Elementary Mathematics 2020 (I)

Set A

Questions 81–100 of 100

81Mensuration • Plane Figures • Quadrilateral

Very easyIdeal time: about 25 s

If the diagonals of a rhombus are $x$ and $y$ , then what is its area ?

\frac{x y}{2}

\frac{x y}{4}

x y

x^{2} - y^{2}

Show answer & solution

Option A

\frac{x y}{2}

Approach

A rhombus has perpendicular diagonals, so its area is half their product.

Area from perpendicular diagonals

Step-by-step solution

1
$the diagonals of a rhombus are perpendicular$
Why: This is what separates a rhombus from a general parallelogram, and it is what makes the half-product formula apply.
The diagonals of a rectangle are equal and bisect each other
2
$Area = \frac{1}{2} x y$
Why: For any quadrilateral with perpendicular diagonals $d_{1}$ and $d_{2}$ the area is $\frac{1}{2} d_{1} d_{2}$ .
Area from perpendicular diagonals

Common mistakes

Option B

Halved twice — used the half-diagonals $\frac{x}{2}$ and $\frac{y}{2}$ and then halved their product again.

Option C

Dropped the factor $\frac{1}{2}$ , treating the rhombus like a rectangle of sides $x$ and $y$ .

Option D

No plausible route reaches $x^{2} - y^{2}$ .

Skills used

Area from perpendicular diagonals The diagonals of a rectangle are equal and bisect each other

82Geometry • Triangle • Pythagoras Theorem

ModerateIdeal time: about 1.5 min

The lengths of sides of a triangle are $3 x, 4 y, 5 3 z$ , where $3 x < 4 y < 5 3 z$ . If one of the angles is $9 0^{\circ}$ , then what are the minimum integral values of $x, y, z$ respectively ?

1, 2, 3

2, 3, 4

1, 1, 1

3, 4, 5

Show answer & solution

Option C

1, 1, 1

Approach

The coefficients $3, 4, 5$ already form a Pythagorean triple, so look for the values that leave the three sides as exactly $3$ , $4$ and $5$ .

Pythagoras' theorem

Step-by-step solution

1
$(3 x)^{2} + (4 y)^{2} = (5 3 z)^{2}$
Why: The ordering $3 x < 4 y < 5 3 z$ makes $5 3 z$ the longest side, so it faces the right angle and is the hypotenuse.
Pythagoras' theorem The greater side faces the greater angle
2
$9 x^{2} + 16 y = 25 z^{2/3}$
Why: Square each side: $(4 y)^{2} = 16 y$ and $(5 3 z)^{2} = 25 z^{2/3}$ .
Power of a power, and the move to a common base
3
$x = y = z = 1$ gives $9 + 16 = 25$
Why: $1$ is the smallest value each of $x, y, z$ can take, and it makes the three sides exactly $3, 4, 5$ — the triple the coefficients were pointing at. Since no smaller positive integers exist, this is the minimum.

Shortcut

Read the coefficients: $3, 4, 5$ is already a right triangle, so the multipliers $x$ , $y$ , $3 z$ should all be $1$ .

Common mistakes

Option D

Took the triple $3, 4, 5$ off the coefficients and wrote it down as the values of $x, y, z$ themselves.

Option A

Read the strict inequalities as forcing $x, y, z$ to be distinct and increasing, so took the first three integers; $9 + 32 = 41 \neq = 25 \cdot 3^{2/3}$ .

Option B

No plausible route reaches $2, 3, 4$ .

Skills used

Pythagoras' theorem The greater side faces the greater angle Power of a power, and the move to a common base

83Geometry • Triangle • Triangle Centres

Very easyIdeal time: about 20 s

What is the maximum number of circum-circles that a triangle can have ?

1

2

3

DInfinite

Show answer & solution

Option A

1

Approach

A circle through all three vertices has its centre equidistant from them, and only one point of the plane is.

The circumcentre is the point equidistant from the vertices

Step-by-step solution

1
$the centre must satisfy O A = O B = O C$
Why: A circum-circle passes through all three vertices, so its centre is equidistant from them and the common distance is the radius.
A circle is the locus of points at a fixed distance from a fixed point
2
$that point is the circumcentre, and it is unique$
Why: The perpendicular bisectors of the three sides meet at exactly one point, so there is one possible centre and one possible radius — hence one circle.
The circumcentre is the point equidistant from the vertices

Common mistakes

Option C

Counted one circle per side or per pair of vertices, as though each needed its own.

Option D

Confused the circum-circle with the infinitely many circles through just two of the vertices.

Option B

No plausible route reaches $2$ .

Skills used

The circumcentre is the point equidistant from the vertices A circle is the locus of points at a fixed distance from a fixed point

84Mensuration • Plane Figures • Circle

EasyIdeal time: about 50 s

If an arc of a circle of radius $6$ cm subtends a central angle measuring $3 0^{\circ}$ , then which one of the following is an approximate length of the arc ?

3.14

cmB

2.15

cmC

2.14

cmD

2

Show answer & solution

Option A

3.14

Approach

Take the $3 0^{\circ}$ share of the full circumference.

