Elementary Mathematics 2017 (I)

Set A

Questions 61–80 of 100

61Trigonometry • Trigonometric Identity • Elimination of the Angle

HardIdeal time: about 3 min

If $x sin θ = y cos θ = \frac{2 z t a n θ}{1 - t a n ^{2} θ}$ , then what is $4 z^{2} (x^{2} + y^{2})$ equal to?

(x^{2} + y^{2})^{3}

(x^{2} - y^{2})^{2}

(x^{2} - y^{2})^{3}

(x^{2} + y^{2})^{2}

Show answer & solution

Option B

(x^{2} - y^{2})^{2}

Approach

Name the common value $p$ , read $sin θ$ , $cos θ$ and $tan θ$ off the first two equalities, and eliminate $θ$ with the Pythagorean identity.

Pythagorean trigonometric identities

Step-by-step solution

1
$x sin θ = y cos θ = \frac{2 z t a n θ}{1 - t a n ^{2} θ} = p$
Why: A chain of equalities is handled by naming the common value once.
Pythagorean trigonometric identities
2
$sin θ = \frac{p}{x}, cos θ = \frac{p}{y} \Rightarrow tan θ = \frac{y}{x}$
Why: The first two equalities give both ratios in terms of $p$ , and their quotient is $tan θ$ with $p$ cancelling.
Reciprocal and quotient relations between the ratios
3
$\frac{p ^{2}}{x ^{2}} + \frac{p ^{2}}{y ^{2}} = 1 \Rightarrow p^{2} (x^{2} + y^{2}) = x^{2} y^{2}$
Why: $sin^{2} θ + cos^{2} θ = 1$ is what removes $θ$ from the first two equalities.
4
$\frac{2 z \cdot \frac{y}{x}}{1 - \frac{y ^{2}}{x ^{2}}} = \frac{2 z x y}{x ^{2} - y ^{2}} = p \Rightarrow 2 z x y = p (x^{2} - y^{2})$
Why: Substituting $tan θ = \frac{y}{x}$ into the third expression and clearing the $x^{2}$ in the denominator.
Difference of two squares
5
$4 z^{2} x^{2} y^{2} = p^{2} (x^{2} - y^{2})^{2} \Rightarrow 4 z^{2} x^{2} y^{2} (x^{2} + y^{2}) = x^{2} y^{2} (x^{2} - y^{2})^{2}$
Why: Squaring, then multiplying by $x^{2} + y^{2}$ so that $p^{2} (x^{2} + y^{2})$ can be replaced by $x^{2} y^{2}$ from step 3.
6
$4 z^{2} (x^{2} + y^{2}) = (x^{2} - y^{2})^{2}$
Why: Cancelling the common factor $x^{2} y^{2}$ , which is non-zero because $sin θ$ and $cos θ$ appear in denominators of the given ratios.

Common mistakes

Option A

Used $x^{2} + y^{2}$ in place of $x^{2} - y^{2}$ and carried a spurious third power.

Option C

Got the right base $x^{2} - y^{2}$ but cubed it, as if the elimination had been done on three equalities rather than squared once.

Option D

Right power, wrong sign inside: kept $x^{2} + y^{2}$ from the Pythagorean step instead of the $x^{2} - y^{2}$ from the $tan θ$ step.

Skills used

Pythagorean trigonometric identities Reciprocal and quotient relations between the ratios Difference of two squares

62Trigonometry • Trigonometric Identity

EasyIdeal time: about 45 s

If $cos θ_{1} + cos θ_{2} + cos θ_{3} = 3$ , then what is $sin θ_{1} + sin θ_{2} + sin θ_{3}$ equal to?

0

1

2

3

Show answer & solution

Option A

0

Approach

A sum of three cosines can only reach $3$ if each is at its maximum, which pins every angle.

Sine and cosine lie between

- 1

and

1

Step-by-step solution

1
$cos θ_{i} \leq 1 for each i, cos θ_{1} + cos θ_{2} + cos θ_{3} = 3$
Why: Three quantities each at most $1$ can total $3$ only by all being exactly $1$ ; any shortfall in one cannot be made up by the others.
Sine and cosine lie between $- 1$ and $1$
2
$cos θ_{i} = 1 \Rightarrow sin^{2} θ_{i} = 1 - cos^{2} θ_{i} = 0 \Rightarrow sin θ_{i} = 0$
Why: The Pythagorean identity converts each cosine value straight into its sine without needing the angle itself.
Pythagorean trigonometric identities
3
$sin θ_{1} + sin θ_{2} + sin θ_{3} = 0 + 0 + 0 = 0$
Why: All three sines are zero, so their sum is zero.

