Elementary Mathematics 2017 (II)

Set A

Questions 61–80 of 100

61Mensuration • Plane Figures • Circle

HardIdeal time: about 2.5 min

Segment QR of length r is a tangent at Q to a circle of radius r with centre at P. What is the area of the part of the triangle PQR, which is outside the circular region ?

\frac{π r ^{2}}{16}

\frac{r ^{2}}{2} - \frac{π r ^{2}}{8}

\frac{r ^{2}}{2} - \frac{π r ^{2}}{16}

\frac{r ^{2}}{4} - \frac{π r ^{2}}{8}

Show answer & solution

Option B

\frac{r ^{2}}{2} - \frac{π r ^{2}}{8}

Approach

The tangency makes $△ P QR$ right-angled and isosceles; subtract the $4 5^{\circ}$ sector at $P$ from its area.

Area and arc length of a circular sector

Step-by-step solution

1
$P Q ⊥ QR$ and $P Q = QR = r$
Why: A tangent is perpendicular to the radius at the point of contact, and both lengths are given as $r$ .
The two tangents from an external point
2
Area of $△ P QR = \frac{1}{2} \times r \times r = \frac{r ^{2}}{2}$
Why: Half base times height, with the two perpendicular sides serving as base and height.
Area of a triangle from base and height
3
$tan ∠ QP R = \frac{QR}{P Q} = 1 \Rightarrow ∠ QP R = 4 5^{\circ}$
Why: The triangle is right-angled and isosceles, so the angle at the centre $P$ is $4 5^{\circ}$ .
Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles
4
The part of the triangle inside the circle is the sector of radius $r$ at $P$ with angle $4 5^{\circ}$ : area $= \frac{45}{360} π r^{2} = \frac{π r ^{2}}{8}$
Why: $P$ is the centre, so every point of the triangle within distance $r$ of $P$ lies in the sector cut off by $P Q$ and $P R$ ; $P Q = r$ reaches the circle exactly, and $P R = r 2 > r$ crosses it.
Area and arc length of a circular sector
5
Required area $= \frac{r ^{2}}{2} - \frac{π r ^{2}}{8}$
Why: Remove the circular part from the whole triangle.

Shortcut

The triangle is half a square of side $r$ and the circular part is one eighth of the circle, so the answer is $\frac{r ^{2}}{2} - \frac{π r ^{2}}{8}$ at a glance.

Common mistakes

Option A

No natural route reaches $\frac{π r ^{2}}{16}$ .

Option C

Halved the sector angle, using $22 \cdot 5^{\circ}$ and a sector area of $\frac{π r ^{2}}{16}$ .

Option D

Took the triangle's area as $\frac{r ^{2}}{4}$ , halving it a second time.

Skills used

The two tangents from an external point Area and arc length of a circular sector Area of a triangle from base and height Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles

62Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 55 s

In a triangle ABC, AD is perpendicular on BC. If $∠ B A C = 9 0^{\circ}$ , AB = c, BC = a, CA = b and AD = p, then which one of the following is correct ?

p = ab c

p^{2} = b c

p = \frac{b c}{a}

p = \frac{ab}{c}

Show answer & solution

Option C

p = \frac{b c}{a}

Approach

Compute the area of the right triangle two ways — from the legs, and from the hypotenuse with its altitude.

Area of a triangle from base and height

Step-by-step solution

1
$∠ B A C = 9 0^{\circ}$ , so the legs are $A B = c$ and $C A = b$ , and area $= \frac{1}{2} b c$
Why: The two sides enclosing the right angle serve directly as base and height.
Area of a triangle from base and height
2
Taking $B C = a$ as the base with altitude $A D = p$ : area $= \frac{1}{2} a p$
Why: $A D$ is perpendicular to $B C$ , so it is the height corresponding to the base $B C$ .
3
$\frac{1}{2} b c = \frac{1}{2} a p \Rightarrow p = \frac{b c}{a}$
Why: One triangle has one area, so the two expressions must agree.

Shortcut

For a right triangle the altitude to the hypotenuse is always (product of the legs) over the hypotenuse.

Common mistakes

Option A

No natural route reaches $p = ab c$ ; it is not even dimensionally a length.

Option B

Confused the altitude with the geometric-mean relation $p^{2} = B D \cdot D C$ , substituting the sides $b$ and $c$ for the two segments of the hypotenuse.

Option D

Took $a$ and $b$ as the legs and $c$ as the hypotenuse, ignoring that the right angle is at $A$ so the hypotenuse is $B C = a$ .

Skills used

Area of a triangle from base and height

63Geometry • Circle • Chords

ModerateIdeal time: about 1.5 min

AB and CD are parallel chords of a circle $3$ cm apart. If AB = $4$ cm, CD = $10$ cm, then what is the radius of the circle ?

7

cmB

19

cmC

29

cmD

14

Show answer & solution

Option C

29

Approach

Drop the perpendicular from the centre to each chord, write Pythagoras for both, and use the $3$ cm gap to link the two distances.

