Elementary Mathematics 2017 (I)

Set A

Questions 81–100 of 100

81Geometry • Circle • Chords

ModerateIdeal time: about 1.5 min

Two parallel chords of a circle whose diameter is $13$ cm are respectively $5$ cm and $12$ cm in length. If both the chords are on the same side of the diameter, then the distance between these chords is

5.5

cmB

5

cmC

3.5

cmD

3

Show answer & solution

Option C

3.5

Approach

Get each chord's distance from the centre from a right triangle on its half-chord and the radius; on the same side those distances subtract.

The perpendicular from the centre bisects a chord

Step-by-step solution

1
$r = \frac{13}{2} = 6.5 cm$
Why: The radius is half the given diameter.
2
$d_{1} = 6. 5^{2} - 2. 5^{2} = 42.25 - 6.25 = 6 cm$
Why: The perpendicular from the centre bisects the $5$ cm chord, so the right triangle has legs $2.5$ and $d_{1}$ with hypotenuse $6.5$ .
The perpendicular from the centre bisects a chord Pythagoras' theorem
3
$d_{2} = 6. 5^{2} - 6^{2} = 42.25 - 36 = 2.5 cm$
Why: The same construction on the $12$ cm chord, whose half is $6$ .
4
$d_{1} - d_{2} = 6 - 2.5 = 3.5 cm$
Why: Both chords lie on the same side of the centre, so the gap between them is the difference of the two distances, not their sum.

Common mistakes

Option A

No route reaches $5.5$ cm here.

Option B

No route reaches $5$ cm here.

Option D

Slipped on $6.25$ , reading it as $3$ rather than $2.5$ .

Skills used

The perpendicular from the centre bisects a chord Pythagoras' theorem

82Mensuration • Solids • Cone

ModerateIdeal time: about 1.5 min

If the radius of a right circular cone is increased by $p %$ without increasing its height, then what is the percentage increase in the volume of the cone?

p^{2}

2 p^{2}

\frac{p ^{2}}{100}

p (2 + \frac{p}{100})

Show answer & solution

Option D

p (2 + \frac{p}{100})

Approach

With the height fixed the volume goes as $r^{2}$ , so multiply by $(1 + \frac{p}{100})^{2}$ and subtract $1$ .

Percentage change as a multiplier

Step-by-step solution

1
$V = \frac{1}{3} π r^{2} h$
Why: Only $r$ changes, so $V$ is proportional to $r^{2}$ and the constants drop out of any ratio.
Volume of a right circular cone
2
$r \mapsto r (1 + \frac{p}{100})$
Why: An increase of $p %$ is the multiplier $1 + \frac{p}{100}$ .
Percentage change as a multiplier
3
$\frac{V ^{'}}{V} = (1 + \frac{p}{100})^{2} = 1 + \frac{2 p}{100} + \frac{p ^{2}}{10000}$
Why: Expand with the square-of-a-sum identity; the cross term is what a careless answer loses.
Square of a sum and of a difference
4
$(\frac{V ^{'}}{V} - 1) \times 100 = 2 p + \frac{p ^{2}}{100} = p (2 + \frac{p}{100})$
Why: Subtracting $1$ leaves the fractional increase; multiplying by $100$ turns it into a percentage.

Common mistakes

Option A

No route reaches $p^{2}$ here.

Option B

No route reaches $2 p^{2}$ here.

Option C

Kept only the square term $\frac{p ^{2}}{100}$ of the expansion and dropped the $2 p$ .

Skills used

Percentage change as a multiplier Volume of a right circular cone Square of a sum and of a difference

83Mensuration • Plane Figures • Circle

EasyIdeal time: about 1 min

A copper wire when bent in the form of a square encloses an area of $121$ cm $^{2}$ . If the same wire is bent in the form of a circle, it encloses an area equal to

121

^{2}

144

^{2}

154

^{2}

168

^{2}

Show answer & solution

Option C

154

^{2}

Approach

Bending the wire preserves its length, so the square's perimeter becomes the circle's circumference.

Area and circumference of a circle

Step-by-step solution

1
$a^{2} = 121 \Rightarrow a = 11 cm$
Why: The square's area gives its side.
Perimeter and area of a square
2
$P = 4 a = 44 cm$
Why: That side gives the perimeter, which is the length of the wire.
3
$2 π r = 44 \Rightarrow r = \frac{44 \times 7}{2 \times 22} = 7 cm$
Why: The same wire bent into a circle has that length as its circumference.
Area and circumference of a circle
4
$π r^{2} = \frac{22}{7} \times 49 = 154 cm^{2}$
Why: The radius gives the enclosed area.

Common mistakes

Option A

Assumed bending the wire preserves the enclosed area rather than the length.

Option B

No route reaches $144$ cm $^{2}$ here.

Option D

No route reaches $168$ cm $^{2}$ here.

Skills used

Area and circumference of a circle Perimeter and area of a square

84Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

ABC is a triangle and D is a point on the side BC. If BC $= 12$ cm, BD $= 9$ cm and $∠$ ADC $= ∠$ BAC, then the length of AC is equal to

5

cmB

6

cmC

8

cmD

9

Show answer & solution

Option B

6

Approach

The given equal angles plus the shared angle at $C$ make triangles $A D C$ and $B A C$ similar, which turns $A C$ into the mean proportional between $B C$ and $D C$ .

