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

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.

Answer: C

The radius is \(r_{2} = 7 - 4\sqrt{2}\) cm, about 1.34 cm.

Put the corner W at the origin with WX along the x-axis and WV along the y-axis, so the square runs from 0 to 4 in each direction. Circle \(C_{1}\) touches VW and WX, so its centre is \(O_{1} = (1, 1)\). Circle \(C_{2}\) touches VY and XY, the two sides meeting at Y, so \(O_{2} = (4 - r_{2},\ 4 - r_{2})\).

Both centres lie on the diagonal WY, and since the circles touch externally, \(O_{1}O_{2} = 1 + r_{2}\):

\[ \begin{aligned} \sqrt{2}\,(3 - r_{2}) &= 1 + r_{2} \\ r_{2}(1 + \sqrt{2}) &= 3\sqrt{2} - 1 \\ r_{2} &= (3\sqrt{2} - 1)(\sqrt{2} - 1) \\ &= 6 - 3\sqrt{2} - \sqrt{2} + 1 = 7 - 4\sqrt{2} \end{aligned} \]

The other values are impossible: \(4 - 3\sqrt{2}\) is negative, and \(1 + 2\sqrt{2} \approx 3.83\) and \(5 + 3\sqrt{2} \approx 9.24\) give circles wider than the 4 cm square.

Answer: B

The ratio is 7/2.

Let each chalk-stick have radius \(r\). In the hexagonal packing, the outer sticks touch the central one, so their centres are \(2r\) from the axis and their outer edges \(3r\) from it. The container's radius is therefore \(3r\).

\[ \begin{aligned} \text{occupied} &= 7\pi r^{2} \\ \text{container} &= \pi(3r)^{2} = 9\pi r^{2} \\ \text{empty} &= 2\pi r^{2} \end{aligned} \]

The common length cancels, so occupied : empty = 7 : 2. 9/2 compares the container with the empty space, and 5/2 and 3 do not follow from the radius \(3r\).

Answer: A

The radius is \(r_3 = \frac{1}{2}(-3 + \sqrt{17})\) cm.

Circles that touch externally have their centres separated by the sum of the radii, so the sides of triangle \(O_1O_3O_2\) are \(O_1O_3 = 2 + r_3\), \(O_2O_3 = 1 + r_3\) and \(O_1O_2 = 2 + 1 = 3\). The right angle is at \(O_3\), so by Pythagoras:

\[ \begin{aligned} (2 + r)^2 + (1 + r)^2 &= 3^2 \\ 2r^2 + 6r + 5 &= 9 \\ r^2 + 3r - 2 &= 0 \\ r &= \frac{-3 + \sqrt{17}}{2} \end{aligned} \]

taking the positive root (about 0.56 cm). The option with \(+3\) is the negative root with its sign dropped and is far too large; \(-2\) and \(2\sqrt{17}\) do not satisfy the quadratic.

Answer: D

The cone turns through \(\frac{10\pi}{13}\) radians about P.

The cone rolls on its slant side with the apex P fixed, so each point of the base rim traces a circle on the ground of radius equal to the slant height:

\[ \begin{aligned} l &= \sqrt{5^{2} + 12^{2}} = 13\ \text{cm} \\ \text{arc} &= 2\pi r = 10\pi\ \text{cm} \\ \theta &= \frac{10\pi}{13} \end{aligned} \]

Q touches the ground again after the base has rolled once, laying down a length equal to its circumference along that circle. Using the height 12 in place of the slant height gives \(10\pi/12 = 5\pi/6\), not offered; \(5\pi/12\) and \(5\pi/24\) come from further slips, and \(24\pi/5\) inverts the ratio.

Answer: A

The diameter is \(\frac{1}{\sqrt{2}}\).

Joining the midpoints of a unit square's sides gives a square (a rhombus with right angles) whose side is the distance between adjacent midpoints:

\[ \begin{aligned} s &= \sqrt{\left(\tfrac{1}{2}\right)^{2} + \left(\tfrac{1}{2}\right)^{2}} \\ &= \frac{1}{\sqrt{2}} \end{aligned} \]

The largest circle inside a square touches all four sides, so its diameter equals the side, \(1/\sqrt{2}\).

\(1/(2\sqrt{2})\) is the radius rather than the diameter. \(\sqrt{2}\) is the diagonal of the original square, and \(2\sqrt{2}\) is larger than the square itself.

Answer: B

The relation is Area of SPQT = Area of PXWR − Area of RUVZ.

PQ is tangent to the circle at Q, so the radius RQ is perpendicular to PQ and triangle PQR is right-angled at Q. By Pythagoras, \(PR^{2} = PQ^{2} + QR^{2}\).

The square PXWR is built on PR, so its area is \(PR^{2}\). The square SPQT is built on PQ, area \(PQ^{2}\). The square RUVZ has sides RU and RZ, both radii, so its area is \(QR^{2}\) (QR is also a radius). Hence

\[ PQ^{2} = PR^{2} - QR^{2} \]

that is SPQT = PXWR − RUVZ. The three areas cannot all be equal, since PR is the hypotenuse and longer than both PQ and QR; the other two options subtract the wrong pair.

Answer: C

\(r_{2} = 7 - 4\sqrt{2}\) cm.

Each circle is tangent to the two sides at its corner, so its centre lies on the diagonal WY, at a distance \(r\sqrt{2}\) from the corner. The diagonal is \(4\sqrt{2}\), and along it lie the corner distance of \(C_{1}\), both radii, and the corner distance of \(C_{2}\):

\[ \begin{aligned} r_{1}\sqrt{2} + r_{1} + r_{2} + r_{2}\sqrt{2} &= 4\sqrt{2} \\ (r_{1} + r_{2})(\sqrt{2} + 1) &= 4\sqrt{2} \\ r_{1} + r_{2} &= 4\sqrt{2}(\sqrt{2} - 1) = 8 - 4\sqrt{2} \\ r_{2} &= 7 - 4\sqrt{2} \approx 1.34 \end{aligned} \]

\(4 - 3\sqrt{2}\) is negative and \(1 + 2\sqrt{2} \approx 3.83\) and \(5 + 3\sqrt{2}\) are too big to fit in a 4 cm square beside a 1 cm circle.