The perpendicular from the centre bisects a chord

Step-by-step solution

1
The perpendicular from the centre bisects each chord, giving half-chords $2$ cm and $5$ cm
Why: A chord is bisected by the perpendicular from the centre, so $A B = 4$ gives $2$ and $C D = 10$ gives $5$ .
The perpendicular from the centre bisects a chord
2
$r^{2} = 2^{2} + d_{1}^{2}$ and $r^{2} = 5^{2} + d_{2}^{2}$ , with $d_{1}, d_{2}$ the distances from the centre
Why: Each half-chord, its distance from the centre and the radius form a right triangle.
Pythagoras' theorem
3
$A B$ is the shorter chord, so it is the farther from the centre; with both chords on the same side, $d_{1} = d_{2} + 3$
Why: Among chords of one circle, the shorter lies farther from the centre, which fixes which distance is the larger.
4
$4 + (d_{2} + 3)^{2} = 25 + d_{2}^{2} \Rightarrow 6 d_{2} + 13 = 25 \Rightarrow d_{2} = 2, d_{1} = 5$
Why: Expand and cancel $d_{2}^{2}$ , leaving a linear equation.
Square of a sum and of a difference
5
$r^{2} = 5^{2} + 2^{2} = 29 \Rightarrow r = 29$ cm
Why: Substitute back into either Pythagoras relation.
denotes the non-negative root

Shortcut

Try $d_{2} = 2$ : then $r^{2} = 29$ and $d_{1} = 29 - 4 = 5$ , which is indeed $3$ more than $2$ .

Common mistakes

Option A

No natural route reaches $7$ cm.

Option B

Slip in the expansion or in pairing half-chord with distance, giving $r^{2} = 19$ instead of $29$ .

Option D

No natural route reaches $14$ cm.

Skills used

The perpendicular from the centre bisects a chord Pythagoras' theorem Square of a sum and of a difference

denotes the non-negative root

64Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

The diagonals of a cyclic quadrilateral ABCD intersect at P and the area of the triangle APB is $24$ square cm. If AB = $8$ cm and CD = $5$ cm, then what is the area of the triangle CPD ?

24

square cmB

15

square cmC

12.5

square cmD

9.375

square cm

Show answer & solution

Option D

9.375

square cm

Approach

Show $△ A P B$ and $△ D P C$ are similar with ratio $\frac{A B}{D C}$ , then scale the area by the square of that ratio.

Step-by-step solution

1
$∠ A P B = ∠ D P C$ (vertically opposite) and $∠ B A C = ∠ B D C$ (angles in the same segment on $B C$ )
Why: $A B C D$ is cyclic, so the two angles standing on the chord $B C$ from $A$ and from $D$ are equal.
Angles on a straight line and vertically opposite angles The angle at the centre is twice the angle at the circumference
2
$△ A P B \sim △ D P C$ by AA
Why: Two pairs of equal angles are enough for similarity, and the correspondence pairs $A B$ with $D C$ .
Similar triangles have proportional sides
3
$\frac{[ A P B ]}{[ D P C ]} = (\frac{A B}{D C})^{2} = (\frac{8}{5})^{2} = \frac{64}{25}$
Why: Areas of similar triangles are in the square of the ratio of corresponding sides.
4
$[C P D] = 24 \times \frac{25}{64} = \frac{600}{64} = 9 \cdot 375$ square cm
Why: Invert the ratio to pass from the known area to the unknown one.

Shortcut

Area scales as the square of the side ratio: $24 \times (\frac{5}{8})^{2} = 24 \times \frac{25}{64} = 9 \cdot 375$ .

Common mistakes

Option A

Assumed the two triangles are congruent, or that the diagonals cut the quadrilateral into equal-area pieces.

Option B

Scaled by the side ratio itself, $24 \times \frac{5}{8} = 15$ , instead of by its square.

Option C

No natural route reaches $12 \cdot 5$ square cm.

Skills used

65Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 50 s

In an equilateral triangle ABC, BD is drawn perpendicular to AC. What is $BD^{2}$ equal to ?

AD^{2}

2 AD^{2}

3 AD^{2}

4 AD^{2}

Show answer & solution

Option C

3 AD^{2}

Approach

The perpendicular from $B$ bisects $A C$ , so $A B = 2 A D$ ; then apply Pythagoras in $△ A B D$ .

Pythagoras' theorem

Step-by-step solution

1
Let the side be $a$ ; then $A B = B C = C A = a$
Why: All three sides of an equilateral triangle are equal.
2
$B D ⊥ A C$ with $B A = B C$ , so $D$ is the midpoint of $A C$ and $A D = \frac{a}{2}$
Why: In an isosceles triangle the altitude from the apex to the base bisects that base — here $△ A B D$ and $△ C B D$ are congruent by RHS.
Congruence criteria for triangles Base angles of an isosceles triangle
3
$B D^{2} = A B^{2} - A D^{2} = a^{2} - \frac{a ^{2}}{4} = \frac{3 a ^{2}}{4}$
Why: $△ A B D$ is right-angled at $D$ , with $A B$ as hypotenuse.
Pythagoras' theorem
4
$A D^{2} = \frac{a ^{2}}{4}$ , so $B D^{2} = 3 \times \frac{a ^{2}}{4} = 3 A D^{2}$
Why: Expressing both in terms of $a^{2}$ makes the ratio immediate.
Area and altitude of an equilateral triangle

Shortcut

$A B = 2 A D$ , so $B D^{2} = (2 A D)^{2} - A D^{2} = 3 A D^{2}$ .

Common mistakes

Option A

Assumed the altitude equals half the side, i.e. $B D = A D$ , treating $△ A B D$ as isosceles.

Option B

Treated $△ A B D$ as a right isosceles triangle with $A D$ as a leg and $B D$ as the hypotenuse.

Option D

Used $A B = 2 A D$ but reported $A B^{2} = 4 A D^{2}$ , forgetting to subtract $A D^{2}$ .

