Geometry (Lines, Angles, Triangles, Polygons) Questions and Answers

Answer: A

The value of x is 1.

In the two triangles, AC = DF, BC = EF, and the included angles ∠C and ∠F are equal. By the SAS (side-angle-side) congruence rule, triangle ABC is congruent to triangle DEF, so the third sides are equal: AB = DE.

\[ \begin{aligned} 2x - 1 &= 5x - 4 \\ 3 &= 3x, \quad x = 1 \end{aligned} \]

Check: AB = 2(1) - 1 = 1 and DE = 5(1) - 4 = 1. Any other value of x would make AB and DE unequal, which the congruence rules out.

Answer: C

The least and greatest values of P are 21 and 29.

By the triangle inequality, the third side c must be greater than the difference of the other two sides and less than their sum:

\[ 10 - 5 \lt c \lt 10 + 5, \quad 5 \lt c \lt 15 \]

Since P = 15 + c is an integer, c is an integer from 6 to 14. So P runs from 15 + 6 = 21 to 15 + 14 = 29.

19 and 20 would need a third side of 4 or 5, which cannot close the triangle, and 28 or 27 as the greatest value stops short of c = 14, which is allowed.

Answer: B

The true statement is that a regular polygon of 10 sides has 10 lines of symmetry.

A regular polygon with n sides has exactly n lines of symmetry: for even n, lines through opposite vertices and through midpoints of opposite sides. So a regular decagon has 10.

The others are false. A regular hexagon has 6 lines of symmetry, not only 4. A circle has infinitely many, since every diameter is a line of symmetry. An angle has exactly one line of symmetry, its bisector, not two.

Answer: A

The third angle is 100°.

The angles of a triangle add up to 180°:

\[ 180^\circ - (50^\circ + 30^\circ) = 100^\circ \]

80° is the sum of the two given angles, not the third one, and 40° and 60° do not make the total 180°.

Answer: C

The correct match is (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i).

The flat face of a blackboard represents part of a plane. A line extends endlessly in both directions, so it has no definite length. A ray starts at one end point and goes on in one direction. A line segment has two end points and a fixed length.

Options 1 and 2 give the ray two end points or the line one end point, and option 4 makes the blackboard a segment.

Answer: C

The value of \(7y + 5\) is 26.

The diagonals of a parallelogram bisect each other, so OP = OR and OQ = OS:

\[ \begin{aligned} x + 7 &= 20 \Rightarrow x = 13 \\ x + y &= 16 \Rightarrow y = 3 \\ 7y + 5 &= 21 + 5 = 26 \end{aligned} \]

21 is 7y without the 5. Equating OP with OQ instead of with OR, as if all half-diagonals were equal, is the usual slip behind the other values.

Answer: B

The value is 19.

Euler's formula for a polyhedron gives the number of vertices:

\[ \begin{aligned} F + V - E &= 2 \\ 7 + x - 15 &= 2 \Rightarrow x = 10 \\ 2F + 3E - 4x &= 14 + 45 - 40 = 19 \end{aligned} \]

A wrong value of x, for example from writing F + E - V = 2, leads to the other options.

Answer: A

The value of \(4x + 6^{\circ}\) is 154°.

Interior angles on the same side of a transversal cutting two parallel lines are supplementary:

\[ \begin{aligned} (2x + 15) + (3x - 20) &= 180 \\ 5x - 5 &= 180 \Rightarrow x = 37 \\ 4x + 6 &= 148 + 6 = 154 \end{aligned} \]

148° is 4x without adding the 6. Taking the two angles as equal, as for alternate angles, gives x = 35 and 146°.

Answer: B

The value of \(3x + 25^{\circ}\) is 85°.

Since AM = CM, ∠ACM = ∠A = x, and since BM = CM, ∠BCM = ∠B = 70°. So ∠C = x + 70°. Using the angle sum of triangle ABC:

\[ \begin{aligned} x + 70 + (x + 70) &= 180 \\ 2x &= 40 \Rightarrow x = 20 \\ 3x + 25 &= 60 + 25 = 85 \end{aligned} \]

The shortcut: M is equidistant from A, B and C, so it is the midpoint of the hypotenuse and ∠C = 90°, giving x = 20 directly.

Answer: D

∠ADC is 100°.

\[ \begin{aligned} \angle C &= 180 - 50 - 70 = 60 \\ \angle ACD &= 60 / 2 = 30 \\ \angle ADC &= 180 - 50 - 30 = 100 \end{aligned} \]

In triangle ADC the angles at A and C are 50° and 30°, so the third is 100°. 30° is ∠ACD itself, and 70° and 50° are the given angles, not ∠ADC.