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SAT II Math I : SAT Subject Test in Math I

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Example Questions

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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Example Question #2 : Finding Sides

In triangle $\Delta GHJ$ , $m\angle G = 50^{\circ }$ and $m \angle H = 65^{\circ }$ .

Which of the following statements is true about the lengths of the sides of $\Delta GHJ$ ?

Possible Answers:

HJ < GH < GJ

HJ < GH = GJ

GH < HJ< GJ

HJ< GH < GJ

HJ< GJ < GH

Correct answer:

HJ < GH = GJ

Explanation:

In a triangle, the shortest side is opposite the angle of least measure; the longest side is opposite the angle of greatest measure. Therefore, if we order the angles, we can order their opposite sides similarly.

Since the measures of the three interior angles of a triangle must total $180^{\circ }$ ,

$m \angle J + m \angle G + m \angle H = 180^{\circ }$

$m \angle J + 50 ^{\circ }+ 65^{\circ } = 180^{\circ }$

$m \angle J + 115^{\circ } = 180^{\circ }$

$m \angle J =65^{\circ }$

$\angle G$ has the least degree measure, so its opposite side, $\overline{HJ}$ , is the shortest. $\angle H \cong \angle J$ , so by the Isosceles Triangle Theorem, their opposite sides $\overline{GJ}$ and $\overline{GH}$ are congruent. Therefore, the correct choice is

HJ < GH = GJ .

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Example Question #471 : Sat Subject Test In Math I

Which of the following describes a triangle with sides one kilometer, 100 meters, and 100 meters?

Possible Answers:

The triangle is obtuse and scalene.

The triangle is obtuse and isosceles, but not equilateral.

The triangle is acute and equilateral.

The triangle cannot exist.

The triangle is acute and isosceles, but not equilateral.

Correct answer:

The triangle cannot exist.

Explanation:

One kilometer is equal to 1,000 meters, so the triangle has sides of length 100, 100, and 1,000. However,

100 + 100 = 200 < 1,000

That is, the sum of the least two sidelengths is not greater than the third. This violates the Triangle Inequality, and this triangle cannot exist.

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Example Question #6 : Finding Sides

Right_triangle

Note: Figure NOT drawn to scale.

Refer to the above diagram. Evaluate .

Possible Answers:

$N = 3\frac{6}{13}$

$N = 4\frac{1}{2}$

The correct answer is not among the other responses.

N = 3

N = 4

Correct answer:

The correct answer is not among the other responses.

Explanation:

The altitude perpendicular to the hypotenuse of a right triangle divides that triangle into two smaller triangles similar to each other and the large triangle. Therefore, the sides are in proportion. The hypotenuse of the triangle is equal to

$\sqrt{5^{2}+12^{2}} = \sqrt{25+144} = \sqrt{169} =13$

Therefore, we can set up, and solve for in, a proportion statement involving the shorter side and hypotenuse of the large triangle and the larger of the two smaller triangles:

$\frac{N}{12} = \frac{5}{13}$

$\frac{N}{12} \times 12 = \frac{5}{13} \times 12$

$N = \frac{60}{13} = 4\frac{8}{13}$

This is not one of the choices.

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Example Question #472 : Sat Subject Test In Math I

What is the sum of three sides of a square if the fourth side has a length of ?

Possible Answers:

Correct answer:

Explanation:

All of the sides of a square have the same length.

That means all four sides of this square have a length of .

The sum of three of them would then be

5+5+5=15 .

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Example Question #1 : Quadrilaterals

Screen_shot_2015-03-06_at_6.20.29_pm

The area of the rectangle is , what is the area of the kite?

Possible Answers:

Correct answer:

Explanation:

The area of a kite is half the product of the diagonals.

$A=\frac{p \cdot q}{2}$

The diagonals of the kite are the height and width of the rectangle it is superimposed in, and we know that because the area of a rectangle is base times height.

Therefore our equation becomes:

$A=\frac{l \cdot w}{2}$ .

We also know the area of the rectangle is $A=l \cdot w=80$ . Substituting this value in we get the following:

$A=\frac{l \cdot w}{2}=\frac{80}{2}=40$

Thus,, the area of the kite is .

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Example Question #1 : How To Find The Length Of The Diagonal Of A Kite

Kite vt act

Using the kite shown above, find the length of the red (vertical) diagonal.

Possible Answers:

15.5

19.5

17.5

Correct answer:

Explanation:

In order to solve this problem, first observe that the red diagonal line divides the kite into two triangles that each have side lengths of and Notice, the hypotenuse of the interior triangle is the red diagonal. Therefore, use the Pythagorean theorem: a^2+b^2=c^2 , where the length of the red diagonal.

The solution is:

8^2+15^2=c^2

64+225=c^2

c^2=289

$c=\sqrt{289}=\sqrt{17\times 17}=17$

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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 #1 : Finding Sides

Example Question #31 : Geometry

Example Question #1 : Finding Sides

Example Question #4 : Finding Sides

Example Question #2 : Finding Sides

Example Question #471 : Sat Subject Test In Math I

Example Question #6 : Finding Sides

Example Question #472 : Sat Subject Test In Math I

Example Question #1 : Quadrilaterals

Example Question #1 : How To Find The Length Of The Diagonal Of A Kite

All SAT II Math I Resources

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