Geometry & Mensuration: previous-year questions

Answer: (B) decrease by 4%.

New area $= (0.8L) \times (1.2B) = 0.96\,LB$, i.e. 96% of the original, a decrease of 4%. General rule: when one dimension increases by $x\%$ and the other decreases by $x\%$, the area always falls by $\frac{x^2}{100}\%$; here $\frac{20^2}{100} = 4\%$. For unequal changes of $a\%$ and $b\%$ the net change is $a + b + \frac{ab}{100}$, which gives $20 - 20 - \frac{400}{100} = -4\%$. Option A is the usual trap of assuming that equal percentage increase and decrease cancel out. The same $\frac{x^2}{100}$ rule gives the net loss when two articles are sold at one price with equal profit and loss percentages.

Answer: (A) $7\sqrt{3}$ cm

$\pi r^2 = 154$ gives $r^2 = 49$, so the circumradius is 7 cm. For an equilateral triangle the side equals $r\sqrt{3}$, so every side (and hence the 'longest') is $7\sqrt{3}$ cm. 7 cm is the radius itself, not the side, and 3·5 cm and 28 cm do not fit the formula.

Answer: (A) $1 : \sqrt{3}$

Equating the curved surfaces, $2\pi r^2 = \pi r l$, so the slant height $l = 2r$. Then $h = \sqrt{l^2 - r^2} = \sqrt{3}r$, so $r : h = 1 : \sqrt{3}$. Option B reverses the ratio, and C and D come from using the wrong relation between $l$ and $h$.

Answer: (A) 5

Interior and exterior angles add up to $180^\circ$, so in the ratio 3 : 2 the exterior angle is $72^\circ$. The number of sides is $360^\circ \div 72^\circ = 5$. A 10-sided polygon would have exterior angles of $36^\circ$, giving a ratio of 4 : 1.

Answer: (A) 8 cm

The first square's area is $\frac{d^2}{2} = \frac{32}{2} = 16$ cm², so the second square's area is 32 cm². Its diagonal satisfies $\frac{d^2}{2} = 32$, so $d = 8$ cm. $8\sqrt{2}$ cm would correspond to doubling the diagonal (four times the area).

Answer: (C) $\frac{\pi}{3}^c$

The two angles add to 65°20′3″ + 54°39′57″ = 120°0′0″, so the third angle is 180° − 120° = 60°. In radians (circular measure) 60° is $\frac{\pi}{3}$. $\frac{2\pi}{3}$ is the sum of the two given angles, not the third one.

Answer: (C) 16 cm

Each chord of 12 cm is at a distance $\sqrt{10^2 - 6^2} = 8$ cm from the centre. Two distinct equal parallel chords lie on opposite sides of the centre, so they are 8 + 8 = 16 cm apart. 12, 14 and 18 cm do not come from this calculation.

Answer: (B) 9·4 m

The ladder is the hypotenuse and the 4·7 m distance is adjacent to the 60° angle, so length = 4·7 ÷ $\cos 60^\circ$ = 4·7 ÷ $\frac{1}{2}$, i.e. the ladder is 9·4 m long. Option A, 4·7 m, is only the base distance.

Answer: (B) 55°

Where the external bisectors of B and C meet, $\angle BOC = 90^\circ - \frac{1}{2}\angle A$. With $\angle A = 70^\circ$ this is $90^\circ - 35^\circ = 55^\circ$. 125° would be the result for internal bisectors, and 45°, 65° and 75° do not fit the formula.

Answer: (B) ± 16

By the distance formula $(-12)^2 + y^2 = 20^2$, so $y^2 = 400 - 144 = 256$ and $y = \pm 16$. This is the 12-16-20 right triangle, so ± 15 and ± 14 do not fit.

Answer: (B) 19%

Area varies as $r^2$, so the new area is $0.9^2 = 0.81$ of the old one. The area falls by 19%. 20% simply doubles the 10% and ignores the $0.1^2$ term.

Answer: (B) 6 cm

With D on BC, triangles BAD and BCA share angle B and have ∠BAD = ∠BCA, so they are similar. Hence $\frac{BD}{BA} = \frac{AD}{CA} = \frac{3}{8}$, giving BD = 16 × $\frac{3}{8}$ = 6 cm. 3 cm just repeats AD.

Answer: (C) $22\frac{2}{5}$ cm

11 litres = 11000 cm³ and the base area is $\frac{22}{7} \times 12.5^2 = \frac{6875}{14}$ cm². The drop is $11000 \times \frac{14}{6875} = 22.4$ cm = $22\frac{2}{5}$ cm. Using the diameter instead of the radius would give a much smaller drop.

Answer: (D) 6 cm

The increase is $4\pi[(r + 2)^2 - r^2] = 16\pi(r + 1) = 352$. With $\pi = \frac{22}{7}$, $r + 1 = 352 \times \frac{7}{352} = 7$, so $r = 6$ cm. A radius of 5 cm would give an increase of only about 302 cm².

Answer: (C) 52 cm

The diagonals of a rhombus bisect each other at right angles, so each side is $\sqrt{12^2 + 5^2} = 13$ cm. The perimeter is 4 × 13 = 52 cm. 13 cm is just one side.

Answer: (A) $\sqrt{3} : 1$

Equal volumes give $r_1^2 h_1 = r_2^2 h_2$, so $\frac{r_1^2}{r_2^2} = \frac{h_2}{h_1} = 3$. Hence $r_1 : r_2 = \sqrt{3} : 1$; the shorter cylinder must be wider. Option B reverses the ratio, and 3 : 1 forgets the square root.

Answer: (A) 14 cm

Foil area = 57·75 ÷ 0·10 = 577·5 cm², which is the curved surface $\pi r l$ with r = 10·5 cm, so $l = 577.5 \div 33 = 17.5$ cm. Height = $\sqrt{17.5^2 - 10.5^2} = \sqrt{196} = 14$ cm. 28 cm and 7 cm are double and half of this.

Answer: (C) $\frac{9}{4}$ cm

The sphere displaces $\frac{4}{3}\pi \times 3^3 = 36\pi$ cm³ of water. Spread over the base $\pi \times 4^2 = 16\pi$ cm², the rise is $\frac{36}{16} = \frac{9}{4}$ cm. Option B inverts this fraction.

Answer: (B) 200%

With the radius fixed, the cone's volume $\frac{1}{3}\pi r^2 h$ is proportional to its height. Making the height three times the original makes the volume three times too, an increase of 200%. 300% confuses 'three times' with 'an increase of three times'.

Answer: (C) 20 ft

Pool plus wall has area $\frac{36}{25}$ of the pool, so the outer radius is $\frac{6}{5}$ of the pool radius R. Then R + 4 = $\frac{6R}{5}$, giving R = 20 ft. With R = 16 ft the wall's share would be about 0·56 of the pool.