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

Answer: D

The ratio is 1 : 6.

The points divide BC into four equal parts, so PR, which spans two of them, is half of BC. The centroid G lies one-third of the way up from BC, so its height above BC is one-third of the height \(h\) of the triangle:

\[ \begin{aligned} [\text{PGR}] &= \tfrac{1}{2} \cdot \tfrac{BC}{2} \cdot \tfrac{h}{3} \\ &= \tfrac{1}{6} \cdot \tfrac{1}{2} \cdot BC \cdot h \\ &= \tfrac{1}{6}[\text{ABC}] \end{aligned} \]

1 : 3 is the ratio for the triangle GBC, with the whole of BC as base. 1 : 2 and 1 : 4 forget either the halved base or the one-third height.

Answer: B

The largest area is \(x^{2}\) square units.

The largest triangle in a semicircle of radius \(x\) takes the diameter as its base, of length \(2x\), and its third vertex at the top of the arc, at the greatest possible height \(x\):

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

\(\tfrac{1}{2}x^{2}\) is what a triangle with base \(x\) would give, and \(2x^{2}\) and \(4x^{2}\) forget the factor of one half or double the height; none can fit inside the semicircle, whose own area is only \(\tfrac{\pi}{2}x^{2}\), about \(1.57x^{2}\).

Answer: D

The ratio of the radii is 5 : 4.

Arc length equals radius times the angle in radians, \(l = r\theta\). For the same arc length, the radius is inversely proportional to the angle:

\[ \begin{aligned} r_{1} \times 60 &= r_{2} \times 75 \\ \frac{r_{1}}{r_{2}} &= \frac{75}{60} = \frac{5}{4} \end{aligned} \]

The circle in which the arc makes the smaller angle (60°) has the larger radius. 4 : 5 is the ratio of the angles, the common mistake of treating radius as proportional to angle.

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.