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

Answer: A

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

Let the distance from P to QR be h. ST is parallel to QR at distance h/2 from it, so it is also h/2 from P. Triangles PST and PQR are therefore similar with ratio 1 : 2, and areas of similar figures scale as the square of the ratio:

\[ \begin{aligned} \frac{[PST]}{[PQR]} &= \left(\frac{1}{2}\right)^{2} = \frac{1}{4} \\ [SQRT] &= [PQR] - [PST] = \frac{3}{4}[PQR] \\ \frac{[PST]}{[SQRT]} &= \frac{1/4}{3/4} = \frac{1}{3} \end{aligned} \]

1/4 is the ratio of the small triangle to the whole triangle, not to the trapezium. 1/2 confuses the ratio of lengths with the ratio of areas, and 2/5 matches no step of the working.

Answer: C

Each angle of a rectangle is a right angle.

A rectangle is a quadrilateral whose four angles are all equal. Since the angles of any quadrilateral add up to 360°, each angle is 360°/4 = 90°, a right angle.

An acute angle is less than 90° and an obtuse angle more than 90° (but less than 180°), so neither can be an angle of a rectangle. A straight angle is exactly 180°, which would flatten the figure into a line.

Answer: A

The shadow is 1.6 m long.

After 4 seconds the object is 1.2 × 4 = 4.8 m from the lamp post. Let the shadow length be x. The lamp, the top of the object and the tip of the shadow lie on one line, so the triangles formed by the lamp post and by the object are similar:

\[ \begin{aligned} \frac{3.6}{4.8 + x} &= \frac{0.9}{x} \\ 3.6x &= 0.9x + 4.32 \\ 2.7x &= 4.32 \\ x &= 1.6\ \text{m} \end{aligned} \]

Check: the lamp is four times as high as the object, so the shadow is one third of the distance walked, 4.8/3 = 1.6 m. 4.8 is the distance walked (and is printed in cm), and 0.9 m is the object's height.

Answer: B

The values are x = 30° and y = 40°.

RP passes through the centre, so it is a diameter, and an angle in a semicircle is a right angle. So angle RSP = 90° and angle RQP = 90°.

\[ \begin{aligned} \text{In } \triangle RSP: y &= 180^\circ - 90^\circ - 50^\circ = 40^\circ \\ \text{In } \triangle RQP: x &= 180^\circ - 90^\circ - 60^\circ = 30^\circ \end{aligned} \]

The first option gives the same two numbers in the wrong order (x = 40°, y = 30°), the commonest mistake here. 50° and 40° reuse the given angle instead of working out x.

Answer: D

A parallelogram with perpendicular diagonals is a rhombus.

The diagonals of a parallelogram bisect each other. If they are also perpendicular, each diagonal is the perpendicular bisector of the other, which makes all four sides equal: the parallelogram is a rhombus.

A square also has perpendicular diagonals, but it is a special rhombus with right angles, which the condition does not require; the general answer is rhombus. A rectangle's diagonals are equal but not perpendicular unless it is a square, and a trapezium is not a parallelogram.

Answer: C

Angle B is 40°.

Since AB = AC, the triangle is isosceles and the base angles B and C are equal. The angles of a triangle sum to 180°:

\[ \begin{aligned} \angle B + \angle C &= 180^{\circ} - 100^{\circ} = 80^{\circ} \\ \angle B &= \frac{80^{\circ}}{2} = 40^{\circ} \end{aligned} \]

50° comes from halving 100° instead of the remaining 80°, and 60° and 45° would not give a total of 180° with equal base angles.

Answer: A

The hypotenuse is 4 cm.

A right-angled isosceles triangle with equal sides \(a\) has hypotenuse \(a\sqrt{2}\), so

\[ \begin{aligned} 2a + a\sqrt{2} &= 4 + 4\sqrt{2} \\ a\sqrt{2}(\sqrt{2} + 1) &= 4(1 + \sqrt{2}) \\ a\sqrt{2} &= 4 \end{aligned} \]

Since \(a\sqrt{2}\) is the hypotenuse, it is 4 cm, and each leg is \(2\sqrt{2}\) cm. Check: \(2\sqrt{2} + 2\sqrt{2} + 4 = 4\sqrt{2} + 4\).

\(2\sqrt{2}\) cm is the length of each equal side, not the hypotenuse. 5 cm and 6 cm do not satisfy the perimeter.

Answer: A

The street is about 15.2 m wide.

The ladder's foot stays at one point, so the street's width is the sum of the two horizontal distances from the foot to each wall:

\[ \begin{aligned} \text{side 1} &= \sqrt{13^2 - 8^2} \\ &= \sqrt{105} \approx 10.25 \\ \text{side 2} &= \sqrt{13^2 - 12^2} = 5 \\ \text{width} &\approx 15.25 \text{ m} \end{aligned} \]

The paper's 15·2 m is this value cut to one decimal. 14 m and 14·6 m are both shorter than 10.25 + 5, and 12 m is the height of the second window, not a width.

Answer: B

The distance is 13 cm.

A tangent is perpendicular to the radius at the point of contact, so the radius, the tangent and the line from P to the centre form a right-angled triangle. The radius is half the diameter, 5 cm:

\[ OP = \sqrt{5^{2} + 12^{2}} = \sqrt{169} = 13\ \text{cm} \]

Using the diameter, 10 cm, as a side instead of the radius gives about 15.6 cm, near the 15 cm option; 12 cm is just the tangent length, and 10 cm the diameter.

Answer: D

Let the smaller side be b and the diagonal d = 3b. The other side l satisfies \[l^{2}=d^{2}-b^{2}=9b^{2}-b^{2}=8b^{2},\] so \(l=2\sqrt{2}\,b\).

The ratio of length to breadth is \(2\sqrt{2}:1\), option D. The ratio \(3:1\) is diagonal to side, \(\sqrt{3}:1\) and \(\sqrt{2}:1\) do not satisfy the Pythagoras relation here.

Check with numbers: if the smaller side is 1, the diagonal is 3 and the larger side is the square root of 8, which is 2 times the square root of 2, so the ratio is 2.828 to 1.