Common mistakes

Option B

$1$ would require exactly one sine to be non-zero, impossible once every cosine is forced to $1$ .

Option C

$2$ would require two non-zero sines, impossible for the same reason.

Option D

Read the answer off the pattern of the question, expecting the sine sum to mirror the cosine sum at $3$ .

Skills used

Sine and cosine lie between

- 1

and

1

Pythagorean trigonometric identities

63Trigonometry • Trigonometric Equation

Very easyIdeal time: about 40 s

What is the value of $θ$ which satisfies the equation $cos θ + tan θ = 1$ ?

0^{\circ}

3 0^{\circ}

4 5^{\circ}

6 0^{\circ}

Show answer & solution

Option A

0^{\circ}

Approach

Test each of the four printed standard angles directly in $cos θ + tan θ = 1$ .

Exact trigonometric ratios of the standard angles

Step-by-step solution

1
$θ = 0^{\circ} : cos 0 + tan 0 = 1 + 0 = 1. ✓$
Why: The first candidate already satisfies the equation exactly.
Exact trigonometric ratios of the standard angles
2
$θ = 3 0^{\circ} : \frac{3}{2} + \frac{1}{3} \approx 0.866 + 0.577 = 1.443. θ = 4 5^{\circ} : \frac{2}{2} + 1 \approx 1.707. θ = 6 0^{\circ} : 0.5 + 3 \approx 2.232.$
Why: None of the remaining standard angles comes close to satisfying the equation, confirming $0^{\circ}$ as the unique match among the options.

Common mistakes

Option B

Estimates $cos 3 0^{\circ} + tan 3 0^{\circ}$ loosely as $\approx 1$ by rounding too aggressively.

Option C

Picks $4 5^{\circ}$ as the 'natural' angle where $tan θ = 1$ , without checking $cos θ$ also needs to vanish.

Option D

Similarly guesses $6 0^{\circ}$ without direct substitution.

Skills used

Exact trigonometric ratios of the standard angles

64Trigonometry • Trigonometric Identity

ModerateIdeal time: about 1.5 min

What is the value of $sin x \frac{1}{1 + c o s x} + \frac{1}{1 - c o s x}$ ?

2

22

2 tan x

0

Show answer & solution

Option A

2

Approach

Add the two fractions over the common denominator $1 - cos^{2} x$ , which the Pythagorean identity turns into $sin^{2} x$ .

Pythagorean trigonometric identities

Step-by-step solution

1
$\frac{1}{1 + c o s x} + \frac{1}{1 - c o s x} = \frac{( 1 - c o s x ) + ( 1 + c o s x )}{( 1 + c o s x ) ( 1 - c o s x )}$
Why: The two denominators are conjugates, so their product is the natural common denominator.
Difference of two squares
2
$= \frac{2}{1 - c o s ^{2} x} = \frac{2}{s i n ^{2} x}$
Why: The cosine terms cancel in the numerator, and $1 - cos^{2} x = sin^{2} x$ .
Pythagorean trigonometric identities
3
$\frac{2}{s i n ^{2} x} = \frac{2}{∣ s i n x ∣}$
Why: The root of a quotient splits, and the principal square root of $sin^{2} x$ is $∣ sin x ∣$ .
$ab = a b$ and $a / b = a / b$ denotes the non-negative root
4
$sin x \cdot \frac{2}{∣ s i n x ∣} = 2 (sin x > 0)$
Why: The $sin x$ outside cancels the one under the root, leaving a constant.

Common mistakes

Option B

Doubled the numerator when adding the fractions, getting $\frac{8}{s i n ^{2} x}$ under the root.

Option C

Took $(1 + cos x) (1 - cos x)$ as $cos^{2} x$ rather than $1 - cos^{2} x$ , leaving $\frac{2 s i n x}{c o s x}$ .

Option D

Read the numerator sum $(1 - cos x) + (1 + cos x)$ as $0$ , cancelling the $1$ s rather than the cosines.

Skills used

Pythagorean trigonometric identities Difference of two squares

ab = a b

and

a / b = a / b

denotes the non-negative root

65Trigonometry • Trigonometric Identity

Very easyIdeal time: about 30 s

What is $\frac{c o s ^{4} A - s i n ^{4} A}{c o s ^{2} A - s i n ^{2} A}$ equal to?

cos^{2} A - sin^{2} A

cos A - sin A

1

2

Show answer & solution

Option C

1

Approach

Factor the numerator as a difference of squares; the denominator is one of the factors and cancels.

