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

Answer: A

The letter with no line of symmetry is L.

No line divides L into two mirror-image halves, vertically, horizontally or diagonally.

A and M each have a vertical line of symmetry, and X has both a vertical and a horizontal one.

Answer: C

The perimeter is 6.

An equilateral triangle of side \(a\) has area \(\frac{\sqrt{3}}{4}a^{2}\). Setting this equal to \(\sqrt{3}\):

\[ \begin{aligned} \frac{\sqrt{3}}{4}a^{2} &= \sqrt{3} \\ a^{2} &= 4, \quad a = 2 \end{aligned} \]

The perimeter is \(3a = 6\). 2 is the side itself, 4 is \(a^{2}\) or two sides, and 8 does not follow from the area.

Answer: B

RS is 28/5 cm.

The area of a parallelogram is any side times the perpendicular distance to it from the opposite side. PT is the height on base QR, and PV is the height on base RS.

\[ \begin{aligned} QR \times PT &= RS \times PV \\ 7 \times 4 &= RS \times 5 \\ RS &= \frac{28}{5} = 5.6\ \text{cm} \end{aligned} \]

QR equals PS = 7 cm because opposite sides of a parallelogram are equal. 20/7 is PT times PV divided by PS, which mixes the two heights, and 35/4 is PS times PV over PT, which pairs each height with the wrong base.

Answer: A

The ratio is \(\frac{1}{3}\).

Let the distance from P to QR be \(h\). The gap between ST and QR is \(h/2\), so ST is at a distance \(h/2\) from P, halfway down. Because ST is parallel to QR, triangle PST is similar to triangle PQR with ratio \(1:2\), and areas scale as the square of that ratio:

\[ \begin{aligned} \text{Area}(PST) &= \tfrac{1}{4}\,\text{Area}(PQR) \\ \text{Area}(SQRT) &= \tfrac{3}{4}\,\text{Area}(PQR) \\ \text{ratio} &= \frac{1/4}{3/4} = \frac{1}{3} \end{aligned} \]

\(\frac{1}{4}\) is the ratio of triangle PST to the whole triangle PQR, not to the trapezium, and \(\frac{1}{2}\) is the ratio of the lengths ST and QR, not of the areas.

Answer: A

The radius is \(r_{3} = \frac{1}{2}(-3 + \sqrt{17})\) cm.

When two circles touch externally, the distance between their centres is the sum of their radii. So \(O_{1}O_{2} = 2 + 1 = 3\), \(O_{1}O_{3} = 2 + r_{3}\) and \(O_{2}O_{3} = 1 + r_{3}\). The right angle at \(O_{3}\) makes \(O_{1}O_{2}\) the hypotenuse of triangle \(O_{1}O_{3}O_{2}\):

\[ \begin{aligned} (2 + r)^{2} + (1 + r)^{2} &= 3^{2} \\ 2r^{2} + 6r + 5 &= 9 \\ r^{2} + 3r - 2 &= 0 \end{aligned} \]

\[ \begin{aligned} r &= \frac{-3 \pm \sqrt{9 + 8}}{2} \\ &= \frac{-3 \pm \sqrt{17}}{2} \end{aligned} \]

A radius must be positive, so \(r_{3} = (-3 + \sqrt{17})/2 \approx 0.56\) cm, which fits the small circle in the figure. \((3 + \sqrt{17})/2\) has the wrong sign on 3, \((-2 + \sqrt{17})/2\) has the wrong linear coefficient, and \((-3 + 2\sqrt{17})/2 \approx 2.6\) would make \(C_{3}\) larger than \(C_{1}\).

Answer: D

The area of the rectangle is 4 cm².

Look at the triangle PLS. Measured against the square, it has base PS (a side, 2 cm) and its third vertex L on the opposite side QR, so its height is also 2 cm:

\[ \tfrac{1}{2} \times 2 \times 2 = 2\ \text{cm}^{2} \]

Measured against the rectangle, the same triangle has base PL (a side of the rectangle) and its third vertex S on the opposite side MN. A triangle on one side of a rectangle with its apex on the opposite side has half the rectangle's area. So the rectangle's area is 2 × 2 = 4 cm², the same as the square's, wherever L lies on QR.

2 is the triangle's area, 8 doubles the square, and 2√2 is the length of the square's diagonal, not an area.

Answer: A

\(\angle DEC + \angle BFC\) equals \(\angle BCD - \angle BAD\).

In the figure A, D, E lie on one line and A, B, F on another, and EB and DF cross at C. Write \(A = \angle BAD\), \(E = \angle DEC\) and \(F = \angle BFC\).

Triangle AEB gives the angle at B, triangle AFD the angle at D, and the angles of quadrilateral ABCD sum to 360°:

\[ \begin{aligned} \angle ABC &= 180^\circ - A - E \\ \angle ADC &= 180^\circ - A - F \\ \angle BCD &= 360^\circ - A - \angle ABC - \angle ADC \\ &= A + E + F \end{aligned} \]

So \(E + F = \angle BCD - \angle BAD\). The options that add \(\angle BAD\) instead of subtracting it overshoot by \(2\angle BAD\), and \(\angle CBA + \angle ADC\) equals \(360^\circ - 2A - E - F\), not \(E + F\).

Answer: A

The angle θ is 36 degrees.

Five equal segments joined this way form a regular pentagram, whose vertices lie on a regular pentagon. The angle at a point of the star is an inscribed angle standing on one side of that pentagon, which subtends 360/5 = 72 degrees at the centre:

\[ \theta = \frac{1}{2} \times 72^{\circ} = 36^{\circ} \]

Check: the five point angles of any five-pointed star add up to 180 degrees, and 5 × 36 = 180. 72 degrees is the central angle, 108 degrees is the interior angle of the pentagon, and 45 has no place in a five-fold figure.

Answer: A

The outline is a rhombus.

A plane through both apexes contains the common axis of the two cones. It cuts each cone in an isosceles triangle whose base is a diameter of the shared circular base. The two triangles share that diameter as a common side and point in opposite directions, so together they form a quadrilateral with all four sides equal to the slant height: a rhombus.

A single triangle would come from cutting one cone only. An ellipse arises when a plane cuts a cone without passing through its apex, and a hexagon cannot come from two cones.

Answer: A

The answer is 3 squares, with 4 triangles each.

The dodecagon is made of 12 central triangles with apex angle 30 degrees, so its area is

\[ 12 \times \tfrac{1}{2} r^2 \sin 30^{\circ} = 3r^2, \]

exactly three squares of side r. Twelve triangles shared among 3 squares means 4 per square.

The pieces do fit that way. Triangles 1, 4 and 7 are equilateral with side r (area \(\frac{\sqrt{3}}{4}r^2\)), each thin edge triangle 10, 11, 12 has area \(\frac{2 - \sqrt{3}}{4}r^2\), and the six 30-degree triangles have area \(\frac{1}{4}r^2\) each. One equilateral, one thin and two 30-degree triangles total \(r^2\). 4 squares would need more area than the dodecagon has, and 3 or 2 triangles per square would leave pieces over.