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SAT II Math I : Other 2-Dimensional 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 #1 : Other 2 Dimensional Geometry

Which of the following describes a triangle with sides of length 10 inches, 1 foot, and 2 feet?

Possible Answers:

This is an obtuse triangle.

More information is needed to answer this question.

This is an acute triangle.

This is a right triangle.

This triangle cannot exist.

Correct answer:

This triangle cannot exist.

Explanation:

One foot is equal to 12 inches, so the triangle would have sides 10, 12, and 24 inches. Since

10 + 12 = 22 < 24 ,

the triangle violates the Triangle Inequality, which states that the sum of the lengths of the two smaller sides must exceed the length of the third. The triangle cannot exist.

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Example Question #2 : Other 2 Dimensional Geometry

Which of the following describes a triangle with sides of length nine yards, thirty feet, and 360 inches?

Possible Answers:

The triangle is obtuse and isosceles.

The triangle is obtuse and scalene.

The triangle is acute and scalene.

The triangle cannot exist.

The triangle is acute and isosceles.

Correct answer:

The triangle is acute and isosceles.

Explanation:

Nine yards is equal to $9 \times 36 = 324$ inches.

30 feet is equal to $30 \times 12 = 360$ inches.

In terms of inches, the triangle has sides of length 324, 360, 360; this exists since

324 + 360 > 360

and this is an isosceles triangle, since two sides have the same length.

Also,

$324^{2} + 360^{2} > 360^{2}$ ,

making the triangle acute.

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Example Question #3 : Other 2 Dimensional Geometry

Thingy_5

Refer to the above diagram. Which of the following choices gives a set of collinear points?

Possible Answers:

Z,G,H

A,C,H

A,F,G

Z,D,E

Y,C,X

Correct answer:

Z,D,E

Explanation:

Collinear points are points that are contained in the same line. Of the four choices, only Z,D,E fit the description, since all are on Line .

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Example Question #4 : Other 2 Dimensional Geometry

You are given triangles $\Delta MNO$ and $\Delta PQR$ , with $\angle N \cong \angle Q$ . Which of these statements, along with what you are given, is enough to prove that $\Delta MNO \sim \Delta PQR$ ?

Possible Answers:

None of the other responses is correct.

$\Delta MNO$ and $\Delta PQR$ have the same perimeter.

$\frac{MN}{PQ} = \frac{MO}{PR}$

MN = PQ

$\angle M \cong \angle P$

Correct answer:

$\angle M \cong \angle P$

Explanation:

MN = PQ gives us the congruence of two corresponding angles and one corresponding side; this is not enough to establish similarity.

The perimeters of the triangles are irrelevant to their similarity, so $\Delta MNO$ and $\Delta PQR$ having the same perimeter does not help to establish similarity, with or without what is given.

$\frac{MN}{PQ} = \frac{MP}{PR}$ establishes the proportionality of two nonincluded sides of the angles known to be congruent. However, there is no statement that establishes similarity as a result of this.

$\angle M \cong \angle P$ , along with $\angle N \cong \angle Q$ , sets up the conditions of the Angle-Angle Similarity Postulate, which states that if two triangles have two pairs of congruent angles between them, the triangles are similar. $\angle M \cong \angle P$ is the correct choice.

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Example Question #77 : 2 Dimensional Geometry

Regular Octagon ABCDEFGH has perimeter 80. $\overline{AB }$ has as its midpoint; segment $\overline{ME}$ is drawn. To the nearest tenth, give the length of $\overline{ME}$ .

Possible Answers:

30.4

24.1

26.1

24.7

21.8

Correct answer:

24.7

Explanation:

Below is the regular Octagon ABCDEFGH , with the referenced midpoint and segment $\overline{ME}$ . Note that perpendiculars have also been constructed from and to meet $\overline{BE}$ at and , respectively.

Octagon 2

$\bigtriangleup MBE$ is a right triangle with legs $\overline{MB}$ and $\overline{ BE}$ and hypotenuse $\overline{ME}$ .

The perimeter of the regular octagon is 80, so the length of each side is one-eighth of 80, or 10. Consequently,

$MB = \frac{1}{2} AB = \frac{1}{2} \cdot 10 = 5$

To find the length of $\overline{ BE}$ , we can break it down as

BE = BX + XY + YE

Quadrilateral CDYX is a rectangle, so XY = CD = 10 .

$\bigtriangleup BXC$ is a 45-45-90 triangle with leg $\overline{BX}$ and hypotenuse $\overline{BC}$ ; by the 45-45-90 Triangle Theorem,

$BX = \frac{BC }{\sqrt{2}} \approx \frac{10}{1.4142} \approx 7.0711$

For similar reasons, $EY \approx 7.0711$ .

Therefore,

$BE = BX + XY + YE \approx 7.0711 + 10 + 7.0711 \approx 24.1421$

can now be evaluated using the Pythagorean Theorem:

$ME = \sqrt{(MB)^{2}+ ( BE ) ^{2}}$

Substituting and evaluating:

$ME = \sqrt{5 ^{2}+24.1421^{2}}$

$\approx \sqrt{25 +582.8427}$

$\approx \sqrt{ 607.8427}$

$\approx 24.7$ ,

the correct choice.

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SAT II Math I : Other 2-Dimensional Geometry

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

All SAT II Math I Resources

Example Questions

Example Question #1 : Other 2 Dimensional Geometry

Example Question #2 : Other 2 Dimensional Geometry

Example Question #3 : Other 2 Dimensional Geometry

Example Question #4 : Other 2 Dimensional Geometry

Example Question #77 : 2 Dimensional Geometry

All SAT II Math I Resources

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