Area and arc length of a circular sector

Step-by-step solution

1
$arc = \frac{θ}{360} \times 2 π r = \frac{30}{360} \times 2 π (6)$
Why: An arc is the same fraction of the circumference as its central angle is of $36 0^{\circ}$ .
Area and arc length of a circular sector
2
$= \frac{1}{12} \times 12 π = π$
Why: $3 0^{\circ}$ is a twelfth of the circle, and $2 π (6) = 12 π$ , so the twelfth is exactly $π$ .
Unitary method: find one unit, then scale
3
$π \approx 3.14$ cm
Why: The question asks for an approximate length, so give $π$ as a decimal.
Area and circumference of a circle

Shortcut

In radians, $3 0^{\circ} = \frac{π}{6}$ and arc $= r θ = 6 \times \frac{π}{6} = π$ .

Common mistakes

Option C

Reached $π$ but wrote the decimal with the wrong leading digit, $2.14$ for $3.14$ .

Option B

No plausible route reaches $2.15$ .

Option D

No plausible route reaches $2$ .

Skills used

Area and arc length of a circular sector Unitary method: find one unit, then scale Area and circumference of a circle

85Geometry • Triangle • Pythagoras Theorem

EasyIdeal time: about 1.5 min

A ladder $5 m$ long is placed in a room so as to reach a point $4.8 m$ high on a wall and on turning the ladder over to the opposite side of the wall without moving the base, it reaches a point $1.4 m$ high. What is the breadth of the room?

5.8 m

6 m

6.2 m

7.5 m

Show answer & solution

Option C

6.2 m

Approach

Each position of the ladder is a right triangle with the same hypotenuse $5$ ; find the two foot-to-wall distances and add them.

Pythagoras' theorem

Step-by-step solution

1
$d_{1} = 5^{2} - 4. 8^{2} = 25 - 23.04 = 1.96 = 1.4$
Why: The ladder, the wall and the floor form a right triangle with the ladder as hypotenuse; $d_{1}$ is the base's distance from the first wall.
Pythagoras' theorem
2
$d_{2} = 5^{2} - 1. 4^{2} = 25 - 1.96 = 23.04 = 4.8$
Why: The base has not moved, so the same hypotenuse $5$ applies to the opposite wall, where the ladder reaches $1.4$ m.
3
$breadth = d_{1} + d_{2} = 1.4 + 4.8 = 6.2$ m
Why: The base sits between the two walls, so the room's breadth is the sum of its distances from them.
Cut a compound plane figure into standard pieces

Shortcut

$(1.4, 4.8, 5)$ is the $(7, 24, 25)$ triple scaled by $\frac{1}{5}$ , so the two legs simply swap when the ladder is turned over.

Common mistakes

Option A

No plausible route reaches $5.8$ m.

Option B

No plausible route reaches $6$ m.

Option D

No plausible route reaches $7.5$ m.

Skills used

Pythagoras' theorem Cut a compound plane figure into standard pieces

86Mensuration • Plane Figures • Composite and Shaded Regions

EasyIdeal time: about 1 min

What is the area of the largest square plate cut from a circular disk of radius one unit?

4 square units

22 square units

π square units

2 square units

Show answer & solution

Option D

2 square units

Approach

The largest square that fits in a circle is the inscribed one, whose diagonal is the diameter.

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$diagonal = 2 r = 2$
Why: A square can be no wider than the circle, and the widest chord is the diameter; the inscribed square reaches it with its diagonal.
Side of a regular polygon inscribed in a circle
2
$a 2 = 2 \Rightarrow a = 2$
Why: A square of side $a$ has diagonal $a 2$ by Pythagoras.
Pythagoras' theorem
3
$Area = a^{2} = (2)^{2} = 2$
Why: Area of a square is the side squared, and the surd disappears at once.
Perimeter and area of a square

Shortcut

With perpendicular diagonals of length $2$ each, the area is $\frac{1}{2} \times 2 \times 2 = 2$ — no need to find the side.

Common mistakes

Option A

Took the diameter $2$ as the square's side, giving $2^{2} = 4$ — a square too big to fit in the circle.

Option B

Multiplied the side by the diagonal, $2 \times 2$ , instead of squaring the side.

Option C

Gave the area of the circle itself rather than of the square cut from it.

Skills used

Side of a regular polygon inscribed in a circle Pythagoras' theorem Perimeter and area of a square

87Mensuration • Solids • Sphere

Very hardIdeal time: about 3 min

Out of $4$ identical balls of radius $r$ , $3$ balls are placed on a plane such that each ball touches the other two balls. The $4^{th}$ ball is placed on them such that this ball touches all the three balls. What is the distance of centre of $4^{th}$ ball from the plane?

2 \frac{2}{3} r unit

\frac{3 + 2 2}{2} r unit

\frac{r}{3 - 2 2} unit

\frac{3 + 2 2}{3} r unit

Show answer & solution

Option D

\frac{3 + 2 2}{3} r unit

Approach

The four centres form a regular tetrahedron of edge $2 r$ ; add its height to the $r$ by which the bottom centres already stand above the plane.

