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

Answer: B

The value of 3x - 12 is 9.

Euler's formula for polyhedra links vertices, edges and faces:

\[ \begin{aligned} V - E + F &= 2 \\ 10 - 15 + x &= 2 \Rightarrow x = 7 \\ 3x - 12 &= 21 - 12 = 9 \end{aligned} \]

A polyhedron with 10 vertices, 15 edges and 7 faces is a pentagonal prism. 18 would follow from x = 10, which comes from misreading the formula as F = V, and 7 is the value of x itself rather than of 3x - 12.

Answer: D

The value of 3p - 50° is 190°.

Supplementary angles add up to 180°. With the larger angle L and the smaller p:

\[ \begin{aligned} L + p &= 180^{\circ},\quad L - p = 20^{\circ} \\ 2p &= 160^{\circ} \Rightarrow p = 80^{\circ} \\ 3p - 50^{\circ} &= 240^{\circ} - 50^{\circ} = 190^{\circ} \end{aligned} \]

Using the larger angle (100°) gives 250°, option C. 310° and 270° do not come from either angle.

Answer: D

The true statement is \(\Delta ABC \cong \Delta FED\), by AAS.

In triangle ABC, \(\angle B = 90^{\circ}\), \(\angle A = 40^{\circ}\), so \(\angle C = 50^{\circ}\), and BC (opposite A) is 8 cm. In triangle DEF, \(\angle E = 90^{\circ}\), \(\angle F = 40^{\circ}\), so \(\angle D = 50^{\circ}\), and DE (opposite F) is 8 cm.

Matching equal angles gives A with F, B with E and C with D, and the equal sides BC and ED are each opposite the 40° angle. Two angles and a non-included side are equal, so the triangles are congruent by AAS in the order ABC and FED.

RHS needs the hypotenuses to be equal, which is not given, and the correspondence DEF or DFE pairs unequal angles.

Answer: B

The triangle can be obtuse and scalene when \(\angle R = 15^{\circ}\) and PR \(\gt\) 18 cm.

With \(\angle P = 55^{\circ}\) and \(\angle R = 15^{\circ}\), \(\angle Q = 110^{\circ}\), an obtuse angle. The longest side is opposite the largest angle, so PR (opposite Q) is longer than QR = 18 cm, as stated, and all three angles differ, so the sides do too.

With \(\angle R = 25^{\circ}\), PQ = QR would force \(\angle R = \angle P\), a contradiction. \(\angle R = 65^{\circ}\) gives \(\angle Q = 60^{\circ}\), not obtuse. \(\angle R = 35^{\circ}\) gives a right angle at Q.

Answer: C

11, 60, 63 is not a Pythagorean triplet.

A Pythagorean triplet satisfies \(a^{2} + b^{2} = c^{2}\) with c the largest number. Check each:

\[ \begin{aligned} 7^{2} + 24^{2} &= 49 + 576 = 625 = 25^{2} \\ 8^{2} + 15^{2} &= 64 + 225 = 289 = 17^{2} \\ 13^{2} + 84^{2} &= 169 + 7056 = 7225 = 85^{2} \\ 11^{2} + 60^{2} &= 121 + 3600 = 3721 \end{aligned} \]

3721 is \(61^{2}\), not \(63^{2} = 3969\). So 11, 60, 61 is a triplet, but 11, 60, 63 is not.

Answer: C

The angle is 60°.

Let the angle be x. Its supplement is 180° - x and its complement is 90° - x. The supplement is four times the complement:

\[ \begin{aligned} 180 - x &= 4(90 - x) \\ 180 - x &= 360 - 4x \\ 3x &= 180, \quad x = 60 \end{aligned} \]

Check: the supplement is 120° and the complement is 30°, and 120 = 4 × 30. For 30° the ratio is 150 : 60, for 45° it is 135 : 45 = 3, and for 75° it is 105 : 15 = 7, so none of those works.

Answer: A

The triangle must be obtuse.

The angles of a triangle add up to 180°:

\[ \begin{aligned} x + (3x + 20) + 6x &= 180 \\ 10x &= 160, \quad x = 16 \end{aligned} \]

The angles are 16°, 3(16) + 20 = 68° and 6(16) = 96°. One angle is greater than 90°, so the triangle is obtuse.

It is not acute (which needs all angles under 90°) or right (which needs a 90° angle). It is not isosceles either, since all three angles are different.

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.