AnswersFall 24 GMS Geometry Concepts and Connections BSolving for Angle Measures of Right Triangles

Central Angles Answers

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Which best explains why all equilateral triangles are similar?

A
All equilateral triangles can be mapped onto each other using dilations.
B
All equilateral triangles can be mapped onto each other using rigid transformations.
C
All equilateral triangles can be mapped onto each other using combinations of dilations and rigid transformations.
D
All equilateral triangles are congruent and therefore similar, with side lengths in a 1:1 ratio.
5

Triangle V U W is shown. The length of side W V is 6 centimeters, the length of side W U is 3 StartRoot 3 EndRoot centimeters, and the length of side U V is 3 centimeters.

Question illustration
A
m∠V = 30°, m∠U = 60°, m∠W = 90°
B
m∠V = 90°, m∠U = 60°, m∠W = 30°
C
m∠V = 30°, m∠U = 90°, m∠W = 60°
D
m∠V = 60°, m∠U = 90°, m∠W = 30°
6

On a coordinate plane, Rectangles A B C D and E F G H are shown. The length of side A B is 6 units and the length of side B C is 3 units. The length of side E F is 8 units and the length of side F G is 4 units.

Question illustration
A
Yes, because corresponding sides are parallel and have lengths in the ratio
Option A
B
Yes, because both figures are rectangles and all rectangles are similar.
C
No, because the center of dilation is not at (0, 0).
D
No, because corresponding sides have different slopes.
8

In the diagram, DG = 12, GF = 4, EH = 9, and HF = 3.

Question illustration
A
∠DFE is 4 times greater than ∠GFH.
B
∠FHG is the measure of ∠FED.
Option B
C
∠DFE is congruent to ∠GFH.
D
∠FHG is congruent to ∠EFD.
12

Which equation can be used to find the length of ?

Question illustration
A
(10)sin(40o) = AC
B
(10)cos(40o) = AC
C
= AC
Option C
D
= AC
Option D
13

Consider △RST and △RYX.

Question illustration
A
StartFraction R Y Over Y S EndFraction = StartFraction R X Over X T EndFraction = StartFraction X Y Over T S EndFraction
Option A
B
StartFraction R Y Over R S EndFraction = StartFraction R X Over R T EndFraction = StartFraction X Y Over T S EndFraction
Option B
C
StartFraction R Y Over R S EndFraction = StartFraction R X Over R T EndFraction = StartFraction R S Over R Y EndFraction
Option C
D
StartFraction R Y Over R X EndFraction = StartFraction R S Over R T EndFraction = StartFraction X Y Over T S EndFraction
Option D
14

In which triangle is the value of x equal to tan−1? (Images may not be drawn to scale.)

Question illustration
A
A right triangle is shown. The length of the hypotenuse is 5.2 and the length of the side adjacent to the right angle is 3.1. The angle between the 2 sides is x.
Option A
B
A right triangle is shown. The length of the hypotenuse is 5.2 and the length of the side adjacent to the right angle is 3.1. The angle opposite to side with length 3.1 is x.
Option B
C
A right triangle is shown. The length of 2 sides are 5.2 and 3.1. The angle opposite to side with length 5.2 is x.
Option C
D
A right triangle is shown. The length of 2 sides are 5.2 and 3.1. The angle opposite to side with length 3.1 is x.
Option D
15

Triangles R S T and V U T are connected at point T. Angles R S T and V U T are right angles. The length of side R S is 12 and the length of side S T is 16. The length of side T U is 8 and the length of U V is 6.

Question illustration
A
StartFraction R S Over V U EndFraction = StartFraction S T Over U T EndFraction and angle S is-congruent-to angle U
Option A
B
StartFraction R S Over V U EndFraction = StartFraction S T Over U T EndFraction = StartFraction R T Over V T EndFraction
Option B
C
StartFraction R S Over V U EndFraction = StartFraction T U Over T S EndFraction and angle S is-congruent-to angle U
Option C
D
StartFraction R S Over V U EndFraction = StartFraction T U Over T S EndFraction = StartFraction R T Over V T EndFraction
Option D
16

Triangles L M N and P O N connect at point N. Angles L M N and N O P are congruent.

Question illustration
A
because both triangles appear to be equilateral
B
because∠MNL and ∠ONP are congruent angles
C
because one pair of congruent corresponding angles is sufficient to determine similar triangles
D
because both triangles appear to be isosceles, ∠MLN ≅ ∠LMN, and ∠NOP ≅ ∠OPN

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