Side of a regular polygon inscribed in a circle

Step-by-step solution

1
$each bottom centre is r above the plane$
Why: A ball resting on a plane touches it at one point, so its centre is one radius up.
The two tangents from an external point
2
$the three bottom centres form an equilateral triangle of side 2 r$
Why: Two balls that touch have their centres $r + r$ apart, and each of the three touches the other two.
Distance between the centres of touching circles
3
$circumradius R = \frac{2 r}{3}$
Why: An equilateral triangle inscribed in a circle of radius $R$ has side $R 3$ , so $R = side / 3$ . The top ball's centre sits directly above this circumcentre, being equidistant from all three.
Side of a regular polygon inscribed in a circle
4
$h = (2 r)^{2} - (\frac{2 r}{3})^{2} = 2 r 1 - \frac{1}{3} = 2 r \frac{2}{3}$
Why: The top centre is $2 r$ from each bottom centre (the balls touch), and that slant, the circumradius and the vertical rise form a right triangle.
Pythagoras' theorem
5
$r + 2 r \frac{2}{3} = r (1 + \frac{2 2}{3}) = \frac{3 + 2 2}{3} r$
Why: Add the $r$ from step $1$ ; writing $2/3$ as $2 / 3$ and taking $3$ as the common denominator puts it in the printed form.
$ab = a b$ and $a / b = a / b$

Common mistakes

Option A

Gave only the rise from the bottom centres to the top centre, $2 r 2/3$ , forgetting the $r$ by which the bottom centres already stand above the plane.

Option B

Reached the right numerator but divided by $2$ instead of $3$ when combining $1 + \frac{2 2}{3}$ over a common denominator.

Option C

No plausible route reaches $\frac{r}{3 - 2 2}$ .

Skills used

Side of a regular polygon inscribed in a circle The two tangents from an external point Distance between the centres of touching circles Pythagoras' theorem

ab = a b

and

a / b = a / b

88Mensuration • Solids

EasyIdeal time: about 1 min

A right circular cylinder just encloses a sphere. If $p$ is the surface area of the sphere and $q$ is the curved surface area of the cylinder, then which one of the following is correct?

p = q

p = 2 q

2 p = q

2 p = 3 q

Show answer & solution

Option A

p = q

Approach

'Just encloses' fixes the cylinder's radius and height in terms of the sphere's radius; then write both surfaces and compare.

Volume and surface area of a sphere

Step-by-step solution

1
$cylinder radius = R, cylinder height = 2 R$
Why: The sphere touches the curved wall all round, so the radii match, and it touches both ends, so the height is a full diameter.
The two tangents from an external point
2
$p = 4 π R^{2}$
Why: Surface area of a sphere of radius $R$ .
Volume and surface area of a sphere
3
$q = 2 π R (2 R) = 4 π R^{2}$
Why: Curved surface of a cylinder is $2 π r h$ ; here $r = R$ and $h = 2 R$ .
Curved and total surface of a cylinder
4
$p = q$
Why: Both come to $4 π R^{2}$ , so the two areas are equal — Archimedes' result.

Common mistakes

Option B

Took the cylinder's height as $R$ rather than the diameter $2 R$ , giving $q = 2 π R^{2}$ and so $p = 2 q$ .

Option C

Used $2 π R^{2}$ — the curved surface of a hemisphere — for the whole sphere, giving $p = 2 π R^{2}$ against $q = 4 π R^{2}$ .

Option D

No plausible route reaches $2 p = 3 q$ .

Skills used

Volume and surface area of a sphere The two tangents from an external point Curved and total surface of a cylinder

89Mensuration • Plane Figures • Quadrilateral

ModerateIdeal time: about 1.5 min

$A B C D$ is a quadrilateral such that $A D = D C = C A = 20 units$ , $B C = 12 units$ and $∠ A B C = 9 0^{\circ}$ . What is the approximate area of the quadrilateral $A B C D$ ?

269 sq units

300 sq units

325 sq units

349 sq units

Show answer & solution

Option A

269 sq units

Approach

$A C$ splits the quadrilateral into an equilateral triangle and a right triangle; find each area and add.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$△ A C D is equilateral with side 20$
Why: $A D = D C = C A = 20$ is given, so the diagonal $A C$ cuts off an equilateral triangle.
Cut a compound plane figure into standard pieces
2
$[A C D] = \frac{3}{4} (20)^{2} = 1003 \approx 173.2$
Why: Area of an equilateral triangle of side $a$ is $\frac{3}{4} a^{2}$ ; take $3 \approx 1.732$ .
Area and altitude of an equilateral triangle
3
$A B = 2 0^{2} - 1 2^{2} = 400 - 144 = 16$
Why: $∠ A B C = 9 0^{\circ}$ , so $A C = 20$ is the hypotenuse of $△ A B C$ and $B C = 12$ one leg — the $(3, 4, 5)$ triple scaled by $4$ .
Pythagoras' theorem
4
$[A B C] = \frac{1}{2} (16) (12) = 96$
Why: In a right triangle the two legs serve as base and height.
Area of a triangle from base and height
5
$[A B C D] = 173.2 + 96 = 269.2 \approx 269$
Why: $B$ and $D$ lie on opposite sides of $A C$ , so the two triangles do not overlap and the areas simply add.

Common mistakes

Option B

Treated $A C$ and $B C$ as the two legs of the right angle, giving $[A B C] = \frac{1}{2} (20) (12) = 120$ and a total near $293$ .

Option C

No plausible route reaches $325$ .

Option D

No plausible route reaches $349$ .

Skills used

Cut a compound plane figure into standard pieces Area and altitude of an equilateral triangle Pythagoras' theorem Area of a triangle from base and height

90Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

A circle with a horizontal diameter marked with points P, Q, R, S in order from left to right. The segments PQ, QR, and RS appear equal. A semi-circle is drawn above the segment QS. The region enclosed between the upper half of the main circle and the upper semi-circle on QS is shaded.

