2

4:3

2:1

4:1

Correct answer:

2:1

Explanation:

The altitude of a right triangle from the vertex of its right triangle to its hypotenuse divides it into two similar triangles.

, as the length of the altitude corresponding to the hypotenuse, is the geometric mean of the lengths of the parts of the hypotenuse it forms; that is, it is the square root of the product of the two:

$BD =\sqrt{ \left ( AD\right ) \left ( DC\right )} = \sqrt{6 \cdot 24} = \sqrt{144} = 12$ .

The ratio of the smaller side of $\Delta BDC$ to that of $\Delta ADB$ is

$\frac{BD}{AD}=\frac{12}{6} = \frac{2}{1}$ or 2:1, making this the similarity ratio. Theratio of the perimeters is always equal to the similarity ratio, so 2:1 is also the ratio of the perimeter of $\Delta BDC$ to that of $\Delta ADB$ .

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Example Question #2 : Perimeter

Thingy

Refer to the above figure. Quadrilateral BCDE is a square. Give the perimeter of Polygon ABCDE in terms of .

Possible Answers:

18x

$\left ( 3 + 3\sqrt{5}\right )x$

12x

$\left ( 3 + 4\sqrt{5}\right )x$

Correct answer:

$\left ( 3 + 3\sqrt{5}\right )x$

Explanation:

$\overline{BE }$ is both one side of Square BCDE and the hypotenuse of $\Delta ABE$ ; its hypotenuse can be calculated from the lengths of the legs using the Pythagorean Theorem:

$BE = \sqrt{(AB)^{2}+ \left ( AE\right )^{2}}$

$BE = \sqrt{x^{2}+ \left ( 2x\right )^{2}} = \sqrt{x^{2}+ 4x ^{2}} = \sqrt{5x ^{2}} = x\sqrt{5}$

Therefore, $BC = CD = DE = x\sqrt{5}$ .

The perimeter of Polygon ABCDE is

P = AB +BC + CD + DE + EA

$P=x + x\sqrt{5} + x\sqrt{5} + x\sqrt{5}+ 2x$

$P=3x + 3x\sqrt{5} = \left ( 3 + 3\sqrt{5}\right )x$

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Example Question #3 : Perimeter

The area of a rectangle is 16. Assuming the length and width are integers, which of the answers is NOT a possible perimeter?

Possible Answers:

$\textup{All of the choices are possible perimeters.}$

Correct answer:

Explanation:

The area of a rectangle is length times width.

A=LW

Determine all the integer combinations that will multiply to an area of 16. The numbers can represent length or width interchangably.

(1,16),(2,8),(4,4)

Write the perimeter formula for a rectangle.

P=2(L+W)

Substitute all the following combinations to determine the perimeters.

P=2(1+16)= 2(17)=34

P=2(2+8)= 2(10)=20

P=2(4+4)= 2(8)=16

The maximum allowable perimeter is 34.

The perimeter that is not possible is .

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SAT II Math I : SAT Subject Test in Math I

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

All SAT II Math I Resources

Example Questions

Example Question #11 : Geometry

Example Question #11 : 2 Dimensional Geometry

Example Question #11 : Area

Example Question #11 : Area

Example Question #451 : Sat Subject Test In Math I

Example Question #22 : Area

Example Question #1 : Perimeter

Example Question #1 : Perimeter

Example Question #2 : Perimeter

Example Question #3 : Perimeter

All SAT II Math I Resources

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