Skills used

Pythagoras' theorem Congruence criteria for triangles Base angles of an isosceles triangle Area and altitude of an equilateral triangle

66Geometry • Circle • Tangents

EasyIdeal time: about 1 min

The distance between the centres of two circles having radii $9$ cm and $4$ cm is $13$ cm. What is the length of the direct common tangent of these circles ?

12

cmB

11

cmC

10

cmD

9.5

Show answer & solution

Option A

12

Approach

Join each centre to its point of contact, complete a rectangle on the two radii, and apply Pythagoras to the right triangle left on the line of centres.

The two tangents from an external point

Step-by-step solution

1
Let the centres be $O_{1}, O_{2}$ with $O_{1} A = 9$ cm, $O_{2} B = 4$ cm and $O_{1} O_{2} = 13$ cm, where $A B$ is the direct common tangent touching at $A$ and $B$ .
Why: A direct (external) common tangent touches both circles on the same side of the line of centres, so the two radii to the points of contact point the same way.
The two tangents from an external point
2
$O_{1} A ⊥ A B$ and $O_{2} B ⊥ A B$ , hence $O_{1} A ∥ O_{2} B$ .
Why: A tangent is perpendicular to the radius at its point of contact; two lines perpendicular to the same line are parallel.
Angles made by a transversal of two parallel lines
3
Drop $O_{2} C ⊥ O_{1} A$ with $C$ on $O_{1} A$ . Then $A B O_{2} C$ has three right angles, so it is a rectangle with $C O_{2} = A B$ and $C A = O_{2} B = 4$ , leaving $O_{1} C = 9 - 4 = 5$ cm.
Why: The rectangle transfers the unknown tangent length onto $C O_{2}$ , a side of a triangle whose other two sides are known.
4
$A B = C O_{2} = O_{1} O_{2}^{2} - O_{1} C^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144 = 12$ cm
Why: Triangle $O_{1} C O_{2}$ is right-angled at $C$ , so Pythagoras delivers the remaining side.
Pythagoras' theorem Length of a common tangent to two circles

Shortcut

Direct common tangent $= d^{2} - (r_{1} - r_{2})^{2} = 169 - 25 = 12$ cm.

Common mistakes

Option B

$11$ cm: no route reaches it; $169 - 25$ is $12$ , not $11$ .

Option C

$10$ cm: no natural miscalculation of $d^{2} - (r_{1} - r_{2})^{2}$ produces it.

Option D

$9.5$ cm sits near $1 3^{2} - 9^{2} \approx 9.38$ , the tangent length from $O_{2}$ to the larger circle, which ignores the smaller radius entirely.

Skills used

The two tangents from an external point Angles made by a transversal of two parallel lines Pythagoras' theorem Length of a common tangent to two circles

67Geometry • Triangle • Triangle Centres

ModerateIdeal time: about 1.5 min

If PL, QM and RN are the altitudes of triangle PQR whose orthocentre is O, then Q is the orthocentre of the triangle

AOPQBOQRCPLRDOPR

Show answer & solution

Option D

OPR

Approach

Identify which two lines of the figure are the altitudes of the candidate triangle and show they cross at $Q$ .

The orthocentre and the orthocentric system

Step-by-step solution

1
$P L ⊥ QR$ , $QM ⊥ R P$ , $R N ⊥ P Q$ , and $O$ is their common point.
Why: $O$ is the orthocentre of $△ P QR$ , so it lies on all three altitudes.
The orthocentre and the orthocentric system
2
In $△ O P R$ , the altitude from $O$ must be perpendicular to $P R$ . The line $QM$ is perpendicular to $P R$ and contains $O$ , so that altitude is the line $O Q$ - and it passes through $Q$ .
Why: A perpendicular to a given line through a given point is unique, so $QM$ is forced to be the altitude from $O$ .
3
In $△ O P R$ , the altitude from $P$ must be perpendicular to $O R$ . But $O R$ lies along $R N$ , which is perpendicular to $P Q$ ; so $P Q ⊥ O R$ , and $P Q$ passes through $P$ - hence $P Q$ is that altitude, and it too passes through $Q$ .
Why: $O$ lies on $R N$ , so the segment $O R$ is part of $R N$ and inherits its direction; a line perpendicular to $R N$ through $P$ is the line $P Q$ .
4
Two altitudes of $△ O P R$ meet at $Q$ , so $Q$ is the orthocentre of $△ O P R$ .
Why: The orthocentre is the common point of the altitudes, and two of them already fix it.

Shortcut

In an orthocentric system ${O, P, Q, R}$ each point is the orthocentre of the triangle formed by the other three, so $Q$ goes with $△ O P R$ .

Common mistakes

Option A

$△ O P Q$ has $R$ , not $Q$ , as its orthocentre - the three vertices of the orthocentric system were permuted.

Option B

$△ O QR$ has $P$ as its orthocentre; $Q$ is a vertex of this triangle, so it could only be the orthocentre if the angle at $Q$ were $9 0^{\circ}$ .

Option C

$P L ⊥ QR$ makes $△ P L R$ right-angled at $L$ , so its orthocentre is $L$ itself, not $Q$ .

Skills used

The orthocentre and the orthocentric system

68Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

In triangle ABC, $∠ C = 9 0^{\circ}$ and CD is the perpendicular from C to AB.

If $(CD)^{- 2} = (BC)^{- 2} + (CA)^{- 2}$ , then which one of the following is correct ?

BC \cdot CD = AB \cdot CA

AB \cdot BC = CD \cdot CA

CA^{2} + CB^{2} = 2 (AD^{2} + CD^{2})

AB \cdot CD = BC \cdot CA

Show answer & solution

Option D

AB \cdot CD = BC \cdot CA

Approach

Write the area of the right triangle twice - once on the two legs, once on the hypotenuse with $C D$ as height - and equate.