Let $P QR S$ be the diameter of a circle of radius $9 cm$ . The length $P Q$ , $QR$ and $R S$ are equal. Semi-circle is drawn with $QS$ as diameter (as shown in the given figure). What is the ratio of the shaded region to that of the unshaded region?

25 : 121

5 : 13

5 : 18

1 : 2

Show answer & solution

Option B

5 : 13

Approach

The shaded piece is the upper half of the big circle less the semi-circle on $QS$ ; the unshaded piece is everything else.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$P S = 18, P Q = QR = R S = 6, QS = 12$
Why: $P S$ is a diameter of the circle of radius $9$ , and the three equal parts make each $6$ , so $QS = 2 \times 6$ .
Area and circumference of a circle
2
$upper half of big circle = \frac{1}{2} π (9)^{2} = 40.5 π$
Why: $P S$ is a diameter, so the arc above it bounds a semi-circle of radius $9$ .
3
$semi-circle on QS = \frac{1}{2} π (6)^{2} = 18 π$
Why: $QS = 12$ is its diameter, so its radius is $6$ .
4
$shaded = 40.5 π - 18 π = 22.5 π$
Why: The figure shades the part of the upper semi-circle lying outside the semi-circle on $QS$ .
Cut a compound plane figure into standard pieces
5
$unshaded = 81 π - 22.5 π = 58.5 π, 22.5 : 58.5 = 5 : 13$
Why: Everything in the circle that is not shaded is unshaded; divide both by $4.5$ to reduce.
Unitary method: find one unit, then scale

Shortcut

Drop $π$ and work in units of a half: shaded $= 81 - 36 = 45$ half-units, whole circle $= 162$ , so shaded : unshaded $= 45 : 117 = 5 : 13$ .

Common mistakes

Option C

Compared the shaded region with the whole circle, $22.5 : 81 = 5 : 18$ , instead of with the unshaded part.

Option A

No plausible route reaches $25 : 121$ .

Option D

No plausible route reaches $1 : 2$ .

Skills used

Cut a compound plane figure into standard pieces Area and circumference of a circle Unitary method: find one unit, then scale

91Mensuration • Plane Figures • Circle

ModerateIdeal time: about 1.5 min

Three identical circles intersect. Their centers A, B, and C form a triangle ABC. The region bounded by the three circles between the centers is shaded.

What is the area of the shaded region in the given figure, if the radius of each of the circles is $2 cm$ ?

43 - 2 π cm^{2}

3 - π cm^{2}

3 - \frac{π}{2} cm^{2}

2 π - 23 cm^{2}

Show answer & solution

Option A

43 - 2 π cm^{2}

Approach

Join the three centres: the shaded piece is that triangle with one $6 0^{\circ}$ sector cut from each corner.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$A B = B C = C A = 2 + 2 = 4$
Why: Each pair of circles touches, so each centre distance is the sum of the two radii; the three centres form an equilateral triangle.
Distance between the centres of touching circles
2
$[A B C] = \frac{3}{4} (4)^{2} = 43$
Why: Area of an equilateral triangle of side $a$ is $\frac{3}{4} a^{2}$ .
Area and altitude of an equilateral triangle
3
$three corner sectors = 3 \times \frac{60}{360} π (2)^{2} = 2 π$
Why: Each angle of an equilateral triangle is $6 0^{\circ}$ , and the circle at that corner has radius $2$ ; the three angles total $18 0^{\circ}$ , so the three sectors make half a circle.
Area and arc length of a circular sector
4
$shaded = 43 - 2 π$
Why: The shaded curvilinear triangle is what is left of $△ A B C$ once the three circular corners are removed.
Cut a compound plane figure into standard pieces

Shortcut

Three $6 0^{\circ}$ sectors of equal radius always make half a circle, so the answer is 'triangle minus half a circle': $43 - \frac{1}{2} π (2)^{2}$ .

Common mistakes

Option D

Used the altitude $\frac{3}{2} a = 23$ in place of the area $\frac{3}{4} a^{2}$ , and subtracted the triangle from the sectors rather than the other way round.

Option C

Halved every length — took the triangle's side as $2$ and the circles' radius as $1$ — which scales the whole answer down by four.

Option B

No plausible route reaches $3 - π$ , which is negative.

Skills used

Cut a compound plane figure into standard pieces Distance between the centres of touching circles Area and altitude of an equilateral triangle Area and arc length of a circular sector

92Mensuration • Plane Figures • Composite and Shaded Regions

ModerateIdeal time: about 2 min

A rectangle inscribed inside a circle. The top side is labeled 6\sqrt{3} \text{ units} and the left side is labeled 6 \text{ units}. The bottom side is 6\sqrt{3} \text{ units} and the right side is 6 \text{ units}. The region of the circle above the top side of the rectangle is shaded.

In the given figure, what is the area of the shaded region?