Difference of two squares

Step-by-step solution

1
$cos^{4} A - sin^{4} A = (cos^{2} A)^{2} - (sin^{2} A)^{2} = (cos^{2} A - sin^{2} A) (cos^{2} A + sin^{2} A)$
Why: The fourth powers are squares of the squares, so the difference-of-squares identity applies with $a = cos^{2} A$ and $b = sin^{2} A$ .
Difference of two squares
2
$\frac{( c o s ^{2} A - s i n ^{2} A ) ( c o s ^{2} A + s i n ^{2} A )}{c o s ^{2} A - s i n ^{2} A} = cos^{2} A + sin^{2} A = 1$
Why: Cancelling the common factor (valid wherever the given expression is defined, i.e. $cos^{2} A \neq = sin^{2} A$ ) leaves the Pythagorean identity.
Pythagorean trigonometric identities

Common mistakes

Option A

Cancelled the wrong factor, keeping $cos^{2} A - sin^{2} A$ instead of $cos^{2} A + sin^{2} A$ .

Option B

Took the square root of the surviving factor as well, as though the whole fraction halved its powers.

Option D

$2$ would need $cos^{2} A + sin^{2} A = 2$ ; no route reaches it.

Skills used

Difference of two squares Pythagorean trigonometric identities

66Trigonometry • Trigonometric Equation

EasyIdeal time: about 1 min

If $7 sin^{2} x + 3 cos^{2} x = 4$ , $0 < x < 9 0^{\circ}$ , then what is the value of $tan x$ ?

2

1

\frac{3}{2}

\frac{1}{3}

Show answer & solution

Option D

\frac{1}{3}

Approach

Use $sin^{2} x + cos^{2} x = 1$ to reduce the equation to one in $sin^{2} x$ , read off the angle, then take its tangent.

Pythagorean trigonometric identities

Step-by-step solution

1
$7 sin^{2} x + 3 (1 - sin^{2} x) = 4$
Why: Replacing $cos^{2} x$ by $1 - sin^{2} x$ leaves a single trigonometric quantity in the equation.
Pythagorean trigonometric identities
2
$4 sin^{2} x + 3 = 4 \Rightarrow sin^{2} x = \frac{1}{4}$
Why: Collecting like terms makes it linear in $sin^{2} x$ .
3
$sin x = \frac{1}{2} \Rightarrow x = 3 0^{\circ}$
Why: On $0 < x < 9 0^{\circ}$ the sine is positive, so only the positive root survives, and $3 0^{\circ}$ is the acute angle whose sine is $\frac{1}{2}$ .
denotes the non-negative root Exact trigonometric ratios of the standard angles
4
$tan x = tan 3 0^{\circ} = \frac{1}{3}$
Why: The question asks for the tangent, not for the angle or its sine.

Shortcut

Write the right side as $4 (sin^{2} x + cos^{2} x)$ ; then $3 sin^{2} x = cos^{2} x$ , so $tan^{2} x = \frac{1}{3}$ in one line.

Common mistakes

Option A

No route reaches $2$ here.

Option B

Assumed the two squares must contribute equally because $4$ lies between $3$ and $7$ , giving $x = 4 5^{\circ}$ .

Option C

Found $x = 3 0^{\circ}$ but reported $cos 3 0^{\circ}$ instead of $tan 3 0^{\circ}$ .

Skills used

Pythagorean trigonometric identities

denotes the non-negative root Exact trigonometric ratios of the standard angles

67Trigonometry • Heights and Distances

EasyIdeal time: about 1 min

An aeroplane flying at a height of $300$ m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are $6 0^{\circ}$ and $4 5^{\circ}$ respectively. What is the height of the lower plane from the ground?

1003

\frac{100}{3}

503

150 (3 + 1)

Show answer & solution

Option A

1003

Approach

The two planes share one vertical line, so one horizontal distance serves both triangles: get it from the upper plane, then use it on the lower.

Angles of elevation and depression

Step-by-step solution

1
$tan 6 0^{\circ} = \frac{300}{d}$
Why: The higher plane subtends the larger angle, so $6 0^{\circ}$ belongs to the plane at $300$ m; $d$ is the horizontal distance from the observer to the foot of the common vertical.
Angles of elevation and depression Sine, cosine and tangent in a right triangle
2
$d = \frac{300}{3} = \frac{300 3}{3} = 1003 m$
Why: Rationalising the denominator turns $\frac{300}{3}$ into $1003$ .
Rationalise a denominator using the conjugate Exact trigonometric ratios of the standard angles
3
$h = d tan 4 5^{\circ} = 1003 \times 1 = 1003 m$
Why: The lower plane stands on the same vertical line, so its elevation $4 5^{\circ}$ is measured over the same distance $d$ .