Step-by-step solution

1
$D C = B C - B D = 12 - 9 = 3 cm$
Why: $D$ lies on $B C$ , so the two pieces add to the whole.
2
$∠ A D C = ∠ B A C, ∠ A C D = ∠ B C A \Rightarrow △ A D C \sim △ B A C$
Why: Two pairs of equal angles is AA similarity; the angle at $C$ is common to both triangles, and the vertices correspond as $A \leftrightarrow B$ , $D \leftrightarrow A$ , $C \leftrightarrow C$ .
Similar triangles have proportional sides
3
$\frac{A C}{B C} = \frac{D C}{A C} \Rightarrow A C^{2} = B C \times D C$
Why: Corresponding sides of similar triangles are in one ratio; written out, this says $A C$ is the mean proportional between the whole $B C$ and the far piece $D C$ .
4
$A C^{2} = 12 \times 3 = 36 \Rightarrow A C = 6 cm$
Why: Take the positive square root, since $A C$ is a length.
denotes the non-negative root

Common mistakes

Option A

Paired the wrong two segments, using $A C^{2} = B D \times D C = 27$ and rounding $5.2$ to $5$ .

Option C

No route reaches $8$ cm here.

Option D

Read off the given $B D = 9$ cm as the answer.

Skills used

denotes the non-negative root

85Mensuration • Solids • Sphere

Very easyIdeal time: about 30 s

If the surface area of a sphere is reduced to one-ninth of the area, its radius reduces to

AOne-fourthBOne-thirdCOne-fifthDOne-ninth

Show answer & solution

Option B

One-third

Approach

Surface area goes as the square of the radius, so a factor $\frac{1}{9}$ on the area is a factor $\frac{1}{3}$ on the radius.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
$S = 4 π r^{2}$
Why: The surface area of a sphere is proportional to $r^{2}$ .
Volume and surface area of a sphere
2
$\frac{S ^{'}}{S} = (\frac{r ^{'}}{r})^{2} = \frac{1}{9}$
Why: The constant $4 π$ cancels in the ratio.
Direct variation as a power: $y \propto x^{n}$
3
$\frac{r ^{'}}{r} = \frac{1}{3}$
Why: Take the positive square root; a radius is positive.
denotes the non-negative root

Common mistakes

Option A

No route reaches one-fourth here.

Option C

No route reaches one-fifth here.

Option D

Carried the area ratio straight across to the radius without taking the square root.

Skills used

Direct variation as a power:

y \propto x^{n}

Volume and surface area of a sphere

denotes the non-negative root

86Geometry • Quadrilateral • Trapezium

EasyIdeal time: about 55 s

In a trapezium ABCD, AB is parallel to CD and the diagonals intersect each other at O. What is the ratio of OA to OC equal to?

ARatio of OB to ODBRatio of BC to CDCRatio of AD to ABDRatio of AC to BD

Show answer & solution

Option A

Ratio of OB to OD

Approach

$A B ∥ C D$ makes triangles $A O B$ and $C O D$ similar, and similarity puts both pairs of diagonal pieces in the same ratio.

Step-by-step solution

1
$∠ O A B = ∠ O C D, ∠ O B A = ∠ O D C$
Why: $A B ∥ C D$ with the diagonals $A C$ and $B D$ as transversals gives equal alternate angles.
Angles made by a transversal of two parallel lines
2
$△ A O B \sim △ C O D$
Why: Two pairs of equal angles is AA similarity, with $A \leftrightarrow C$ , $O \leftrightarrow O$ , $B \leftrightarrow D$ .
Similar triangles have proportional sides
3
$\frac{O A}{O C} = \frac{O B}{O D} = \frac{A B}{C D}$
Why: Corresponding sides of similar triangles share one ratio, so $O A : O C$ is the same as $O B : O D$ .

Common mistakes

Option B

No route makes $O A : O C$ equal to $B C : C D$ in a general trapezium.

Option C

No route makes $O A : O C$ equal to $A D : A B$ in a general trapezium.

Option D

Confused the pieces of the diagonals with the whole diagonals, which are not in that ratio unless the trapezium is isosceles.

Skills used

87Mensuration • Recasting and Melting

ModerateIdeal time: about 2 min

Ice-cream, completely filled in a cylinder of diameter $35$ cm and height $32$ cm, is to be served by completely filling identical disposable cones of diameter $4$ cm and height $7$ cm. The maximum number of persons that can be served in this way is

950

1000

1050

1100

Show answer & solution

Option C

1050

Approach

Compute the cylinder's volume and one cone's volume, then divide.

Recasting preserves volume

Step-by-step solution

1
$V_{cyl} = π (\frac{35}{2})^{2} \times 32 = \frac{22}{7} \times \frac{1225}{4} \times 32 = 30800 cm^{3}$
Why: The tub is a cylinder of radius $17.5$ cm and height $32$ cm; $\frac{1225}{7} = 175$ , so the arithmetic stays exact.
Volume of a right circular cylinder
2
$V_{cone} = \frac{1}{3} π (\frac{4}{2})^{2} \times 7 = \frac{1}{3} \times \frac{22}{7} \times 4 \times 7 = \frac{88}{3} cm^{3}$
Why: Each cone has radius $2$ cm and height $7$ cm, and a cone is one third of the cylinder on the same base.
Volume of a right circular cone
3
$n = \frac{30800}{\frac{88}{3}} = \frac{30800 \times 3}{88} = 1050$
Why: Every cone is filled completely, so the number served is the total volume divided by one cone's volume.
Recasting preserves volume

Common mistakes

Option A

No route reaches $950$ here.