9 (π - 3) sq units

3 (4 π - 33) sq units

3 (3 π - 43) sq units

9 (3 - π) sq units

Show answer & solution

Option B

3 (4 π - 33) sq units

Approach

The rectangle's diagonal is a diameter, so find the radius; the shaded piece is the segment cut off by the top side.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$d = (63)^{2} + 6^{2} = 108 + 36 = 12, so r = 6$
Why: Each corner of the rectangle lies on the circle, and the angle in a semi-circle is $9 0^{\circ}$ , so the diagonal is a diameter.
Perimeter and area of a rectangle The angle at the centre is twice the angle at the circumference
2
$sin \frac{θ}{2} = \frac{3 3}{6} = \frac{3}{2} \Rightarrow \frac{θ}{2} = 6 0^{\circ}, θ = 12 0^{\circ}$
Why: The perpendicular from the centre bisects the chord $63$ , making a right triangle with hypotenuse $r = 6$ and opposite side $33$ .
The perpendicular from the centre bisects a chord Sine, cosine and tangent in a right triangle
3
$sector = \frac{120}{360} π (6)^{2} = 12 π$
Why: The shaded segment sits inside the $12 0^{\circ}$ sector standing on that chord.
Area and arc length of a circular sector
4
$triangle = \frac{1}{2} (6) (6) sin 12 0^{\circ} = 18 \times \frac{3}{2} = 93$
Why: The two radii to the chord's ends enclose $12 0^{\circ}$ , so use $\frac{1}{2} ab sin C$ .
Area of a triangle from two sides and the included angle Exact trigonometric ratios of the standard angles
5
$shaded = 12 π - 93 = 3 (4 π - 33)$
Why: A segment is the sector less the triangle formed by the two radii.
Cut a compound plane figure into standard pieces

Shortcut

The centre is $3$ from the top side and the half-chord is $33$ , so the segment is the $12 0^{\circ}$ sector $12 π$ minus the triangle $\frac{1}{2} (63) (3) = 93$ .

Common mistakes

Option A

Took the central angle as $9 0^{\circ}$ , giving a sector of $9 π$ and a triangle of $18$ , i.e. $9 π - 93$ after the surd was carried through.

Option C

Transposed the $3$ and the $4$ when writing $12 π - 93$ in the factored form $3 (4 π - 33)$ .

Option D

Subtracted the sector from the triangle instead of the triangle from the sector, giving a negative area.

Skills used

Cut a compound plane figure into standard pieces Perimeter and area of a rectangle The angle at the centre is twice the angle at the circumference The perpendicular from the centre bisects a chord Sine, cosine and tangent in a right triangle Area and arc length of a circular sector Area of a triangle from two sides and the included angle Exact trigonometric ratios of the standard angles

93Geometry • Line • Angles at a Point

EasyIdeal time: about 50 s

A horizontal line with an arrow pointing right. From a point on this line, a ray extends upwards and rightwards, forming an angle z. From the same point, another ray extends leftwards and upwards, forming an angle y. The angle between the horizontal line extending left and the left ray is x.

In the given figure, if $\frac{y}{x} = 6$ and $\frac{z}{x} = 5$ , then what is the value of $x$ ?

4 5^{\circ}

3 0^{\circ}

1 5^{\circ}

1 0^{\circ}

Show answer & solution

Option C

1 5^{\circ}

Approach

The three angles sit side by side above one straight line, so they add to $18 0^{\circ}$ .

Angles on a straight line and vertically opposite angles

Step-by-step solution

1
$x + y + z = 18 0^{\circ}$
Why: The two rays divide the half-turn above the straight line into exactly these three angles, and angles on one side of a straight line add to $18 0^{\circ}$ .
Angles on a straight line and vertically opposite angles
2
$y = 6 x, z = 5 x$
Why: The given ratios written as equations, so everything is in one unknown.
Turn a worded condition into a linear equation in one unknown
3
$x + 6 x + 5 x = 12 x = 18 0^{\circ} \Rightarrow x = 1 5^{\circ}$
Why: Substitute and solve the linear equation.

Common mistakes

Option B

Used $36 0^{\circ}$ for angles round a point instead of $18 0^{\circ}$ for angles on a straight line, giving $12 x = 360$ .

Option A

No plausible route reaches $4 5^{\circ}$ .

Option D

No plausible route reaches $1 0^{\circ}$ .

Skills used

Angles on a straight line and vertically opposite angles Turn a worded condition into a linear equation in one unknown

94Mensuration • Plane Figures • Quadrilateral

HardIdeal time: about 2.5 min

$A B C D$ is a trapezium, where $A B$ is parallel to $D C$ . If $A B = 4 cm$ , $B C = 3 cm$ , $C D = 7 cm$ and $D A = 2 cm$ , then what is the area of the trapezium?

22 \frac{2}{3} cm^{2}

22 \frac{3}{2} cm^{2}

223 cm^{2}

\frac{22 2}{3} cm^{2}

Show answer & solution

Option D

\frac{22 2}{3} cm^{2}

Approach

Drop perpendiculars from $A$ and $B$ to $D C$ and use the two slant sides to find the common height.