Shortcut

$\frac{h}{300} = \frac{t a n 4 5 ^{\circ}}{t a n 6 0 ^{\circ}} = \frac{1}{3}$ , so $h = \frac{300}{3} = 1003$ m.

Common mistakes

Option B

Simplified $\frac{300}{3}$ by dividing only the numerator by $3$ .

Option C

Used $tan 6 0^{\circ} = 23$ , which halves the horizontal distance.

Option D

No route reaches $150 (3 + 1)$ here.

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Rationalise a denominator using the conjugate Exact trigonometric ratios of the standard angles

68Trigonometry • Trigonometric Identity • Elimination of the Angle

EasyIdeal time: about 55 s

If $x = a cos θ + b sin θ$ and $y = a sin θ - b cos θ$ , then what is $x^{2} + y^{2}$ equal to?

2 ab

a + b

a^{2} + b^{2}

a^{2} - b^{2}

Show answer & solution

Option C

a^{2} + b^{2}

Approach

Square both expressions and add; the cross terms cancel and $sin^{2} θ + cos^{2} θ = 1$ finishes it.

Square of a sum and of a difference

Step-by-step solution

1
$x^{2} = a^{2} cos^{2} θ + 2 ab sin θ cos θ + b^{2} sin^{2} θ$
Why: Expand $(a cos θ + b sin θ)^{2}$ with the square-of-a-sum identity.
Square of a sum and of a difference
2
$y^{2} = a^{2} sin^{2} θ - 2 ab sin θ cos θ + b^{2} cos^{2} θ$
Why: Expanding $(a sin θ - b cos θ)^{2}$ gives the same cross term with the opposite sign.
3
$x^{2} + y^{2} = a^{2} (cos^{2} θ + sin^{2} θ) + b^{2} (sin^{2} θ + cos^{2} θ)$
Why: Adding kills the cross terms and groups what is left by $a^{2}$ and by $b^{2}$ .
4
$x^{2} + y^{2} = a^{2} + b^{2}$
Why: Each bracket is $1$ by the Pythagorean identity.
Pythagorean trigonometric identities

Shortcut

The pair $(x, y)$ is $(a, b)$ turned through the angle $θ$ , and a rotation does not change length, so $x^{2} + y^{2} = a^{2} + b^{2}$ at sight.

Common mistakes

Option A

Kept the cross terms and dropped the squared terms.

Option B

Added $x$ and $y$ instead of their squares.

Option D

Let the minus sign inside $y$ carry through to $b^{2}$ after squaring.

Skills used

Square of a sum and of a difference Pythagorean trigonometric identities

69Trigonometry • Heights and Distances

ModerateIdeal time: about 1.5 min

From the top of a building $90$ m high, the angles of depression of the top and the bottom of a tree are $3 0^{\circ}$ and $4 5^{\circ}$ respectively. What is the height of the tree?

303

90 - 303

90 + 303

60 + 303

Show answer & solution

Option B

90 - 303

Approach

The depression of the tree's foot fixes the horizontal distance; the depression of its top then fixes how far the tree top sits below the building top.

Angles of elevation and depression

Step-by-step solution

1
$tan 4 5^{\circ} = \frac{90}{d} \Rightarrow d = 90 m$
Why: The foot of the tree is seen at $4 5^{\circ}$ , and the building's full height $90$ m is opposite that angle.
Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles
2
$tan 3 0^{\circ} = \frac{p}{d} \Rightarrow p = \frac{90}{3} = 303 m$
Why: $p$ is the drop from the top of the building to the top of the tree, measured over the same horizontal distance $d$ .
Rationalise a denominator using the conjugate
3
$h = 90 - p = 90 - 303 m$
Why: What is left of the building's height after the drop $p$ is the height of the tree.

Common mistakes

Option A

Stopped at the drop $p$ from the building's top to the tree's top.

Option C

Added the drop to the building's height instead of subtracting it.

Option D

No route reaches $60 + 303$ here.

Skills used

Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles Rationalise a denominator using the conjugate

70Geometry • Triangle • Triangle Inequality

EasyIdeal time: about 45 s

Which one of the following triples does not represent the sides of a triangle?

(3, 4, 5)

(4, 7, 10)

(3, 6, 8)

(2, 3, 6)

Show answer & solution

Option D

(2, 3, 6)

Approach

Apply the triangle inequality to each triple: the two shorter lengths must add to more than the longest.