Area of a triangle from base and height

Step-by-step solution

1
$[△ A B C] = \frac{1}{2} B C \cdot C A$
Why: $∠ C = 9 0^{\circ}$ , so each leg is the perpendicular height to the other.
Area of a triangle from base and height
2
$[△ A B C] = \frac{1}{2} A B \cdot C D$
Why: Taking the hypotenuse $A B$ as base, its perpendicular height is exactly $C D$ .
3
$\frac{1}{2} A B \cdot C D = \frac{1}{2} B C \cdot C A \Rightarrow A B \cdot C D = B C \cdot C A$
Why: One area computed two ways must agree, and the factor $\frac{1}{2}$ cancels.
4
$\frac{1}{C D ^{2}} = \frac{A B ^{2}}{B C ^{2} \cdot C A ^{2}} = \frac{B C ^{2} + C A ^{2}}{B C ^{2} \cdot C A ^{2}} = \frac{1}{C A ^{2}} + \frac{1}{B C ^{2}}$
Why: Squaring and inverting the relation just derived, then replacing $A B^{2}$ by $B C^{2} + C A^{2}$ , reproduces the hypothesis printed in the question: it is an identity in every right triangle, not an extra condition, so the answer is whichever option is likewise always true.
Pythagoras' theorem Negative exponent is a reciprocal

Shortcut

Twice the area is both $A B \cdot C D$ and $B C \cdot C A$ .

Common mistakes

Option A

$B C \cdot C D = A B \cdot C A$ interchanges $A B$ and $B C$ ; the pairing must be (hypotenuse, altitude) against (leg, leg).

Option B

$A B \cdot B C = C D \cdot C A$ pairs the hypotenuse with a leg and the altitude with the other leg; neither product is twice the area.

Option C

$C A^{2} + C B^{2} = 2 (A D^{2} + C D^{2})$ is Apollonius' median relation, valid when $D$ is the MIDPOINT of $A B$ ; here $D$ is the foot of the altitude.

Skills used

Area of a triangle from base and height Pythagoras' theorem Negative exponent is a reciprocal

69Geometry • Quadrilateral • Parallelogram

ModerateIdeal time: about 1.5 min

If a point O in the interior of a rectangle ABCD is joined with each of the vertices A, B, C and D, then $OB^{2} + OD^{2}$ will be equal to

2 OC^{2} + OA^{2}

OC^{2} - OA^{2}

OC^{2} + OA^{2}

OC^{2} + 2 OA^{2}

Show answer & solution

Option C

OC^{2} + OA^{2}

Approach

Resolve $O$ onto a pair of adjacent sides of the rectangle; all four distances then become hypotenuses built from the same four lengths.

Pythagoras' theorem

Step-by-step solution

1
Let the perpendiculars from $O$ to $A B$ and to $A D$ split those sides as $A B = p + q$ and $A D = s + t$ , with $p, q$ measured along $A B$ from $A$ and $B$ , and $s, t$ along $A D$ from $A$ and $D$ .
Why: Adjacent sides of a rectangle are perpendicular, so these four lengths are the legs of every right triangle in sight.
2
$O A^{2} = p^{2} + s^{2}$ , $O B^{2} = q^{2} + s^{2}$ , $O C^{2} = q^{2} + t^{2}$ , $O D^{2} = p^{2} + t^{2}$
Why: Each of the four distances is the hypotenuse of a right triangle whose legs are one of $p, q$ and one of $s, t$ .
Pythagoras' theorem
3
$O B^{2} + O D^{2} = (q^{2} + s^{2}) + (p^{2} + t^{2}) = p^{2} + q^{2} + s^{2} + t^{2}$
Why: Add the pair asked for; every one of the four squares appears exactly once.
4
$O A^{2} + O C^{2} = (p^{2} + s^{2}) + (q^{2} + t^{2}) = p^{2} + q^{2} + s^{2} + t^{2} = O B^{2} + O D^{2}$
Why: The other diagonal pair produces the same four squares, so the two sums are equal - and $O B^{2} + O D^{2} = O C^{2} + O A^{2}$ .
British flag theorem

Shortcut

British flag theorem: for any point $O$ in a rectangle $A B C D$ , $O A^{2} + O C^{2} = O B^{2} + O D^{2}$ .

Common mistakes

Option A

$2 O C^{2} + O A^{2}$ weights one term; the relation is a plain sum of the two distances to the opposite pair of corners.

Option B

$O C^{2} - O A^{2}$ turns the sum into a difference - a sign slip in the British-flag relation.

Option D

$O C^{2} + 2 O A^{2}$ again invents a coefficient; no factor of $2$ survives the cancellation.

Skills used

Pythagoras' theorem British flag theorem

70Mensuration • Solids • Cylinder

ModerateIdeal time: about 1.5 min

A cylinder of height 2x is circumscribed by a sphere of radius 2x such that the circular ends of the cylinder are two small circles on the sphere. What is the ratio of the curved surface area of the cylinder to the surface area of the sphere ?

3 : 4

3 : 3

3 : 2

3 : 1

Show answer & solution

Option A

3 : 4

Approach

The cylinder's radius, half its height and the sphere's radius form a right triangle; get the radius, then take the two surface formulas.