Area of a trapezium

Step-by-step solution

1
$let the foot of the perpendicular from A be p from D$
Why: With $A B ∥ D C$ , the two perpendiculars have the same length $h$ , and the strip between their feet is $A B = 4$ long, so the foot from $B$ is $p + 4$ from $D$ .
Cut a compound plane figure into standard pieces
2
$p^{2} + h^{2} = 2^{2}$ and $(7 - p - 4)^{2} + h^{2} = 3^{2}$
Why: Pythagoras in the two right triangles cut off at each end: $D A = 2$ and $C B = 3$ are their hypotenuses.
Pythagoras' theorem
3
$(3 - p)^{2} - p^{2} = 5 \Rightarrow 9 - 6 p = 5 \Rightarrow p = \frac{2}{3}$
Why: Subtracting the two equations eliminates $h^{2}$ and leaves a linear equation in $p$ .
Eliminate one variable from a linear pair
4
$h^{2} = 4 - \frac{4}{9} = \frac{32}{9} \Rightarrow h = \frac{4 2}{3}$
Why: Back-substitute into $p^{2} + h^{2} = 4$ ; $32 = 42$ .
$ab = a b$ and $a / b = a / b$
5
$Area = \frac{1}{2} (4 + 7) \times \frac{4 2}{3} = \frac{22 2}{3}$
Why: Half the sum of the parallel sides times the perpendicular distance between them.
Area of a trapezium

Shortcut

Draw $B E ∥ A D$ with $E$ on $D C$ : $△ B E C$ has sides $2, 3, 3$ and $E C = 7 - 4 = 3$ . Heron gives its area as $22$ , so $h = \frac{2 \times 2 2}{3} = \frac{4 2}{3}$ .

Common mistakes

Option A

Wrote the height $\frac{4 2}{3}$ as $4 \frac{2}{3}$ , pulling the denominator inside the root.

Option B

Made the same move and then inverted the fraction inside the root, giving $\frac{3}{2}$ .

Option C

No plausible route reaches $223$ .

Skills used

Area of a trapezium Cut a compound plane figure into standard pieces Pythagoras' theorem Eliminate one variable from a linear pair

ab = a b

and

a / b = a / b

95Geometry • Polygon

HardIdeal time: about 2.5 min

A figure showing several intersecting lines forming an eight-pointed star-like shape, with four outward-pointing triangular regions. The outer pointed angles are numbered 1 through 8. Angles 1 and 2 are in one part, 3 and 4 in another, 5 and 6 in another, 7 and 8 in another.

Angles are shown in the given figure. What is value of $∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8$ ?

24 0^{\circ}

36 0^{\circ}

54 0^{\circ}

72 0^{\circ}

Show answer & solution

Option B

36 0^{\circ}

Approach

The eight marked angles are the outer angles of four triangles; the remaining angles of those triangles are the angles of the central quadrilateral.

Angle sum of a triangle

Step-by-step solution

1
$the four lines cut out four triangles and a central quadrilateral$
Why: Each triangle has two of the numbered angles for its outer vertices and its third vertex at a crossing point.
Cut a compound plane figure into standard pieces
2
$sum of all angles in the four triangles = 4 \times 18 0^{\circ} = 72 0^{\circ}$
Why: Every plane triangle has angles summing to $18 0^{\circ}$ .
Angle sum of a triangle
3
$each third angle = the quadrilateral’s angle vertically opposite it$
Why: At each crossing point the triangle's angle and the central quadrilateral's angle are formed by the same two lines on opposite sides, so they are vertically opposite and equal.
Angles on a straight line and vertically opposite angles
4
$∠1 + \dots + ∠8 = 72 0^{\circ} - 36 0^{\circ} = 36 0^{\circ}$
Why: The four third angles together equal the interior angles of the central quadrilateral, which sum to $(4 - 2) \times 18 0^{\circ} = 36 0^{\circ}$ .
Interior and exterior angle sums of a polygon

Common mistakes

Option D

Added the angle sums of the four triangles, $4 \times 18 0^{\circ}$ , and forgot to take out the four angles at the crossing points.

Option A

No plausible route reaches $24 0^{\circ}$ .

Option C

No plausible route reaches $54 0^{\circ}$ .

Skills used

Angle sum of a triangle Cut a compound plane figure into standard pieces Angles on a straight line and vertically opposite angles Interior and exterior angle sums of a polygon

96Geometry • Line • Parallel Lines and Transversals

ModerateIdeal time: about 2 min

Two vertical parallel lines, PQ and RS, with arrows pointing downwards and upwards respectively. A transversal line intersects them at E and G. Another transversal line intersects them at F and H. The two transversals intersect each other at a point B on the right side. The angle at E is labeled 95^\circ below the transversal on the left parallel line. The angle at H is labeled 110^\circ below the transversal on the right parallel line. The angle at B is x.

In the given figure $P Q$ is parallel to $R S$ , $∠ A E F = 9 5^{\circ}$ , $∠ B H S = 11 0^{\circ}$ , and $∠ A B C = x^{\circ}$ . Then what is the value of $x$ ?

15

25

30

35

Show answer & solution

Option B

25

Approach

Carry each given angle across to the line $R S$ with the parallel-line rules, then use the triangle $B G H$ that $R S$ cuts off.