Triangle inequality

Step-by-step solution

1
$3 + 4 = 7 > 5$
Why: The triple $(3, 4, 5)$ passes, so it does represent a triangle.
Triangle inequality
2
$4 + 7 = 11 > 10$
Why: The triple $(4, 7, 10)$ passes as well.
3
$3 + 6 = 9 > 8$
Why: The triple $(3, 6, 8)$ passes.
4
$2 + 3 = 5 < 6$
Why: For $(2, 3, 6)$ the two shorter sides together fall short of the longest, so the two ends can never meet and no triangle exists.

Shortcut

Only the largest side needs testing: scan for a triple where the biggest number is at least the sum of the other two, and $(2, 3, 6)$ is the only one.

Common mistakes

Option A

Picked the familiar right-angled triple, reading the question as asking which one is special.

Option B

Took $4 + 7$ as $10$ and judged the inequality to fail by equality.

Option C

Compared $3 + 6$ with $8$ the wrong way round.

Skills used

Triangle inequality

71Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 55 s

If the perimeter of a rectangle is $10$ cm and the area is $4$ cm $^{2}$ , then its length is

6

cmB

5

cmC

4.5

cmD

4

Show answer & solution

Option D

4

Approach

Turn the perimeter and the area into the sum and the product of the two sides, then solve the quadratic they are the roots of.

Sum and product of roots (Vieta)

Step-by-step solution

1
$2 (l + b) = 10 \Rightarrow l + b = 5$
Why: The perimeter of a rectangle is twice the sum of two adjacent sides.
Perimeter and area of a rectangle
2
$l b = 4$
Why: The area is the product of those same two sides.
3
$t^{2} - 5 t + 4 = 0$
Why: $l$ and $b$ are the two numbers with sum $5$ and product $4$ , so they are the roots of $t^{2} - (sum) t + (product) = 0$ .
Sum and product of roots (Vieta)
4
$(t - 1) (t - 4) = 0 \Rightarrow t = 1 or t = 4$
Why: Splitting the middle term factorises the quadratic, and a product is zero only when a factor is.
Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero
5
$l = 4 cm, b = 1 cm$
Why: The length is the greater of the two dimensions.

Shortcut

Look for two numbers adding to $5$ and multiplying to $4$ : $4$ and $1$ , by inspection.

Common mistakes

Option A

No route reaches $6$ cm here.

Option B

Took the half-perimeter $l + b = 5$ as the length itself.

Option C

No route reaches $4.5$ cm here.

Skills used

Sum and product of roots (Vieta)Perimeter and area of a rectangle Factorise a quadratic by splitting the middle term A product is zero only when a factor is zero

72Geometry • Triangle • Angles of a Triangle

EasyIdeal time: about 45 s

The angles of a triangle are in the ratio $2 : 4 : 3$ . The smallest angle of the triangle is

2 0^{\circ}

4 0^{\circ}

5 0^{\circ}

6 0^{\circ}

Show answer & solution

Option B

4 0^{\circ}

Approach

Write the three angles as $2 k$ , $4 k$ , $3 k$ , use the angle sum of a triangle to find $k$ , then take the smallest term.

Introduce a common factor

k

for a ratio

Step-by-step solution

1
$2 k + 4 k + 3 k = 18 0^{\circ}$
Why: A ratio fixes the angles only up to a common multiplier $k$ , and the three angles of a triangle add to $18 0^{\circ}$ .
Introduce a common factor $k$ for a ratio Angle sum of a triangle
2
$9 k = 18 0^{\circ} \Rightarrow k = 2 0^{\circ}$
Why: The nine parts share the whole $18 0^{\circ}$ .
3
$2 k = 4 0^{\circ}, 4 k = 8 0^{\circ}, 3 k = 6 0^{\circ}$
Why: The smallest angle goes with the smallest ratio term, which is $2$ and not the last one listed.

Shortcut

$18 0^{\circ} \div 9 = 2 0^{\circ}$ per part, so the smallest angle is $2 \times 2 0^{\circ}$ .

Common mistakes

Option A

Stopped at $k$ , the size of one ratio part.

Option C

No route reaches $5 0^{\circ}$ here.

Option D

Took the last listed term $3$ as the smallest, giving $3 k = 6 0^{\circ}$ .

Skills used

Introduce a common factor

k

for a ratio Angle sum of a triangle

73Mensuration • Solids • Cylinder

ModerateIdeal time: about 1.5 min

A ball of radius $1$ cm is put into a cylindrical pipe so that it fits inside the pipe. If the length of the pipe is $14$ m, what is the surface area of the pipe?

2200

square cmB

4400

square cmC

8800

square cmD

17600

square cm

Show answer & solution

Option C

8800

square cm

Approach

The ball just fitting fixes the pipe's radius at $1$ cm; the pipe's surface is then the curved surface of a cylinder, after converting its length to centimetres.