Option B

Rounded the cylinder's volume to $30000$ cm $^{3}$ before dividing.

Option D

No route reaches $1100$ here.

Skills used

Recasting preserves volume Volume of a right circular cylinder Volume of a right circular cone

88Mensuration • Plane Figures • Circle

EasyIdeal time: about 1 min

The radius of a circle is increased so that its circumference increases by $15%$ . The area of the circle will increase by

31.25%

32.25%

33.25%

34.25%

Show answer & solution

Option B

32.25%

Approach

Circumference is proportional to radius, so a $15%$ rise in circumference is a $15%$ rise in radius, and the area multiplies by $1.1 5^{2}$ .

Percentage change as a multiplier

Step-by-step solution

1
$C = 2 π r$
Why: $C$ and $r$ are in direct proportion, so the same percentage rise applies to the radius.
Area and circumference of a circle
2
$r \mapsto 1.15 r$
Why: A rise of $15%$ is the multiplier $1.15$ .
Percentage change as a multiplier
3
$A = π r^{2} \Rightarrow A \mapsto (1.15)^{2} A = 1.3225 A$
Why: Area goes as the square of the radius, so the multiplier is squared.
Square of a sum and of a difference
4
$(1.3225 - 1) \times 100 = 32.25%$
Why: Subtract $1$ to turn the factor into an increase.

Shortcut

$2 \times 15 + \frac{1 5 ^{2}}{100} = 30 + 2.25 = 32.25%$ .

Common mistakes

Option A

Took $\frac{1 5 ^{2}}{100}$ as $1.25$ instead of $2.25$ in the successive-change formula.

Option C

No route reaches $33.25%$ here.

Option D

No route reaches $34.25%$ here.

Skills used

Percentage change as a multiplier Area and circumference of a circle Square of a sum and of a difference

89Geometry • Quadrilateral • Parallelogram

EasyIdeal time: about 45 s

ABCD is a rectangle. The diagonals AC and BD intersect at O. If AB $= 32$ cm and AD $= 24$ cm, then what is OD equal to?

22

cmB

20

cmC

18

cmD

16

Show answer & solution

Option B

20

Approach

The diagonals of a rectangle are equal and bisect each other, so $O D$ is half a diagonal, and the diagonal comes from Pythagoras.

Pythagoras' theorem

Step-by-step solution

1
$B D = A B^{2} + A D^{2} = 3 2^{2} + 2 4^{2}$
Why: $∠ A$ is a right angle in a rectangle, so triangle $A B D$ is right-angled at $A$ with $B D$ as hypotenuse.
Pythagoras' theorem
2
$B D = 1024 + 576 = 1600 = 40 cm$
Why: $(32, 24, 40)$ is $8 \times (4, 3, 5)$ , so the root is exact.
3
$O D = \frac{B D}{2} = 20 cm$
Why: The diagonals of a rectangle bisect each other at $O$ , so $O$ is the midpoint of $B D$ .
The diagonals of a rectangle are equal and bisect each other

Shortcut

$32 : 24 = 4 : 3$ , so the diagonal is $8 \times 5 = 40$ and $O D$ is $20$ .

Common mistakes

Option A

No route reaches $22$ cm here.

Option C

No route reaches $18$ cm here.

Option D

Halved the side $A B = 32$ instead of the diagonal.

Skills used

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

90Mensuration • Plane Figures • Composite and Shaded Regions

Very hardIdeal time: about 3.5 min

A field is divided into four regions as shown in the given figure. What is the area of the field in square metres?

A compound polygon with no lettered vertices, divided into four regions by internal segments. The top edge is horizontal and is split by a mid-point into two parts, each labelled 2 m. A right-angle mark sits at that mid-point, between the top edge and a short vertical segment labelled 1 m that drops from it; a second right-angle mark sits at the top-right corner, between the top edge and the right-hand edge, which is vertical and labelled 3 m. From the foot of the 1 m segment a horizontal segment labelled 2 m runs left to a point on a second, longer vertical line, meeting it at a marked right angle. That second vertical runs from the top-left corner of the top edge down to the bottom apex of the figure, and the part of it lying below the horizontal segment is what the internal label 3 m measures. Two further segments leave the foot of the 1 m segment: one slanting down-left to the bottom apex, and one slanting down-right to the foot of the right-hand edge, so the boundary turns inward at that foot and the figure has a notch there. The left boundary is two slant edges: one labelled 2 m from the top-left corner down-left to the leftmost vertex, and one labelled 3 m from that leftmost vertex down-right to the bottom apex. The four regions are therefore the quadrilateral between the top edge and the horizontal segment, the triangle below that horizontal segment, the quadrilateral to the right of the 1 m segment, and the triangle on the two left-hand slants.