Pythagoras' theorem

Step-by-step solution

1
Half-height $= x$ and sphere radius $= 2 x$ ; let the cylinder's radius be $r$ .
Why: The sphere's centre is the midpoint of the cylinder's axis, so it stands $x$ from each circular end.
2
$r^{2} + x^{2} = (2 x)^{2} \Rightarrow r^{2} = 3 x^{2} \Rightarrow r = 3 x$
Why: The radius to a point of the rim, the half-axis and the sphere's radius to that same point close a right triangle.
Pythagoras' theorem
3
Curved surface of the cylinder $= 2 π r h = 2 π (3 x) (2 x) = 43 π x^{2}$
Why: The cylinder's height is the given $2 x$ ; only the curved surface is asked for, so the two circular ends are excluded.
Curved and total surface of a cylinder
4
Surface of the sphere $= 4 π (2 x)^{2} = 16 π x^{2}$
Why: The sphere's radius is $2 x$ , not the cylinder's $3 x$ .
Volume and surface area of a sphere
5
Ratio $= \frac{4 3 π x ^{2}}{16 π x ^{2}} = \frac{3}{4} = 3 : 4$
Why: $π$ and $x^{2}$ cancel, so the answer is a pure number independent of $x$ .

Common mistakes

Option B

$3 : 3$ puts the cylinder's radius $3 x$ into the sphere's formula, giving $12 π x^{2}$ instead of $16 π x^{2}$ .

Option C

$3 : 2$ uses $2 π R^{2}$ - a hemisphere's curved surface - for the whole sphere, giving $8 π x^{2}$ .

Option D

$3 : 1$ takes the sphere's radius as $x$ rather than $2 x$ , giving $4 π x^{2}$ .

Skills used

Pythagoras' theorem Curved and total surface of a cylinder Volume and surface area of a sphere

71Mensuration • Recasting and Melting

EasyIdeal time: about 1 min

A cylindrical vessel $60$ cm in diameter is partially filled with water. A sphere $30$ cm in diameter is gently dropped into the vessel and is completely immersed. To what further height will the water in the cylinder rise ?

20

cmB

15

cmC

10

cmD

5

Show answer & solution

Option D

5

Approach

Test each option as the rise $h$ by checking whether the displaced cylindrical volume equals the sphere's volume.

Recasting preserves volume

Step-by-step solution

1
Sphere volume $= \frac{4}{3} π (15)^{3} = 4500 π$ .
Why: Sphere radius is half the $30$ cm diameter.
Volume and surface area of a sphere
2
Cylinder base area $= π (30)^{2} = 900 π$ .
Why: Cylinder radius is half the $60$ cm diameter.
Volume of a right circular cylinder
3
Test $h = 5$ : displaced volume $= 900 π \times 5 = 4500 π$ , equal to the sphere's volume.
Why: The rise multiplies the base area to give the volume the sphere pushed aside.
Recasting preserves volume
4
Test $h = 10, 15, 20$ : give $9000 π, 13500 π, 18000 π$ , all too large.
Why: Rules out B, C, A.

Common mistakes

Option A

Uses the sphere's radius ( $15$ ) as the cylinder's radius by mistake: $4500 π / (π \cdot 1 5^{2}) = 20$ .

Option B

No clean single-mistake route reaches $15$ from these numbers.

Option C

Uses both diameters directly as radii, forgetting to halve either: $(4/3) π (30)^{3} / (π \cdot 6 0^{2}) = 10$ .

Skills used

Volume and surface area of a sphere Volume of a right circular cylinder Recasting preserves volume

72Mensuration • Solids • Cone

EasyIdeal time: about 1 min

The vertical angle of a right circular cone is $\frac{π}{2}$ and the slant height is $2 r$ cm. What is the volume of the cone in cubic cm ?

π r^{3}

9 π r^{3}

\frac{π r ^{3}}{3}

3 π r^{3}

Show answer & solution

Option C

\frac{π r ^{3}}{3}

Approach

Halve the vertical angle to get a $4 5^{\circ}$ right triangle inside the cone, read the radius and height off the slant height, then apply the cone volume formula.

Volume of a right circular cone

Step-by-step solution

1
Vertical angle $= \frac{π}{2} = 9 0^{\circ}$ , so the semi-vertical angle is $4 5^{\circ}$ .
Why: The axis of a right circular cone bisects the vertical angle, and $π$ radians is $18 0^{\circ}$ .
Converting between degrees and radians
2
$R = l sin 4 5^{\circ} = 2 r \cdot \frac{1}{2} = r$ and $h = l cos 4 5^{\circ} = 2 r \cdot \frac{1}{2} = r$
Why: The axis, the base radius and the slant height form a right triangle with the semi-vertical angle at the apex, the radius opposite it and the height adjacent.
Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles
3
$V = \frac{1}{3} π R^{2} h = \frac{1}{3} π r^{2} \cdot r = \frac{π r ^{3}}{3}$ cubic cm
Why: Both the base radius and the height have come out equal to $r$ .
Volume of a right circular cone

Shortcut

A $9 0^{\circ}$ vertical angle makes the axial cross-section a right isosceles triangle, so $R = h$ and $l = R 2$ ; with $l = 2 r$ that gives $R = h = r$ at sight.

Common mistakes

Option A

$π r^{3}$ drops the factor $\frac{1}{3}$ , quoting the cylinder on the same base and height.

Option B

$9 π r^{3}$ : no route reaches it.

Option D

$3 π r^{3}$ : no route reaches it.

Skills used

Converting between degrees and radians Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles Volume of a right circular cone

73Mensuration • Solids • Frustum

ModerateIdeal time: about 1.5 min

The radii of the frustum of a right circular cone are in the ratio $2 : 1$ . What is the ratio of the volume of the frustum of the cone to that of the whole cone ?