Curved and total surface of a cylinder

Step-by-step solution

1
$r = 1 cm$
Why: A ball of radius $1$ cm that fits inside the pipe touches the wall all the way round, so the pipe's radius equals the ball's.
2
$h = 14 m = 1400 cm$
Why: The radius is in centimetres and the options are in square centimetres, so the length must be converted before multiplying.
3
$2 π r h = 2 \times \frac{22}{7} \times 1 \times 1400$
Why: A pipe is open at both ends, so only the curved surface counts; there are no circular end faces to add.
Curved and total surface of a cylinder
4
$= 8800 cm^{2}$
Why: $\frac{22}{7} \times 2800 = 22 \times 400 = 8800$ .

Common mistakes

Option A

No route reaches $2200$ square cm here.

Option B

Used $π r h$ , dropping the factor $2$ from the curved-surface formula.

Option D

Took the pipe's radius to be the ball's diameter, $2$ cm.

Skills used

Curved and total surface of a cylinder

74Mensuration • Plane Figures • Circle

EasyIdeal time: about 50 s

The areas of two circular fields are in the ratio $16 : 49$ . If the radius of the bigger field is $14$ m, then what is the radius of the smaller field?

4

8

9

10

Show answer & solution

Option B

8

Approach

Areas of circles vary as the square of the radius, so take the square root of the area ratio and scale the known radius.

Direct variation as a power:

y \propto x^{n}

Step-by-step solution

1
$\frac{A _{1}}{A _{2}} = (\frac{r _{1}}{r _{2}})^{2} = \frac{16}{49}$
Why: Area is $π r^{2}$ , so the constant $π$ cancels and the area ratio is the square of the radius ratio.
Area and circumference of a circle Direct variation as a power: $y \propto x^{n}$
2
$\frac{r _{1}}{r _{2}} = \frac{4}{7}$
Why: Take the positive square root of both sides; a radius is positive.
denotes the non-negative root
3
$r_{1} = \frac{4}{7} \times 14 = 8 m$
Why: The bigger field has radius $14$ m, so the smaller one is $\frac{4}{7}$ of it.

Common mistakes

Option A

Quoted the ratio term $4$ as the radius, without scaling by $\frac{14}{7}$ .

Option C

No route reaches $9$ m here.

Option D

No route reaches $10$ m here.

Skills used

Direct variation as a power:

y \propto x^{n}

Area and circumference of a circle

denotes the non-negative root

75Geometry • Quadrilateral • Parallelogram

ModerateIdeal time: about 1.5 min

Let ABCD be a rectangle. Let P, Q, R, S be the mid-points of sides AB, BC, CD, DA respectively. Then the quadrilateral PQRS is a

ASquareBRectangle, but need not be a squareCRhombus, but need not be a squareDParallelogram, but need not be a rhombus

Show answer & solution

Option C

Rhombus, but need not be a square

Approach

By the midpoint theorem each side of $P QR S$ is half a diagonal of the rectangle, and a rectangle's diagonals are equal, so all four sides are equal.

Midpoint theorem

Step-by-step solution

1
$P Q ∥ A C, P Q = \frac{1}{2} A C$
Why: In triangle $A B C$ the points $P$ and $Q$ are the midpoints of $A B$ and $B C$ , so the midpoint theorem applies.
Midpoint theorem
2
$R S = \frac{1}{2} A C, QR = \frac{1}{2} B D, S P = \frac{1}{2} B D$
Why: The same argument in the other three corner triangles gives the remaining three sides.
3
$A C = B D \Rightarrow P Q = QR = R S = S P$
Why: The diagonals of a rectangle are equal, so the two different half-diagonals are equal and all four sides of $P QR S$ match: it is a rhombus.
The diagonals of a rectangle are equal and bisect each other
4
$∠ P QR = 9 0^{\circ} ⟺ A C ⊥ B D$
Why: $P Q ∥ A C$ and $QR ∥ B D$ , so $P QR S$ is right-angled only when the rectangle's diagonals are perpendicular, which happens only when $A B C D$ is a square.

Common mistakes

Option A

Took the special case of a square $A B C D$ as the general one.

Option B

Swapped the result for its dual: it is the midpoint quadrilateral of a rhombus that is a rectangle.

Option D

Stopped at the general Varignon parallelogram without using that a rectangle's diagonals are equal.