6 + \frac{3}{4} 5

5 + \frac{3}{2} 3

9 + \frac{3}{4} 15

7 + 22

Show answer & solution

Option C

9 + \frac{3}{4} 15

Approach

Cut the field along the segments already drawn into a rectangle, two triangles and a trapezium, and add the four areas.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
Top edge $A$ , $B$ , $C$ with $A B = B C = 2$ ; $B E = 1$ down from $B$ ; $E F = 2$ left to $F$ on the vertical through $A$ ; $F G = 3$ down to the bottom apex $G$ ; $C D = 3$ down from $C$ ; $H$ at the far left with $H A = 2$ and $H G = 3$
Why: Name the corners from the drawing. The right angles marked at $B$ and at $C$ make $B E$ and $C D$ vertical against the horizontal top edge, and the right angle at $F$ makes $E F$ horizontal; the $3$ m label sits beside the part of the internal vertical BELOW $F$ , so it measures $F G$ and not the whole of $A G$ .
2
$[A B E F] = 2 \times 1 = 2 m^{2}$
Why: $A B E F$ has two vertical sides $A F$ and $B E$ and two horizontal sides $A B$ and $F E$ , so it is a rectangle of sides $2$ m and $1$ m; this also fixes $A F = B E = 1$ m.
Perimeter and area of a rectangle
3
$[F E G] = \frac{1}{2} \times 2 \times 3 = 3 m^{2}$
Why: Triangle $F E G$ is right-angled at $F$ , with the horizontal $F E = 2$ m as base and the vertical $F G = 3$ m as height.
Area of a triangle from base and height
4
$[B C D E] = \frac{1}{2} (1 + 3) \times 2 = 4 m^{2}$
Why: $B E$ and $C D$ are both vertical, so they are the parallel sides, of lengths $1$ m and $3$ m; the horizontal $B C = 2$ m is the distance between them.
Area of a trapezium
5
$A G = A F + F G = 1 + 3 = 4, s = \frac{2 + 3 + 4}{2} = \frac{9}{2}, [H A G] = \frac{9}{2} \cdot \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} 15$
Why: The left region is the triangle on $H A = 2$ m, $H G = 3$ m and the whole internal vertical $A G = 4$ m; no height is marked for it, so Heron's formula is what is left.
Heron's formula
6
$2 + 3 + 4 + \frac{3}{4} 15 = 9 + \frac{3}{4} 15 m^{2}$
Why: The four regions tile the field and do not overlap, so their areas add.
Cut a compound plane figure into standard pieces

Common mistakes

Option A

No route reaches $6 + \frac{3}{4} 5$ here.

Option B

No route reaches $5 + \frac{3}{2} 3$ here.

Option D

Read the $3$ m label as the whole internal vertical $A G$ , which makes the left triangle a $(2, 3, 3)$ one of area $22$ and shortens $F G$ to $2$ m, and then counted the right-hand region as a triangle on the $3$ m edge rather than a trapezium, leaving $2 + 2 + 3 = 7$ .

Skills used

Cut a compound plane figure into standard pieces Perimeter and area of a rectangle Area of a triangle from base and height Area of a trapezium Heron's formula

91Geometry • Circle • Touching Circles

HardIdeal time: about 2 min

In the figure given below, D is the diameter of each circle. What is the diameter of the shaded circle?

Four equal circles of diameter D arranged in a 2 \times 2 square, each touching its two neighbours. A small hatched (shaded) circle sits in the central gap, tangent to all four.

D (2 - 1)

D (2 + 1)

D (2 + 2)

D (2 - 2)

Show answer & solution

Option A

D (2 - 1)

Approach

The four centres form a square of side $D$ ; the shaded circle sits at its centre and touches each big circle externally.

Distance between the centres of touching circles

Step-by-step solution

1
$R = \frac{D}{2}, centre-to-centre = 2 R = D$
Why: Neighbouring circles touch externally, so the distance between their centres is the sum of their radii, and the four centres form a square of side $D$ .
Distance between the centres of touching circles
2
$half-diagonal = \frac{D 2}{2} = \frac{D}{2}$
Why: The shaded circle's centre is the centre of that square, half a diagonal from each of the four centres, and the diagonal of a square of side $D$ is $D 2$ by Pythagoras.
Pythagoras' theorem
3
$\frac{D 2}{2} = R + r = \frac{D}{2} + r$
Why: The shaded circle touches each big circle externally, so that half-diagonal is the sum of the two radii.
4
$r = \frac{D 2}{2} - \frac{D}{2} = \frac{D ( 2 - 1 )}{2} \Rightarrow 2 r = D (2 - 1)$
Why: Subtract, then double: the question asks for the diameter, not the radius.
Rationalise a denominator using the conjugate

Common mistakes

Option B

Added the radii instead of subtracting, which gives a circle bigger than the four it sits between.

Option C

No route reaches $D (2 + 2)$ here.

Option D

Turned the radius $\frac{D ( 2 - 1 )}{2}$ into a diameter by multiplying by $2$ rather than by $2$ .

Skills used

Distance between the centres of touching circles Pythagoras' theorem Rationalise a denominator using the conjugate

92Geometry • Triangle • Similarity

ModerateIdeal time: about 1.5 min

In the figure given below, AC is parallel to ED and AB $=$ DE $= 5$ cm and BC $= 7$ cm. What is the area ABDE : area BDE : area BCD equal to?