1 : 8

1 : 4

3 : 4

7 : 8

Show answer & solution

Option D

7 : 8

Approach

Restore the whole cone: the cone cut off the top is similar to it with half the radius, so it takes one-eighth of the volume and the frustum keeps the rest.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
Let the frustum's radii be $2 a$ (base) and $a$ (top), since they are in the ratio $2 : 1$ .
Why: A ratio fixes the two radii only up to a common factor, which will cancel.
Introduce a common factor $k$ for a ratio
2
The cut is parallel to the base, so the small cone removed is similar to the whole cone with linear ratio $\frac{a}{2 a} = \frac{1}{2}$ .
Why: A plane parallel to the base cuts off a cone similar to the original; the top radius of the frustum is the base radius of that small cone.
Similar triangles have proportional sides
3
$\frac{V _{small}}{V _{whole}} = (\frac{1}{2})^{3} = \frac{1}{8}$
Why: Volume varies as the cube of the linear scale, because $\frac{1}{3} π R^{2} H$ carries one factor of the scale in each of $R$ , $R$ and $H$ .
Direct variation as a power: $y \propto x^{n}$ Volume of a right circular cone
4
$\frac{V _{frustum}}{V _{whole}} = 1 - \frac{1}{8} = \frac{7}{8} = 7 : 8$
Why: The frustum is what is left of the whole cone after the small cone is taken away.

Shortcut

Radii $2 : 1$ makes the cut-off cone a half-scale copy, so it is $\frac{1}{8}$ of the whole and the frustum is $\frac{7}{8}$ .

Common mistakes

Option A

$1 : 8$ is the small cone that was cut off, not the frustum that remains.

Option B

$1 : 4$ is $(\frac{1}{2})^{2}$ , the ratio of the two circular faces, quoted as if it were a volume ratio.

Option C

$3 : 4$ subtracts the squared ratio instead of the cubed one: $1 - \frac{1}{4}$ .

Skills used

Introduce a common factor

k

y \propto x^{n}

Volume of a right circular cone

74Mensuration • Solids • Cone

EasyIdeal time: about 55 s

From a solid cylinder whose height is $8$ cm and of base radius $6$ cm, a conical cavity of height $8$ cm and of base radius $6$ cm is formed by hollowing out. What is the inner surface area of the cavity ?

6 π

square cmB

8 π

square cmC

10 π

square cmD

60 π

square cm

Show answer & solution

Option D

60 π

square cm

Approach

The wall of the cavity is exactly the curved surface of the removed cone, so find the slant height and apply $π r l$ .

Slant height and curved surface of a cone

Step-by-step solution

1
$l = r^{2} + h^{2} = 6^{2} + 8^{2} = 36 + 64 = 100 = 10$ cm
Why: A slant surface is measured along the slant height, not the vertical height, so $l$ must be found first.
Slant height and curved surface of a cone
2
Inner surface of the cavity $= π r l = π \times 6 \times 10 = 60 π$ square cm
Why: Only the cone's curved surface bounds the hollow: the circular face at the top is the opening rather than a wall, and the cylinder's own surfaces are outside the cavity.
Volume and surface of a composite solid

Shortcut

$r = 6$ , $h = 8$ gives $l = 10$ at sight; curved surface $= π r l = 60 π$ .

Common mistakes

Option A

$6 π$ is $π r$ - the slant height was dropped from $π r l$ .

Option B

$8 π$ uses the vertical height $8$ in place of the product $r l$ .

Option C

$10 π$ is $π l$ - the radius was dropped instead.

Skills used

Slant height and curved surface of a cone Volume and surface of a composite solid

75Mensuration • Combination of Solids

EasyIdeal time: about 1.5 min

A tent has been constructed which is in the form of a right circular cylinder surmounted by a right circular cone whose axis coincides with the axis of the cylinder. If the radius of the base of the cylinder is $50$ m, the height of the cylinder is $10$ m and the total height of the tent is $15$ m, then what is the capacity of the tent in cubic metres ?

37500 π

\frac{87500 π}{3}

\frac{26500 π}{3}

25000 π

Show answer & solution

Option B

\frac{87500 π}{3}

Approach

Split the tent into the cylinder and the cone above it, taking the cone's height as the difference of the two given heights, and add the volumes.

Volume and surface of a composite solid

Step-by-step solution

1
Cone height $= 15 - 10 = 5$ m; both pieces have radius $50$ m.
Why: The stated $15$ m is the height of the whole tent, so the cone occupies only what is left above the cylinder; the cone stands on the cylinder's top circle and therefore shares its radius.
Volume and surface of a composite solid
2
$V_{cylinder} = π r^{2} h = π (50)^{2} (10) = 25000 π$ cubic m
Why: The lower part is a right circular cylinder of radius $50$ m and height $10$ m.
Volume of a right circular cylinder
3
$V_{cone} = \frac{1}{3} π r^{2} h = \frac{1}{3} π (50)^{2} (5) = \frac{12500 π}{3}$ cubic m
Why: The upper part is a cone on the same base with height $5$ m.
Volume of a right circular cone
4
$V = 25000 π + \frac{12500 π}{3} = \frac{75000 π + 12500 π}{3} = \frac{87500 π}{3}$ cubic m
Why: Capacity is a volume, so the two pieces simply add - unlike surface area, where the shared circle would be excluded.

Common mistakes

Option A

$37500 π$ takes the cone's height as the total height $15$ m instead of $15 - 10 = 5$ m.

Option C

$\frac{26500 π}{3}$ : no route reaches it; it is smaller even than the cylinder alone.

