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

Answer: A

Only one statement is correct.

Statement 1 is the triangle inequality, which holds for every triangle: the sum of any two sides exceeds the third.

Statement 2 is reversed. A square is a special rectangle (one with all sides equal); a rectangle in general does not have all sides equal.

Statement 3 gives the circumference, \(2\pi r\), not the area, which is \(\pi r^{2}\).

So only statement 1 is right, and 'Only two' requires accepting one of the two common confusions.

Answer: B

The distance is \(2\sqrt{5}\) cm.

The perpendicular from the centre bisects the chord, so it meets a radius and half the chord (4 cm) in a right triangle:

\[ d = \sqrt{6^{2} - 4^{2}} = \sqrt{20} = 2\sqrt{5} \]

\(\sqrt{5}\) would come from halving the result. \(2\sqrt{7}\) is \(\sqrt{36 - 8}\), using the whole chord without squaring, and \(\sqrt{7}\) is half of that.

Answer: C

The value of ∠ BCD is 80°.

Angles in the same segment of a circle are equal. ∠ BAC and ∠ BDC both stand on the chord BC, so ∠ BDC = ∠ BAC = 45°. In triangle BCD the angles add up to 180°:

\[ \begin{aligned} \angle BCD &= 180^\circ - (55^\circ + 45^\circ) \\ &= 80^\circ \end{aligned} \]

90° would assume BD is a diameter, which nothing in the question says. 65° and 70° do not follow from the given angles.

Answer: A

The value of ∠ POS is 105°.

∠ POR and ∠ ROQ lie on the straight line PQ, so they add up to 180°. Splitting 180° in the ratio 5 : 7:

\[ \begin{aligned} \angle POR &= \tfrac{5}{12} \times 180^\circ = 75^\circ \\ \angle ROQ &= \tfrac{7}{12} \times 180^\circ = 105^\circ \end{aligned} \]

∠ POS and ∠ ROQ are vertically opposite angles, so ∠ POS = 105°. 75° is ∠ POR (and its opposite ∠ SOQ), not ∠ POS; 150° and 210° double the parts.

Answer: C

The quadrilateral that cannot be constructed is AB = 3, BC = 5, AC = 6, AD = 3, CD = 2.

The diagonal AC splits the quadrilateral into triangles ABC and ADC, and each must satisfy the triangle inequality. In this option, triangle ADC has sides 3, 2 and 6, but 3 + 2 = 5 is less than 6, so the triangle, and hence the quadrilateral, cannot be drawn.

In every other option both triangles are possible: 3, 5, 4 with 6, 6, 4; 2, 4, 4 with 5, 5, 4; and 3, 5, 6 with 3, 5, 6.

Answer: B

The answer is x = 70°, y = 110°.

The 70° angle at C lies between the extensions of sides BC and DC beyond C, so it is vertically opposite the interior angle BCD. Hence angle C = 70°.

In a parallelogram opposite angles are equal, so the interior angle at A is also 70°, which is x. Adjacent angles are supplementary, so the interior angle at D is 180° - 70° = 110°, which is y.

Option A swaps the two values, option C makes adjacent angles equal, and option D gives a pair that does not add up to 180°.

Answer: A

The true statement is \(AB + BC \gt CA\).

By the triangle inequality, the sum of any two sides of a triangle is greater than the third side. That applies in particular to the two shorter sides compared with the longest, CA.

If AB + BC were less than or equal to CA, the three points could not form a triangle. The Pythagorean relation in option D holds only for a right angle at B, which the given information does not establish.

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.