Points A, B, C lie on a horizontal base line with AB labelled 5 cm and BC labelled 7 cm. Points E and D lie on a horizontal line above it with ED labelled 5 cm; E is above the segment AB and D is above the segment BC. Segments AE, EB, BD and DC are drawn, forming triangle ABE, triangle EBD and triangle BCD.

10 : 5 : 7

8 : 4 : 7

2 : 1 : 2

8 : 4 : 5

Show answer & solution

Option A

10 : 5 : 7

Approach

Every triangle here has a base on one of the two parallel lines and its apex on the other, so all share the same height and their areas are proportional to their bases.

Area of a triangle from base and height

Step-by-step solution

1
$[A B E] = \frac{1}{2} \times A B \times h = \frac{5 h}{2}$
Why: $A C ∥ E D$ , so the distance $h$ between the two lines is the height of every one of these triangles; $E$ is the apex above the base $A B = 5$ cm.
Area of a triangle from base and height
2
$[E B D] = \frac{1}{2} \times E D \times h = \frac{5 h}{2}$
Why: Here the base $E D = 5$ cm lies on the upper line and the apex $B$ on the lower one; the height is the same $h$ .
3
$[B C D] = \frac{1}{2} \times B C \times h = \frac{7 h}{2}$
Why: Base $B C = 7$ cm on the lower line, apex $D$ on the upper.
4
$[A B D E] = [A B E] + [E B D] = \frac{5 h}{2} + \frac{5 h}{2} = 5 h$
Why: The segment $E B$ splits the quadrilateral $A B D E$ into exactly those two triangles.
5
$[A B D E] : [B D E] : [B C D] = 5 h : \frac{5 h}{2} : \frac{7 h}{2} = 10 : 5 : 7$
Why: Multiply through by $\frac{2}{h}$ to clear the common height.

Common mistakes

Option B

No route reaches $8 : 4 : 7$ here.

Option C

Got $[A B D E] = 2 [B D E]$ correctly but then assumed the remaining triangle $B C D$ equals the quadrilateral.

Option D

No route reaches $8 : 4 : 5$ here.

Skills used

Area of a triangle from base and height

93Geometry • Quadrilateral • Parallelogram

EasyIdeal time: about 1 min

In the figure given below, PQRS is a parallelogram. PA bisects angle P and SA bisects angle S. What is angle PAS equal to?

Parallelogram PQRS with P at the top left, Q at the top right, R at the bottom right and S at the bottom left. An interior point A lies near the left side; segments PA and SA are drawn, meeting at A.

6 0^{\circ}

7 5^{\circ}

9 0^{\circ}

10 0^{\circ}

Show answer & solution

Option C

9 0^{\circ}

Approach

$∠ P$ and $∠ S$ are co-interior angles of the parallelogram and so add to $18 0^{\circ}$ ; their halves add to $9 0^{\circ}$ , leaving $9 0^{\circ}$ for the third angle of triangle $P A S$ .

Angles made by a transversal of two parallel lines

Step-by-step solution

1
$∠ S P Q + ∠ P S R = 18 0^{\circ}$
Why: $P Q ∥ S R$ with $P S$ as transversal, so the two interior angles on the same side of $P S$ are supplementary.
Angles made by a transversal of two parallel lines
2
$∠ A P S = \frac{1}{2} ∠ S P Q, ∠ A S P = \frac{1}{2} ∠ P S R$
Why: $P A$ bisects the angle at $P$ and $S A$ bisects the angle at $S$ .
3
$∠ A P S + ∠ A S P = \frac{1}{2} (∠ S P Q + ∠ P S R) = 9 0^{\circ}$
Why: Half of a supplementary pair is a complementary pair.
4
$∠ P A S = 18 0^{\circ} - 9 0^{\circ} = 9 0^{\circ}$
Why: The three angles of triangle $P A S$ add to $18 0^{\circ}$ .
Angle sum of a triangle

Shortcut

The bisectors of two co-interior angles always meet at right angles -- the answer is $9 0^{\circ}$ whatever the parallelogram.

Common mistakes

Option A

Assumed the parallelogram has angles $6 0^{\circ}$ and $12 0^{\circ}$ and then quoted a half-angle rather than $∠ P A S$ .

Option B

No route reaches $7 5^{\circ}$ here.

Option D

No route reaches $10 0^{\circ}$ here.

Skills used

Angles made by a transversal of two parallel lines Angle sum of a triangle

94Geometry • Triangle • Angle Bisector

EasyIdeal time: about 1 min

In the figure given below, $∠$ A $= 8 0^{\circ}$ and $∠$ ABC $= 6 0^{\circ}$ . BD and CD bisect angles B and C respectively. What are the values of $x$ and $y$ respectively?

Triangle ABC with A at the apex, B at the bottom left and C at the bottom right. An interior point D lies below A; segments BD and CD are drawn. The angle at D in triangle BDC is labelled y^\circ. At vertex C two adjacent angles are each labelled x^\circ, formed by the bisector CD.

10

and

130

10

and

125

20

and

130

20

and

125

Show answer & solution

Option C

20

and

130

Approach

Get $∠ C$ from the angle sum of triangle $A B C$ , halve it for $x$ , then use the angle sum in triangle $B D C$ for $y$ .