Option D

$25000 π$ is the cylinder by itself, with the cone forgotten.

Skills used

Volume and surface of a composite solid Volume of a right circular cylinder Volume of a right circular cone

76Mensuration • Solids • Cylinder

ModerateIdeal time: about 2 min

Two rectangular sheets of sizes $2 π \times 4 π$ and $π \times 5 π$ are available. A hollow right circular cylinder can be formed by joining a pair of parallel sides of any sheet. What is the maximum possible volume of the circular cylinder that can be formed this way?

4 π^{2}

8 π^{2}

1.25 π^{2}

6.25 π^{2}

Show answer & solution

Option B

8 π^{2}

Approach

Roll each sheet both ways, express the volume in terms of which side becomes the circumference, and take the largest of the four.

Volume of a right circular cylinder

Step-by-step solution

1
Rolling a sheet $a \times b$ about the pair of sides of length $b$ gives $2 π R = a$ , so $R = \frac{a}{2 π}$ and height $b$ , whence $V = π R^{2} b = \frac{a ^{2} b}{4 π}$ .
Why: The side that is bent round becomes the circumference of the base; the side joined to itself becomes the height. Since $a$ enters squared and $b$ only linearly, the LONGER side should be the one bent round.
Volume of a right circular cylinder Area and circumference of a circle
2
Sheet $2 π \times 4 π$ : $\frac{( 2 π ) ^{2} ( 4 π )}{4 π} = 4 π^{2}$ or $\frac{( 4 π ) ^{2} ( 2 π )}{4 π} = 8 π^{2}$
Why: Both admissible rollings of the first sheet; the larger comes from bending the $4 π$ side.
3
Sheet $π \times 5 π$ : $\frac{( π ) ^{2} ( 5 π )}{4 π} = \frac{5 π ^{2}}{4} = 1.25 π^{2}$ or $\frac{( 5 π ) ^{2} ( π )}{4 π} = \frac{25 π ^{2}}{4} = 6.25 π^{2}$
Why: The same two rollings for the second sheet.
4
$max {4 π^{2}, 8 π^{2}, 1.25 π^{2}, 6.25 π^{2}} = 8 π^{2}$
Why: All four cylinders are constructible, so the maximum asked for is simply the largest of the four values.

Shortcut

$V = \frac{a ^{2} b}{4 π}$ , so bend the longer side: the sheets give $\frac{( 4 π ) ^{2} ( 2 π )}{4 π} = 8 π^{2}$ and $\frac{( 5 π ) ^{2} ( π )}{4 π} = 6.25 π^{2}$ ; the larger is $8 π^{2}$ .

Common mistakes

Option A

$4 π^{2}$ is the first sheet rolled the other way, bending the shorter side $2 π$ into the circumference.

Option C

$1.25 π^{2}$ is the smallest of the four - the second sheet bent about its shorter side.

Option D

$6.25 π^{2}$ is the best the second sheet can manage, but the first sheet does better.

Skills used

Volume of a right circular cylinder Area and circumference of a circle

77Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

In a triangle ABC, the medians AD and BE intersect at G. A line DF is drawn parallel to BE such that F is on AC. If AC = $9$ cm, then what is CF equal to?

2.25

cmB

3

cmC

4.5

cmD

6

Show answer & solution

Option A

2.25

Approach

Work inside $△ B E C$ : $D$ bisects $B C$ and $D F ∥ B E$ , so the converse of the midpoint theorem makes $F$ bisect $E C$ .

Midpoint theorem

Step-by-step solution

1
$B E$ is a median, so $E$ is the midpoint of $A C$ and $E C = \frac{A C}{2} = \frac{9}{2} = 4.5$ cm.
Why: A median runs from a vertex to the midpoint of the opposite side.
2
In $△ B E C$ , $D$ is the midpoint of $B C$ and $D F ∥ B E$ with $F$ on the side $E C$ .
Why: $A D$ is a median, so $D$ bisects $B C$ ; the line through $D$ parallel to $B E$ meets $A C$ at a point $F$ that lies between $E$ and $C$ , i.e. on the third side of $△ B E C$ .
3
$F$ is the midpoint of $E C$ , so $C F = \frac{E C}{2} = \frac{4.5}{2} = 2.25$ cm.
Why: Converse of the midpoint theorem: a line through the midpoint of one side, parallel to a second side, bisects the third.
Midpoint theorem

Shortcut

$F$ is the midpoint of $E C$ and $E$ the midpoint of $A C$ , so $C F = \frac{A C}{4} = \frac{9}{4} = 2.25$ cm.

Common mistakes

Option B

$3$ cm is $\frac{A C}{3}$ , obtained by importing the centroid's $2 : 1$ ratio into a length measured along $A C$ .

Option C

$4.5$ cm is $E C$ itself: the candidate stopped at the midpoint $E$ without halving a second time.

Option D

$6$ cm is $\frac{2}{3} A C$ , the centroid ratio again applied to the wrong segment.

Skills used

Midpoint theorem

78Geometry • Triangle • Similarity

EasyIdeal time: about 1.5 min

In a triangle PQR, X is a point on PR and Y is a point on QR such that PR = $10$ cm, RX = $4$ cm, YR = $2$ cm, QR = $5$ cm. Which one of the following is correct?

AXY is parallel to PQBPQ =

2

XYCPX = QYDPQ =

3

Show answer & solution

Option A

XY is parallel to PQ

Approach

Compare the two ratios cut off at the common vertex $R$ ; equal ratios force similarity, and similarity forces the parallelism.