Angles made by a transversal of two parallel lines

Step-by-step solution

1
$∠ A E P = 18 0^{\circ} - 9 5^{\circ} = 8 5^{\circ}$
Why: $∠ A E F = 9 5^{\circ}$ is measured from $E A$ down to $E F$ along $P Q$ ; $E$ lies on the straight line $P Q$ , so the angle on the other side of $E$ is its supplement.
Angles on a straight line and vertically opposite angles
2
$∠ B GS = 8 5^{\circ}$
Why: $P Q ∥ R S$ and $A B$ is a transversal, so the angle the line $A B$ makes below $G$ on $R S$ equals the one it makes below $E$ on $P Q$ — corresponding angles.
Angles made by a transversal of two parallel lines
3
$∠ B H G = 18 0^{\circ} - 11 0^{\circ} = 7 0^{\circ}$
Why: $∠ B H S = 11 0^{\circ}$ and $H$ lies on the straight line $R S$ , so the angle from $H B$ up to $H R$ is its supplement.
4
$x = 18 0^{\circ} - 8 5^{\circ} - 7 0^{\circ} = 2 5^{\circ}$
Why: $G$ , $H$ and $B$ form a triangle whose angle at $G$ is $8 5^{\circ}$ , at $H$ is $7 0^{\circ}$ and at $B$ is $x$ .
Angle sum of a triangle

Shortcut

$x$ is the exterior angle at $B$ of the pair of lines: $x = 11 0^{\circ} - 8 5^{\circ} = 2 5^{\circ}$ , the difference of the two angles each line makes with the parallels.

Common mistakes

Option A

Carried $9 5^{\circ}$ straight across to $G$ instead of its supplement $8 5^{\circ}$ , giving $18 0^{\circ} - 9 5^{\circ} - 7 0^{\circ} = 1 5^{\circ}$ .

Option C

No plausible route reaches $30$ .

Option D

No plausible route reaches $35$ .

Skills used

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

97Geometry • Line • Parallel Lines and Transversals

ModerateIdeal time: about 2 min

Two horizontal parallel lines, the top one contains points A and B, the bottom one contains points E, C, and D. A line passes through E and A, extending to H. Another line passes through C and B, extending upwards. Another line passes through D and A. The angle between EA and AC is 40^\circ. The angle at D is 55^\circ. The angle between AH and AB is x.

In the given figure $A B$ is parallel to $C D$ and $A C$ is parallel to $B D$ . If $∠ E A C = 4 0^{\circ}$ , $∠ F D G = 5 5^{\circ}$ , $∠ H A B = x^{\circ}$ , then what is the value of $x$ ?

85

80

75

65

Show answer & solution

Option A

85

Approach

$A B D C$ is a parallelogram, so $∠ B A C$ equals $∠ B D C$ ; then $∠ H A B$ is what is left of the straight line $H A E$ .

Angles of a parallelogram

Step-by-step solution

1
$∠ B D C = ∠ F D G = 5 5^{\circ}$
Why: $D B$ and $D G$ are opposite rays, and so are $D C$ and $D F$ , so the two angles are vertically opposite at $D$ .
Angles on a straight line and vertically opposite angles
2
$A B ∥ C D$ and $A C ∥ B D$ , so $A B D C$ is a parallelogram
Why: Both pairs of opposite sides are parallel, which is the definition.
3
$∠ B A C = ∠ B D C = 5 5^{\circ}$
Why: $∠ A$ and $∠ D$ are opposite angles of the parallelogram $A B D C$ , and opposite angles are equal.
Angles of a parallelogram
4
$∠ B A E = ∠ B A C + ∠ C A E = 5 5^{\circ} + 4 0^{\circ} = 9 5^{\circ}$
Why: Ray $A C$ lies between ray $A B$ and ray $A E$ , so the two angles sit side by side.
5
$x = 18 0^{\circ} - 9 5^{\circ} = 8 5^{\circ}$
Why: In the figure $A H$ is the upward continuation of $E A$ , so $H$ , $A$ , $E$ are collinear and $∠ H A B$ and $∠ B A E$ are a linear pair.

Common mistakes

Option B

No plausible route reaches $80$ .

Option C

No plausible route reaches $75$ .

Option D

No plausible route reaches $65$ .

Skills used

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

98Mensuration • Plane Figures • Circle

EasyIdeal time: about 1.5 min

A large semi-circle with diameter AC. A point F lies on AC. A semi-circle is drawn below the segment AF, and another semi-circle is drawn below the segment FC. Point B is on the large semi-circle, E is on the left small semi-circle, D is on the right small semi-circle.

In the given figure, there are three semi circles $A B C$ , $A E F$ and $C D F$ . The distance between $A$ and $C$ is $28 units$ and $F$ is the mid-point of $A C$ . What is the total area of the three semi circles?

924 square units

824 square units

624 square units

462 square units

Show answer & solution

Option D

462 square units

Approach

Get each semi-circle's radius from $A C = 28$ and add the three half-areas.

Area and circumference of a circle

Step-by-step solution

1
$A C = 28 \Rightarrow large radius = 14; A F = F C = 14 \Rightarrow small radii = 7$
Why: $A C$ is the diameter of the big semi-circle, and $F$ bisects it, so $A F$ and $F C$ are the diameters of the two small ones.
Area and circumference of a circle
2
$big = \frac{1}{2} π (14)^{2} = 98 π$
Why: A semi-circle is half of $π r^{2}$ .
3
$two small = 2 \times \frac{1}{2} π (7)^{2} = 49 π$
Why: Each small semi-circle is $24.5 π$ and there are two of them.
4
$98 π + 49 π = 147 π = 147 \times \frac{22}{7} = 462$
Why: The question asks for the total of the three, not for the region they enclose. $147 \div 7 = 21$ and $21 \times 22 = 462$ .
Cut a compound plane figure into standard pieces

Shortcut

Halving a semi-circle's radius quarters its area, so the two small ones together are half the big one: total $= \frac{3}{2} \times 98 π = 147 π = 462$ .