Angle sum of a triangle

Step-by-step solution

1
$∠ A C B = 18 0^{\circ} - 8 0^{\circ} - 6 0^{\circ} = 4 0^{\circ}$
Why: The three angles of triangle $A B C$ add to $18 0^{\circ}$ .
Angle sum of a triangle
2
$x = \frac{1}{2} \times 4 0^{\circ} = 2 0^{\circ}$
Why: $C D$ bisects $∠ A C B$ into the two angles marked $x$ in the figure.
Internal angle bisector of a triangle
3
$∠ D B C = \frac{1}{2} \times 6 0^{\circ} = 3 0^{\circ}$
Why: $B D$ bisects $∠ A B C$ .
4
$y = 18 0^{\circ} - 3 0^{\circ} - 2 0^{\circ} = 13 0^{\circ}$
Why: Angle sum in triangle $B D C$ , whose other two angles are the two half-angles.

Shortcut

$D$ is the incentre, so $y = 9 0^{\circ} + \frac{∠ A}{2} = 9 0^{\circ} + 4 0^{\circ} = 13 0^{\circ}$ , and $x = \frac{18 0 ^{\circ} - 8 0 ^{\circ} - 6 0 ^{\circ}}{2} = 2 0^{\circ}$ .

Common mistakes

Option A

Halved $∠ C$ twice, taking $x$ as a quarter of $4 0^{\circ}$ .

Option B

Made both errors: $x$ quartered and $y$ mis-subtracted.

Option D

Slipped in $18 0^{\circ} - 3 0^{\circ} - 2 0^{\circ}$ , giving $12 5^{\circ}$ for $y$ .

Skills used

Angle sum of a triangle Internal angle bisector of a triangle

95Geometry • Triangle

HardIdeal time: about 2 min

In the figure given below, PQR is a non-isosceles right-angled triangle, right angled at Q. If LM and QT are parallel and QT $=$ PT, then what is $∠$ RLM equal to?

Right-angled triangle PQR with the right angle marked at Q at the bottom left, P at the top and R at the bottom right. T lies on the hypotenuse PR and M lies on PR below T. L lies on the base QR between Q and R. Segment QT joins Q to T, and segment LM joins L to M, drawn parallel to QT.

∠

PQTB

∠

LRMC

∠

RMLD

∠

QPT

Show answer & solution

Option B

∠

LRM

Approach

$QT = P T$ forces $T$ to be the midpoint of the hypotenuse, so $T Q = T R$ and triangle $T QR$ is isosceles; the parallel $L M$ then copies $∠ R QT$ down to $∠ R L M$ .

The circumcentre is the point equidistant from the vertices

Step-by-step solution

1
$T P = T Q \Rightarrow T P = T Q = T R$
Why: $T$ lies on $P R$ and is equidistant from $P$ and $Q$ . In a triangle right-angled at $Q$ the point of the hypotenuse equidistant from all three vertices is its midpoint, the circumcentre -- so $T$ is that midpoint and $T R$ joins the list.
The circumcentre is the point equidistant from the vertices
2
$∠ T QR = ∠ T R Q$
Why: Triangle $T QR$ has $T Q = T R$ , so its base angles are equal.
Base angles of an isosceles triangle
3
$∠ R L M = ∠ R QT$
Why: $L M ∥ QT$ cut by the transversal $QR$ , with $L$ between $Q$ and $R$ , gives equal corresponding angles at $L$ and at $Q$ .
Angles made by a transversal of two parallel lines
4
$∠ R L M = ∠ T R Q = ∠ L R M$
Why: $L$ lies on $QR$ and $M$ on $R P$ , so $∠ L R M$ and $∠ T R Q$ are the same angle at $R$ .

Common mistakes

Option A

$∠ P QT = ∠ QP T$ , which equals $∠ R L M$ only when the triangle is isosceles -- and the question says it is not.

Option C

Took the third angle of triangle $R L M$ ; it equals $∠ R T Q$ , not the two equal base angles.

Option D

$∠ QP T$ is the same angle as $∠ P QT$ , so this fails for the same reason as A.

Skills used

The circumcentre is the point equidistant from the vertices Base angles of an isosceles triangle Angles made by a transversal of two parallel lines

96Geometry • Quadrilateral • Parallelogram

HardIdeal time: about 2 min

In the figure given below, PQ is parallel to RS and PR is parallel to QS. If $∠$ LPR $= 3 5^{\circ}$ and $∠$ UST $= 7 0^{\circ}$ , then what is $∠$ MPQ equal to?

Five straight lines, each drawn with an arrowhead at both ends. Four of them bound a parallelogram PQSR, with P at the top left, Q at the right, S at the bottom and R at the left: the line through P and Q is parallel to the line through R and S, and the line through P and R is parallel to the line through Q and S, each of the four extending beyond both vertices it passes through. The fifth line passes through P alone and carries the point L on it below and to the left of P and the point M on it above and to the right of P, so L, P, M are collinear and the rays PR and PQ both lie between the rays PL and PM; the three angles \angle LPR, \angle RPQ and \angle QPM therefore make up the straight angle at P. The point U is marked on the extension of RS beyond S and the point T on the extension of QS beyond S, so \angle UST is vertically opposite the parallelogram's interior angle \angle RSQ. No angle values are printed in the figure.