Skills used

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

76Geometry • Triangle • Similarity

EasyIdeal time: about 45 s

Let P, Q, R be the mid-points of sides AB, BC, CA respectively of a triangle ABC. If the area of the triangle ABC is $5$ square units, then the area of the triangle PQR is

\frac{5}{3}

square unitsB

\frac{5}{2 2}

square unitsC

\frac{5}{4}

square unitsD

1

square unit

Show answer & solution

Option C

\frac{5}{4}

square units

Approach

The medial triangle is similar to the original with ratio $\frac{1}{2}$ , so its area is a quarter of it.

Midpoint theorem

Step-by-step solution

1
$P Q = \frac{1}{2} C A, QR = \frac{1}{2} A B, R P = \frac{1}{2} B C$
Why: Each side of $P QR$ joins two midpoints of $A B C$ , and the midpoint theorem halves the side it is parallel to.
Midpoint theorem
2
$△ P QR \sim △ A B C, k = \frac{1}{2}$
Why: All three pairs of sides are in the same ratio, which is SSS similarity.
Similar triangles have proportional sides
3
$\frac{[ P QR ]}{[ A B C ]} = k^{2} = \frac{1}{4}$
Why: Areas of similar figures scale as the square of the length ratio.
4
$[P QR] = \frac{1}{4} \times 5 = \frac{5}{4}$
Why: Apply the ratio to the given area of $5$ square units.

Shortcut

The medial triangle and the three corner triangles are four congruent pieces tiling $A B C$ , so each is a quarter: $\frac{5}{4}$ .

Common mistakes

Option A

Divided by $3$ , confusing the medial triangle with the three-way split of the area at the centroid.

Option B

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

Option D

No route reaches $1$ square unit here.

Skills used

Midpoint theorem Similar triangles have proportional sides

77Mensuration • Plane Figures • Quadrilateral

EasyIdeal time: about 55 s

If each of the dimensions of a rectangle is increased by $200%$ , the area is increased by

300%

400%

600%

800%

Show answer & solution

Option D

800%

Approach

Turn each $200%$ increase into a multiplier of $3$ , multiply for the area, then convert the factor back into a percentage increase.

Percentage change as a multiplier

Step-by-step solution

1
$1 + \frac{200}{100} = 3$
Why: An increase of $200%$ multiplies a length by $3$ , not by $2$ .
Percentage change as a multiplier
2
$A^{'} = (3 l) (3 b) = 9 l b = 9 A$
Why: Both dimensions carry the factor $3$ , so the area carries $3^{2}$ .
3
$\frac{A ^{'} - A}{A} \times 100 = (9 - 1) \times 100 = 800%$
Why: The question asks by how much the area increased, which is the factor minus $1$ .

Common mistakes

Option A

Read `increased by $200%$ ' as doubled, so the area came out $4$ times and the increase $300%$ .

Option B

Added the two percentage increases instead of multiplying the two factors.

Option C

No route reaches $600%$ here.

Skills used

Percentage change as a multiplier

78Mensuration • Plane Figures • Circle

HardIdeal time: about 2.5 min

Three circles each of radius $3.5$ cm touch one another. The area subtended between them is

6 (3 π - 2)

square unitsB

6 (2 π - 3)

square unitsC

\frac{49}{8} (23 - π)

square unitsD

\frac{49}{8} (3 - π)

square units

Show answer & solution

Option C

\frac{49}{8} (23 - π)

square units

Approach

Join the three centres to get an equilateral triangle of side $2 r$ ; the gap between the circles is that triangle less the three $6 0^{\circ}$ sectors lying inside it.

Distance between the centres of touching circles

Step-by-step solution

1
$side = 3.5 + 3.5 = 7 cm$
Why: Two circles that touch externally have their centres a sum of radii apart, so the triangle of centres is equilateral with side $7$ cm.
Distance between the centres of touching circles
2
$h = 7^{2} - (\frac{7}{2})^{2} = \frac{7 3}{2}, [△] = \frac{1}{2} \times 7 \times \frac{7 3}{2} = \frac{49 3}{4}$
Why: The altitude of an equilateral triangle bisects its base, so Pythagoras gives the height and then half base times height gives the area.
Pythagoras' theorem Area of a triangle from base and height Area and altitude of an equilateral triangle
3
$3 \times \frac{6 0 ^{\circ}}{36 0 ^{\circ}} \times π (3.5)^{2} = \frac{1}{2} \times π \times \frac{49}{4} = \frac{49 π}{8}$
Why: Each angle of the equilateral triangle is $6 0^{\circ}$ , so each circle contributes a sixth of its area; three such sectors make half a circle of radius $3.5$ cm.
Area and arc length of a circular sector Area and circumference of a circle
4
$\frac{49 3}{4} - \frac{49 π}{8} = \frac{49}{8} (23 - π)$
Why: Put both terms over the common denominator $8$ and take out $\frac{49}{8}$ .

