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

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.

Answer: C

The shortest connecting road is 12.0 km.

The shortest road from P to the line QR is the perpendicular, the altitude of the triangle on base QR. Find the area by Heron's formula:

\[ \begin{aligned} s &= \frac{13+14+15}{2} = 21 \\ A &= \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84\ \text{km}^{2} \end{aligned} \]

\[ h = \frac{2A}{QR} = \frac{168}{14} = 12\ \text{km} \]

The foot of this perpendicular lies between Q and R (5 km from Q), so the road meets QR itself. 12.5 and 11.0 are not the altitude, and 10.5 is shorter than any possible road from P.

Answer: A

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

Circles that touch externally have their centres a distance apart equal to the sum of their radii, so \(O_{1}O_{3} = 2 + r_{3}\) and \(O_{2}O_{3} = 1 + r_{3}\), and \(O_{1}O_{2} = 3\). The angle at O₃ is a right angle, so by Pythagoras:

\[ \begin{aligned} (2+r_{3})^{2}+(1+r_{3})^{2} &= 3^{2} \\ 2r_{3}^{2}+6r_{3}-4 &= 0 \\ r_{3}^{2}+3r_{3}-2 &= 0 \end{aligned} \]

\[ \begin{aligned} r_{3} &= \frac{-3 \pm \sqrt{9+8}}{2} \\ &= \frac{-3+\sqrt{17}}{2} \approx 0.56 \end{aligned} \]

The negative root is rejected. \(\frac{1}{2}(3+\sqrt{17}) \approx 3.56\) would be larger than C₁; the other two forms do not satisfy the quadratic.

Answer: C

The shortest road is 12.0 km, the perpendicular from P to QR.

The shortest path from a point to a line is the perpendicular, so we need the altitude from P onto QR. Find the triangle's area with Heron's formula, where \(s = (13 + 14 + 15)/2 = 21\):

\[ \begin{aligned} \text{Area} &= \sqrt{21 \times 8 \times 7 \times 6} \\ &= \sqrt{7056} = 84\ \text{km}^{2} \\ h &= \frac{2 \times 84}{14} = 12\ \text{km} \end{aligned} \]

The foot of the perpendicular falls between Q and R, 5 km from Q (since \(\sqrt{13^{2} - 12^{2}} = 5\)), so the road meets QR itself. No road can be shorter than this perpendicular, which rules out 10.5 and 11.0 km, and 12.5 km is longer than the minimum.