5 5^{\circ}

7 0^{\circ}

7 5^{\circ}

8 0^{\circ}

Show answer & solution

Option C

7 5^{\circ}

Approach

The vertical angle at $S$ gives the parallelogram's angle there, its opposite angle at $P$ equals it, and the three angles at $P$ on one side of the line through $L$ and $M$ close a straight angle.

Angles of a parallelogram

Step-by-step solution

1
$∠ QS R = ∠ U S T = 7 0^{\circ}$
Why: $U$ and $T$ are the rays beyond $S$ , so $∠ U S T$ is vertically opposite the parallelogram's interior angle at $S$ .
Angles on a straight line and vertically opposite angles
2
$∠ R P Q = ∠ QS R = 7 0^{\circ}$
Why: $P Q ∥ R S$ and $P R ∥ QS$ make $P QS R$ a parallelogram, and opposite angles of a parallelogram are equal.
Angles of a parallelogram
3
$∠ L P R + ∠ R P Q + ∠ QP M = 18 0^{\circ}$
Why: $L$ , $P$ and $M$ lie on one straight line, with the rays $P R$ and $P Q$ both on the same side of it, so the three angles at $P$ make a straight angle.
4
$∠ M P Q = 18 0^{\circ} - 7 0^{\circ} - 3 5^{\circ} = 7 5^{\circ}$
Why: Substitute the two known angles.

Common mistakes

Option A

No route reaches $5 5^{\circ}$ here.

Option B

Copied the given $∠ U S T$ , taking $∠ M P Q$ for the parallelogram's opposite angle.

Option D

No route reaches $8 0^{\circ}$ here.

Skills used

Angles of a parallelogram Angles on a straight line and vertically opposite angles

97Geometry • Triangle • Congruence

ModerateIdeal time: about 2 min

In the figure given below, ABC is a triangle with AB $=$ BC and D is an interior point of the triangle ABC such that $∠$ DAC $= ∠$ DCA.

Triangle ABC with B at the apex, A at the bottom left and C at the bottom right. An interior point D lies below B on the line from B towards the base. Segments BD, AD and DC are drawn.

Consider the following statements:

1: Triangle ADC is an isosceles triangle.

2: D is the centroid of the triangle ABC.

3: Triangle ABD is congruent to the triangle CBD.

Which of the above statements are correct?

A1 and 2 onlyB2 and 3 onlyC1 and 3 onlyD1, 2 and 3

Show answer & solution

Option C

1 and 3 only

Approach

Test the three statements separately: the equal base angles make $A D C$ isosceles, that plus $A B = C B$ gives an SSS congruence, but nothing pins $D$ to the centroid.

Congruence criteria for triangles

Step-by-step solution

1
$∠ D A C = ∠ D C A \Rightarrow D A = D C$
Why: Sides opposite equal angles of a triangle are equal, so triangle $A D C$ is isosceles -- statement 1 holds.
Base angles of an isosceles triangle
2
$A B = C B, D A = D C, B D = B D \Rightarrow △ A B D ≅ △ C B D$
Why: Three pairs of equal sides is SSS congruence, the third pair being the shared side $B D$ -- statement 3 holds.
Congruence criteria for triangles
3
$D A = D C \Rightarrow D lies on the perpendicular bisector of A C$
Why: That perpendicular bisector is the whole median from $B$ (since $A B = C B$ too), and the centroid is only one point of it, so $D$ need not be the centroid -- statement 2 fails.
4
$Statements 1 and 3 hold; statement 2 does not.$
Why: Only the option listing exactly $1$ and $3$ can be right.

Common mistakes

Option A

Read the centroid off the figure and missed that the shared side $B D$ completes an SSS congruence.

Option B

Accepted the centroid and overlooked that $∠ D A C = ∠ D C A$ is precisely the isosceles condition.

Option D

Took the figure, in which $D$ happens to be drawn near the centroid, as proof of statement 2.

Skills used

Congruence criteria for triangles Base angles of an isosceles triangle

98Geometry • Triangle • Congruence

EasyIdeal time: about 1 min

In the figure given below, M is the mid-point of AB and $∠$ DAB $= ∠$ CBA and $∠$ AMC $= ∠$ BMD. Then the triangle ADM is congruent to the triangle BCM by

C at the top left and D at the top right, with A at the bottom left, M at the bottom middle and B at the bottom right. Segments CD, CA, CM, CB, DA, DM and AB are drawn, so that CB and DA cross to form an X between the top and bottom edges.

ASAS ruleBSSS ruleCASA ruleDAAA rule

Show answer & solution

Option C

ASA rule

Approach

Strip the common angle $∠ D M C$ out of $∠ A M C = ∠ B M D$ to get $∠ A M D = ∠ B M C$ , then pair that with $A M = B M$ and the given equal angles at $A$ and $B$ .

Congruence criteria for triangles

Step-by-step solution

1
$A M = B M$
Why: $M$ is the midpoint of $A B$ .
2
$∠ D A M = ∠ C B M$
Why: This is the given $∠ D A B = ∠ C B A$ , since $M$ lies on the segment $A B$ .
3
$∠ A M C = ∠ B M D \Rightarrow ∠ A M D = ∠ B M C$
Why: Both of the given angles contain $∠ D M C$ : $∠ A M C = ∠ A M D + ∠ D M C$ and $∠ B M D = ∠ B M C + ∠ C M D$ , so removing the common part leaves the two remaining angles equal.
4
$∠ D A M = ∠ C B M, A M = B M, ∠ A M D = ∠ B M C$
Why: Two angles and the side lying between them match in the two triangles, which is exactly the ASA criterion.
Congruence criteria for triangles

Common mistakes

Option A

Counted $A M = B M$ together with the two angles as though a second pair of sides were known.

