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

Answer: C

The sum of the angles of a triangle is 180°.

Draw a line through one vertex parallel to the opposite side. The three angles at that vertex, which lie along the straight line, are equal to the three angles of the triangle (alternate angles), and angles on a straight line add up to 180°.

60° is one angle of an equilateral triangle, 90° is a right angle, and 360° is the angle sum of a quadrilateral.

Answer: C

The angle is 240 degrees.

One right angle is 90°.

\[ \begin{aligned} 2\tfrac{2}{3} &= \tfrac{8}{3} \\ \tfrac{8}{3} \times 90 &= 8 \times 30 = 240 \end{aligned} \]

210 is \(2\tfrac{1}{3}\) right angles and 330 is \(3\tfrac{2}{3}\), and 285 is not a whole number of thirds of a right angle at all.

Answer: C

The value of 2x + 3y is 22 (printed with a degree sign as 22°).

In a parallelogram opposite sides are parallel, so the diagonal BD makes equal alternate angles with them. AD is parallel to BC, so \(\angle CBD = \angle ADB\); AB is parallel to DC, so \(\angle ABD = \angle CDB\):

\[ \begin{aligned} 12x &= 60 \Rightarrow x = 5 \\ 7y &= 28 \Rightarrow y = 4 \\ 2x + 3y &= 10 + 12 = 22 \end{aligned} \]

Pairing the angles the other way (12x with 28 or 7y with 60) does not give whole-number answers that match any option, which is a useful check on which angles are alternate.

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.