Step-by-step solution

1
$\frac{R X}{R P} = \frac{4}{10} = \frac{2}{5}$ and $\frac{R Y}{R Q} = \frac{2}{5}$
Why: $X$ lies on $P R$ and $Y$ on $QR$ , so both ratios are measured from the shared vertex $R$ along the two sides that meet there.
2
$△ R X Y \sim △ R P Q$ by SAS similarity, the angle at $R$ being common.
Why: Two sides about the common angle are in the same ratio, which is exactly the SAS similarity test.
Similar triangles have proportional sides
3
$∠ R X Y = ∠ R P Q$ , and these are corresponding angles at the transversal $P R$ , so $X Y ∥ P Q$ .
Why: Equal corresponding angles is the converse test for parallel lines.
Angles made by a transversal of two parallel lines
4
$\frac{X Y}{P Q} = \frac{2}{5}$ , i.e. $P Q = \frac{5}{2} X Y$
Why: The similarity ratio applies to the third pair of sides too, which settles the numerical options: the factor is $2.5$ , neither $2$ nor $3$ .

Shortcut

$\frac{R X}{R P} = \frac{R Y}{R Q} = \frac{2}{5}$ , so $X Y ∥ P Q$ by the converse of the basic proportionality theorem.

Common mistakes

Option B

$P Q = 2 X Y$ takes the ratio as $\frac{R X}{X P} = \frac{4}{6}$ , or rounds the true factor $\frac{5}{2}$ down to $2$ .

Option C

$P X = Q Y$ expects the leftover segments to match ( $6$ against $3$ ); the theorem equates ratios, not remainders.

Option D

$P Q = 3 X Y$ rounds the true factor $\frac{5}{2}$ up to $3$ .

Skills used

79Geometry • Line • Parallel Lines and Transversals

ModerateIdeal time: about 1.5 min

Consider the following statements in respect of three straight lines A, B and C on a plane:

1: If A and C are parallel and B and C are parallel; then A and B are parallel.

2: If A is perpendicular to C and B is perpendicular to C; then A and B are parallel.

3: If the acute angle between A and C is equal to the acute angle between B and C; then A and B are parallel.

Which of the above statements are correct?

A1, 2 and 3B1 and 2 onlyC1 and 3 onlyD2 and 3 only

Show answer & solution

Option B

1 and 2 only

Approach

Take the three claims one at a time in the plane, looking for a counterexample to any that is not forced.

Angles made by a transversal of two parallel lines

Step-by-step solution

1
Statement 1: if $A ∥ C$ and $B ∥ C$ then $A ∥ B$ . Were $A$ and $B$ to meet at a point, two distinct lines through that point would both be parallel to $C$ , which the parallel postulate forbids. TRUE.
Why: Parallelism of coplanar lines is transitive, and the question fixes all three lines in one plane.
Angles made by a transversal of two parallel lines
2
Statement 2: if $A ⊥ C$ and $B ⊥ C$ then $A ∥ B$ . Reading $C$ as a transversal of $A$ and $B$ , the corresponding angles are $9 0^{\circ}$ each, hence equal. TRUE.
Why: Equal corresponding angles at a transversal is the converse test for parallel lines.
3
Statement 3: take $C$ horizontal, $A$ inclined at $+ 3 0^{\circ}$ to it and $B$ at $- 3 0^{\circ}$ . Both acute angles equal $3 0^{\circ}$ , yet $A$ and $B$ cross at $6 0^{\circ}$ . FALSE.
Why: An equal acute angle fixes only the SIZE of the inclination, not the side it is taken on: reflecting a line in $C$ preserves the acute angle but reverses the direction.
Angles on a straight line and vertically opposite angles
4
Correct statements: $1$ and $2$ only.
Why: Two of the three survive, so the answer names exactly those two.

Common mistakes

Option A

1, 2 and 3: accepts statement 3, missing the line reflected in $C$ , which makes the same acute angle without being parallel.

Option C

1 and 3 only: rejects the perpendicularity result while accepting the equal-angle one, the exact reverse of the truth.

Option D

2 and 3 only: rejects transitivity of parallelism, which is the least questionable of the three.

Skills used

Angles made by a transversal of two parallel lines Angles on a straight line and vertically opposite angles

80Geometry • Quadrilateral • Parallelogram

EasyIdeal time: about 45 s

The diagonals of a rhombus are of length $20$ cm and $48$ cm. What is the length of a side of the rhombus?

13

cmB

26

cmC

36

cmD

39

Show answer & solution

Option B

26

Approach

The diagonals of a rhombus bisect each other at right angles, so a side is the hypotenuse on the two half-diagonals.

The diagonals of a rectangle are equal and bisect each other

Step-by-step solution

1
Half-diagonals: $\frac{20}{2} = 10$ cm and $\frac{48}{2} = 24$ cm
Why: The diagonals of a rhombus, being those of a parallelogram, bisect each other.
The diagonals of a rectangle are equal and bisect each other
2
Side $= 1 0^{2} + 2 4^{2} = 100 + 576 = 676 = 26$ cm
Why: The diagonals of a rhombus are perpendicular, so the two half-diagonals are the legs of a right triangle whose hypotenuse is a side.
Pythagoras' theorem

Common mistakes

Option A

$13$ cm halves the diagonals twice over, working from $5$ and $12$ instead of $10$ and $24$ .

Option C

$36$ cm: no route reaches it.

Option D

$39$ cm: no route reaches it.

Skills used

The diagonals of a rectangle are equal and bisect each other Pythagoras' theorem