Option B

Only one pair of equal sides is given, so SSS cannot apply.

Option D

AAA establishes similarity, never congruence.

Skills used

Congruence criteria for triangles

99Geometry • Triangle • Congruence

HardIdeal time: about 2 min

ABCD is a square. X is the mid-point of AB and Y is the mid-point of BC.

Consider the following statements:

1: Triangles ADX and BAY are congruent.

2: $∠$ DXA $= ∠$ AYB.

3: DX is inclined at an angle $6 0^{\circ}$ with AY.

4: DX is not perpendicular to AY.

Which of the above statements are correct?

A2, 3 and 4 onlyB1, 2 and 4 onlyC1, 3 and 4 onlyD1 and 2 only

Show answer & solution

Option D

1 and 2 only

Approach

Triangles $A D X$ and $B A Y$ are congruent by SAS; the congruence gives statement 2, and an angle chase on it forces $D X ⊥ A Y$ , which kills both statement 3 and statement 4.

Congruence criteria for triangles

Step-by-step solution

1
$A D = B A = s, ∠ D A X = ∠ A B Y = 9 0^{\circ}, A X = B Y = \frac{s}{2}$
Why: In the square all sides are $s$ and all angles right; $X$ and $Y$ are midpoints, so $A X = B Y = \frac{s}{2}$ . Side-angle-side gives $△ A D X ≅ △ B A Y$ -- statement 1 holds.
Congruence criteria for triangles
2
$∠ D X A = ∠ A Y B$
Why: They are the angles opposite the equal sides $A D$ and $B A$ in that congruence -- statement 2 holds.
3
$∠ A D X + ∠ A X D = 9 0^{\circ}$
Why: Triangle $A D X$ is right-angled at $A$ , so its two acute angles are complementary.
Angle sum of a triangle
4
$∠ B A Y = ∠ A D X \Rightarrow ∠ A X D + ∠ B A Y = 9 0^{\circ}$
Why: Corresponding angles of the congruent triangles. In the small triangle cut off where $A Y$ crosses $D X$ , those two angles are two of the three, so the third is $9 0^{\circ}$ : $D X ⊥ A Y$ .
5
$Statement 3 claims 6 0^{\circ}; statement 4 denies the perpendicularity — both are false.$
Why: Only statements 1 and 2 survive.

Common mistakes

Option A

Rejected the congruence and read the crossing angle off the picture as $6 0^{\circ}$ .

Option B

Got the congruence but never established the perpendicularity, so kept statement 4.

Option C

Kept both false angle statements and dropped the true statement 2.

Skills used

Congruence criteria for triangles Angle sum of a triangle

100Trigonometry • Heights and Distances

ModerateIdeal time: about 1.5 min

From an aeroplane vertically over a straight horizontal road, the angles of depression of two consecutive kilometre-stones on the opposite sides of the aeroplane are observed to be $α$ and $β$ . The height of the aeroplane above the road is

\frac{t a n α + t a n β}{t a n α t a n β}

\frac{t a n α t a n β}{t a n α + t a n β}

\frac{c o t α c o t β}{c o t α + c o t β}

\frac{c o t α + c o t β}{c o t α c o t β}

Show answer & solution

Option B

\frac{t a n α t a n β}{t a n α + t a n β}

Approach

The two ground distances are $h cot α$ and $h cot β$ , and because the stones are on opposite sides of the plane they add to the $1$ km between them.

Angles of elevation and depression

Step-by-step solution

1
$cot α = \frac{d _{1}}{h}, cot β = \frac{d _{2}}{h}$
Why: The plane is vertically above the road, so each stone gives a right triangle in which $h$ is opposite the depression angle and the ground distance is adjacent to it.
Angles of elevation and depression Sine, cosine and tangent in a right triangle
2
$d_{1} + d_{2} = 1 km$
Why: The stones are consecutive kilometre-stones, and they lie on opposite sides of the foot of the vertical, so the two distances add rather than subtract.
3
$h cot α + h cot β = 1 \Rightarrow h = \frac{1}{c o t α + c o t β}$
Why: Substitute and make $h$ the subject.
4
$h = \frac{1}{\frac{1}{t a n α} + \frac{1}{t a n β}} = \frac{t a n α t a n β}{t a n α + t a n β}$
Why: Replace each cotangent by the reciprocal of the tangent and clear the compound fraction.
Reciprocal and quotient relations between the ratios Evaluate a nested fraction from the inside out

Common mistakes

Option A

Inverted the final fraction, writing $cot α + cot β$ for the height instead of its reciprocal.

Option C

Swapped $tan$ for $cot$ throughout, which is $\frac{1}{t a n α + t a n β}$ rather than the height.

Option D

This equals $tan α + tan β$ : the reciprocal was taken on the wrong side of the equation.

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Reciprocal and quotient relations between the ratios Evaluate a nested fraction from the inside out