Common mistakes

Option A

No route reaches $6 (3 π - 2)$ here.

Option B

No route reaches $6 (2 π - 3)$ here.

Option D

Subtracted without first putting $\frac{49 3}{4}$ over the common denominator $8$ .

Skills used

Distance between the centres of touching circles Pythagoras' theorem Area of a triangle from base and height Area and altitude of an equilateral triangle Area and arc length of a circular sector Area and circumference of a circle

79Mensuration • Plane Figures • Polygon

EasyIdeal time: about 45 s

The area of a regular hexagon of side 'a' is equal to

\frac{2}{3} a^{2}

square unitsB

\frac{3 3}{2} a^{2}

square unitsC

\frac{1}{3} a^{2}

square unitsD

\frac{3}{2} a^{2}

square units

Show answer & solution

Option B

\frac{3 3}{2} a^{2}

square units

Approach

Join the centre to the six vertices: the hexagon splits into six equilateral triangles of side $a$ .

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$[hexagon] = 6 \times [△]$
Why: A regular hexagon of side $a$ has circumradius $a$ , so each of the six triangles cut off by joining the centre to the vertices has all three sides equal to $a$ .
Side of a regular polygon inscribed in a circle
2
$h = a^{2} - (\frac{a}{2})^{2} = \frac{3}{2} a$
Why: The altitude of an equilateral triangle bisects the base, leaving a right triangle with legs $\frac{a}{2}$ and $h$ and hypotenuse $a$ .
Pythagoras' theorem
3
$[△] = \frac{1}{2} \times a \times \frac{3}{2} a = \frac{3}{4} a^{2}$
Why: Half base times height.
Area of a triangle from base and height Area and altitude of an equilateral triangle
4
$6 \times \frac{3}{4} a^{2} = \frac{3 3}{2} a^{2}$
Why: Six congruent triangles make the hexagon.

Common mistakes

Option A

No route reaches $\frac{2}{3} a^{2}$ here.

Option C

No route reaches $\frac{1}{3} a^{2}$ here.

Option D

Counted two of the six equilateral triangles instead of all six.

Skills used

Side of a regular polygon inscribed in a circle Pythagoras' theorem Area of a triangle from base and height Area and altitude of an equilateral triangle

80Trigonometry • Heights and Distances

HardIdeal time: about 2.5 min

ABCDEF is a regular polygon. Two poles at C and D are standing vertically and subtend angles of elevation $3 0^{\circ}$ and $6 0^{\circ}$ at A respectively. What is the ratio of the height of the pole at C to that of the pole at D?

1 : 1

1 : 23

23 : 1

2 : 3

Show answer & solution

Option B

1 : 23

Approach

In the regular hexagon $A B C D E F$ the two sightlines $A C$ and $A D$ have known lengths $a 3$ and $2 a$ ; each pole is its sightline times the tangent of its elevation.

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$A C = a 3, A D = 2 a$
Why: A regular hexagon of side $a$ has circumradius $a$ ; $A D$ is a diameter of the circumcircle, and $A$ , $C$ , $E$ form an inscribed equilateral triangle whose side is $a 3$ .
Side of a regular polygon inscribed in a circle
2
$h_{C} = A C tan 3 0^{\circ} = a 3 \times \frac{1}{3} = a$
Why: The pole at $C$ is vertical, so its foot, its top and the observer at $A$ form a right triangle with the horizontal leg $A C$ .
Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles
3
$h_{D} = A D tan 6 0^{\circ} = 2 a \times 3 = 23 a$
Why: The same setup at $D$ , over the longer sightline and the larger angle.
4
$h_{C} : h_{D} = a : 23 a = 1 : 23$
Why: The side $a$ cancels, so the ratio does not depend on the size of the hexagon.

Shortcut

$\frac{h _{C}}{h _{D}} = \frac{A C t a n 3 0 ^{\circ}}{A D t a n 6 0 ^{\circ}} = \frac{3}{2} \cdot \frac{1}{3} = \frac{1}{2 3}$ .

Common mistakes

Option A

Assumed a regular polygon puts $C$ and $D$ at the same distance from $A$ , so the two poles came out equal.

Option C

Wrote the ratio the other way round, giving the pole at $D$ first.

Option D

No route reaches $2 : 3$ here.

Skills used

Side of a regular polygon inscribed in a circle Angles of elevation and depression Sine, cosine and tangent in a right triangle Exact trigonometric ratios of the standard angles