Geometry

10 methods · 37 questions

★ Signature method unlocked: the technique this chapter rewards most often, with 3 worked questions.

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When a question asks for a length or area inside a complex configuration of squares, circles or lines, pure Euclidean geometry can be slow. Instead, place the origin at a convenient corner or centre and assign (x,y) coordinates to the key vertices.

Trigger: An irregular length or a strange shaded area inside a square or along a line, with numerical side lengths given.

Instances:
(i) to find an area, subtract the areas of standard shapes (rectangles, right-angled triangles) whose vertices you just found;
(ii) to find a length, use Pythagoras on the coordinate differences;
(iii) for circle intersections, write down the equation $x^{2} + y^{2} = r^{2}$ and find the intersection algebraically.

Linked questions (3)

NSAA 2019 Section 1, Part A (Mathematics questions) · Q11

The diagram shows two congruent right-angled triangles

P QR

and

T S R

with right angles at

Q

and

S

, respectively.

P Q = T S = 3

QR = S R = 4

P R T

is a straight line.

What is the length, in cm, of

QS

Two congruent right-angled triangles standing on the straight line P R T, meeting at the shared vertex R in the middle of that line. The left triangle has vertices P, Q and R, with Q above the line; its right angle at Q is marked with a small square, side PQ is labelled 3 cm and side QR is labelled 4 cm. The right triangle has vertices T, S and R, with S above the line; its right angle at S is marked with a small square, side TS is labelled 3 cm and side SR is labelled 4 cm. A horizontal segment joins Q to S across the top of the figure. The paper states that the diagram is not to scale.

A $4$
B $32$
C $5.2$
D $42$
E $6.4$
F $8.2$
G $10$

Worked solution

Answer:E

Each triangle has legs

3

and

4

, so its hypotenuse is

5

P R = R T = 5

and

P T = 10

.

Set up coordinates with

P

at the origin and

P T

along the

x

-axis, so

R = (5, 0)

and

T = (10, 0)

.

In a right-angled triangle the altitude from the right angle to the hypotenuse has length

\frac{leg \times leg}{hypotenuse}

, and it cuts the hypotenuse at distance

\frac{leg ^{2}}{hypotenuse}

from the end of that leg. For triangle

P QR

this gives a height of

\frac{3 \times 4}{5} = 2.4

and a foot at distance

\frac{3 ^{2}}{5} = 1.8

from

P

, so

Q = (1.8, 2.4) .

Triangle

T S R

is congruent and is drawn the same way up on the other side of

R

, so by the same computation from

T

S = (10 - 1.8, 2.4) = (8.2, 2.4) .

Both points are at height

2.4

, so

QS

is horizontal and

QS = 8.2 - 1.8 = 6.4.

The answer is E.

NSAA 2017 Section 1, Part A (Mathematics questions) · Q16

The diagram shows a square with side of length

x

cm. A circle is drawn with centre O which lies at the mid-point of one of the sides of the square. This side forms part of a diameter of the circle. The circle passes through two corners of the square as shown.

What is the area, in

cm^{2}

, of the shaded part of the semi-circle?

A circle with a horizontal diameter drawn right across it, the mid-point of that diameter marked by a short tick with the letter O written just below it. A square stands on the upper side of the diameter: its base lies along the diameter with O at the mid-point of the base, and its two upper corners lie on the circle. A double-headed vertical arrow inside the square, close to its right-hand side, is labelled x and spans the full height of the square. The part of the upper half of the circle that lies outside the square is shaded grey, namely a region on each side of the square and a thin region between the top side of the square and the arc. The lower half of the circle is left white.

A $(π - 1) x^{2}$
B $(\frac{π - 2}{2}) x^{2}$
C $(\frac{3 π - 2}{2}) x^{2}$
D $(\frac{3 π - 4}{4}) x^{2}$
E $(\frac{5 π - 4}{4}) x^{2}$
F $(\frac{5 π - 8}{8}) x^{2}$

Worked solution

Answer:F

Put the centre O at the origin with the diameter along the horizontal axis. The base of the square then runs from

(- \frac{x}{2}, 0)

(\frac{x}{2}, 0)

, and its two upper corners are at

(- \frac{x}{2}, x)

and

(\frac{x}{2}, x)

.

Those upper corners lie on the circle, so the radius

r

satisfies

r^{2} = (\frac{x}{2})^{2} + x^{2} = \frac{x ^{2}}{4} + x^{2} = \frac{5 x ^{2}}{4} .

The upper half of the circle therefore has area

\frac{1}{2} π r^{2} = \frac{1}{2} \times π \times \frac{5 x ^{2}}{4} = \frac{5 π x ^{2}}{8} .

The square lies entirely inside that semi-circle and has area

x^{2}

, and the shaded region is exactly what is left of the semi-circle once the square is removed:

\frac{5 π x ^{2}}{8} - x^{2} = (\frac{5 π - 8}{8}) x^{2} .

Numerically the bracket is about

0.96

, against a semi-circle of about

1.96 x^{2}

, so a little under half of the semi-circle is shaded, which matches the picture.

The answer is F.

NSAA 2023 Section 1, Part A (Mathematics questions) · Q5

W X Y Z

is a square of side length 1.

W M : M X = 1 : 2

X N : N Y = 3 : 1

Y P : P Z = 4 : 1

What is the area of triangle

M N P

A square with its corners labelled X at the top left, Y at the top right, Z at the bottom right and W at the bottom left. A point M is marked on the left side between W and X, a point N on the top side between X and Y, and a point P on the right side between Y and Z. The triangle MNP is drawn and shaded grey. No lengths or angles are marked. The paper states that the diagram is not to scale.

A $\frac{1}{3}$
B $\frac{2}{5}$
C $\frac{9}{20}$
D $\frac{1}{30}$
E $\frac{19}{60}$
F $\frac{23}{60}$

Worked solution

Answer:F

Place

W

at the origin, so that the square has

X = (0, 1)

Y = (1, 1)

and

Z = (1, 0)

.

From

W M : M X = 1 : 2

the point

M

is one third of the way up the left side, so

M = (0, \frac{1}{3})

. From

X N : N Y = 3 : 1

the point

N

is three quarters of the way along the top, so

N = (\frac{3}{4}, 1)

. From

Y P : P Z = 4 : 1

the point

P

is four fifths of the way down the right side, so

P = (1, \frac{1}{5})

.

Rather than use a formula, subtract the three surrounding pieces from the unit square.

Triangle

M X N

has its right angle at

X

, with legs

M X = \frac{2}{3}

and

X N = \frac{3}{4}

, so its area is

\frac{1}{2} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{4}

.

Triangle

N Y P

has its right angle at

Y

, with legs

N Y = \frac{1}{4}

and

Y P = \frac{4}{5}

, so its area is

\frac{1}{2} \times \frac{1}{4} \times \frac{4}{5} = \frac{1}{10}

.

What is left below is the trapezium

M W Z P

, with parallel sides

W M = \frac{1}{3}

and

Z P = \frac{1}{5}

a distance 1 apart, so its area is

\frac{1}{2} (\frac{1}{3} + \frac{1}{5}) = \frac{4}{15}

.

These three pieces total

\frac{1}{4} + \frac{1}{10} + \frac{4}{15} = \frac{15 + 6 + 16}{60} = \frac{37}{60},

so triangle

M N P

has area

1 - \frac{37}{60} = \frac{23}{60}

.

The answer is F.