Common mistakes

Option A

Used whole circles instead of semi-circles throughout, giving exactly double: $294 π = 924$ .

Option B

No plausible route reaches $824$ .

Option C

No plausible route reaches $624$ .

Skills used

Area and circumference of a circle Cut a compound plane figure into standard pieces

99Mensuration • Plane Figures • Composite and Shaded Regions

ModerateIdeal time: about 1.5 min

A circle with a rectangle inscribed inside. The rectangle has dimensions 8 \text{ cm} horizontally and 6 \text{ cm} vertically. The region between the circle and the top and bottom sides of the rectangle is shaded. The region between the circle and the left and right sides is unshaded.

What is the approximate area of the shaded region in the figure given?

15.3 cm^{2}

25.5 cm^{2}

28.4 cm^{2}

30.5 cm^{2}

Show answer & solution

Option D

30.5 cm^{2}

Approach

The rectangle's diagonal is a diameter; the shaded region is everything inside the circle but outside the rectangle.

Cut a compound plane figure into standard pieces

Step-by-step solution

1
$d = 8^{2} + 6^{2} = 100 = 10 \Rightarrow r = 5$
Why: The rectangle's corners lie on the circle, so its diagonal is a diameter — the $(3, 4, 5)$ triple doubled.
Perimeter and area of a rectangle Pythagoras' theorem
2
$circle = π (5)^{2} = 25 π \approx 78.54$
Why: Area of a circle of radius $5$ .
Area and circumference of a circle
3
$rectangle = 8 \times 6 = 48$
Why: Length times breadth.
4
$shaded = 25 π - 48 \approx 78.54 - 48 = 30.54 \approx 30.5$
Why: In the figure every part of the circle outside the rectangle is hatched, so the shaded area is the circle less the rectangle.
Cut a compound plane figure into standard pieces

Common mistakes

Option A

Took only half the region — one side of the rectangle's horizontal axis — giving about $15.3$ .

Option B

No plausible route reaches $25.5$ .

Option C

No plausible route reaches $28.4$ .

Skills used

Cut a compound plane figure into standard pieces Perimeter and area of a rectangle Pythagoras' theorem Area and circumference of a circle

100Geometry • Quadrilateral • Parallelogram

HardIdeal time: about 2.5 min

A rectangle ABCD. Diagonals AC and BD intersect at point O.

Consider the following statements with reference to the given figure:

1: The sum of the areas of $△ A O D$ and $△ B O C$ is equal to the sum of the areas of $△ A O B$ and $△ D O C$ .

2: $∠ A O D = ∠ B O C$

3: $A B + B C + C D + D A > A C + B D$

Which of the above statements are correct?

1 and 2 only

2 and 3 only

1 and 3 only

1, 2 and 3

Show answer & solution

Option B

2 and 3 only

Approach

Read the figure as drawn — $A D ∥ B C$ with $A D \neq = B C$ , not a rectangle — and test the three statements on it.

Area of a triangle from base and height

Step-by-step solution

1
$∠ A O D = ∠ B O C$ — statement $2$ is correct
Why: $A$ , $O$ , $C$ are collinear and so are $B$ , $O$ , $D$ , so these two angles are vertically opposite wherever the diagonals cross.
Angles on a straight line and vertically opposite angles
2
$A B + B C > A C, A D + D C > A C, A B + A D > B D, B C + C D > B D$
Why: Triangle inequality in $△ A B C$ , $△ A D C$ , $△ A B D$ and $△ B C D$ in turn.
Triangle inequality
3
$2 (A B + B C + C D + D A) > 2 (A C + B D)$ — statement $3$ is correct
Why: Adding the four inequalities counts each side twice and each diagonal twice; halving gives the perimeter against the sum of the diagonals.
4
$[A O D] = \frac{a ^{2} c}{2 ( a + b )}, [B O C] = \frac{b ^{2} c}{2 ( a + b )}, [A O B] = [D O C] = \frac{ab c}{2 ( a + b )}$
Why: Take $D = (0, 0)$ , $C = (c, 0)$ , $A = (0, a)$ , $B = (c, b)$ as the figure shows, with $A D$ and $B C$ both perpendicular to $D C$ and $b > a$ . The diagonals meet at $(\frac{a c}{a + b}, \frac{ab}{a + b})$ , and each triangle's area follows from base and height.
Area of a triangle from base and height
5
$a^{2} + b^{2} > 2 ab whenever a \neq = b$ — statement $1$ is wrong
Why: $[A O D] + [B O C]$ beats $[A O B] + [D O C]$ by the factor $a^{2} + b^{2}$ against $2 ab$ , and $(a - b)^{2} > 0$ . The two sums are equal only if $A D = B C$ , which the figure does not show.
Square of a sum and of a difference

Common mistakes

Option D

Read the figure as a rectangle, which makes the four triangles equal in area and statement $1$ true as well.

Option A

Read the figure as a rectangle and so accepted statement $1$ , and missed the four-triangle-inequality argument for statement $3$ .

Option C

Read the figure as a rectangle and so accepted statement $1$ , and rejected statement $2$ despite the two angles being vertically opposite.

Skills used

Area of a triangle from base and height Angles on a straight line and vertically opposite angles Triangle inequality Square of a sum and of a difference