IB Questionbank

User interface language: English | Español

Date	May 2021	Marks available	1	Reference code	21M.3.AHL.TZ1.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 1
Command term	Interpret	Question number	2	Adapted from	N/A

Question

This question asks you to examine various polygons for which the numerical value of the area is the same as the numerical value of the perimeter. For example, a $3$ by $6$ rectangle has an area of $18$ and a perimeter of $18$ .

For each polygon in this question, let the numerical value of its area be $A$ and let the numerical value of its perimeter be $P$ .

An $n$ -sided regular polygon can be divided into $n$ congruent isosceles triangles. Let $x$ be the length of each of the two equal sides of one such isosceles triangle and let $y$ be the length of the third side. The included angle between the two equal sides has magnitude $\frac{2 π}{n}$ .

Part of such an $n$ -sided regular polygon is shown in the following diagram.

Consider a $n$ -sided regular polygon such that $A = P$ .

The Maclaurin series for $\tan x$ is $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \dots$

Consider a right-angled triangle with side lengths $a, b$ and $\sqrt{a^{2} + b^{2}}$ , where $a \geq b$ , such that $A = P$ .

Find the side length, $s$ , where $s > 0$ , of a square such that $A = P$ .

[3]

Write down, in terms of $x$ and $n$ , an expression for the area, $A_{T}$ , of one of these isosceles triangles.

[1]

Show that $y = 2 x \sin \frac{π}{n}$ .

[2]

Use the results from parts (b) and (c) to show that $A = P = 4 n \tan \frac{π}{n}$ .

[7]

Use the Maclaurin series for $\tan x$ to find $\lim_{n \to \infty} (4 n \tan \frac{π}{n})$ .

[3]

e.i.

Interpret your answer to part (e)(i) geometrically.

[1]

e.ii.

Show that $a = \frac{8}{b - 4} + 4$ .

[7]

By using the result of part (f) or otherwise, determine the three side lengths of the only two right-angled triangles for which $a, b, A, P \in ℤ$ .

[3]

g.i.

Determine the area and perimeter of these two right-angled triangles.

[1]

g.ii.

Markscheme

$A = s^{2}$ and $P = 4 s$ (A1)

$A = P \Rightarrow s^{2} = 4 s$ (M1)

$s (s - 4) = 0$

$\Rightarrow s = 4 (s > 0)$ A1

Note: Award A1M1A0 if both $s = 4$ and $s = 0$ are stated as final answers.

[3 marks]

$A_{T} = \frac{1}{2} x^{2} \sin \frac{2 π}{n}$ A1

Note: Award A1 for a correct alternative form expressed in terms of $x$ and $n$ only.

For example, using Pythagoras’ theorem, $A_{T} = x \sin \frac{π}{n} \sqrt{x^{2} - x^{2} \sin^{2} \frac{π}{n}}$ or $A_{T} = 2 (\frac{1}{2} (x \sin \frac{π}{n}) (x \cos \frac{π}{n}))$ or $A_{T} = x^{2} \sin \frac{π}{n} \cos \frac{π}{n}$ .

[1 mark]

METHOD 1

uses $\sin θ = \frac{opp}{hyp}$ (M1)

$\frac{\frac{y}{2}}{x} = \sin \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 2

uses Pythagoras’ theorem ${(\frac{y}{2})}^{2} + h^{2} = x^{2}$ and $h = x \cos \frac{π}{n}$ (M1)

${(\frac{y}{2})}^{2} + {(x \cos \frac{π}{n})}^{2} = x^{2} (y^{2} = 4 x^{2} (1 - \cos^{2} \frac{π}{n}))$

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 3

uses the cosine rule (M1)

$y^{2} = 2 x^{2} - 2 x^{2} \cos \frac{2 π}{n} (= 2 x^{2} (1 - \cos \frac{2 π}{n}))$

$= 4 x^{2} \sin^{2} \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

METHOD 4

uses the sine rule (M1)

$\frac{y}{\sin \frac{2 π}{n}} = \frac{x}{\sin (\frac{π}{2} - \frac{π}{n})}$

$y \cos \frac{π}{n} = 2 x \sin \frac{π}{n} \cos \frac{π}{n}$ A1

$y = 2 x \sin \frac{π}{n}$ AG

[2 marks]

$A = P \Rightarrow n A_{T} = n y$ (M1)

Note: Award M1 for equating correct expressions for $A$ and $P$ .

$\frac{1}{2} n x^{2} \sin \frac{2 π}{n} = 2 n x \sin \frac{π}{n} (n x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 n x \sin \frac{π}{n})$

$\frac{1}{2} x^{2} \sin \frac{2 π}{n} = 2 x \sin \frac{π}{n} (x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n})$ A1

uses $\sin \frac{2 π}{n} = 2 \sin \frac{π}{n} \cos \frac{π}{n}$ (seen anywhere in part (d) or in part (b)) (M1)

$x^{2} \sin \frac{π}{n} \cos \frac{π}{n} = 2 x \sin \frac{π}{n}$

attempts to either factorise or divide their expression (M1)

$x \sin \frac{π}{n} (x \cos \frac{π}{n} - 2) = 0$

$x = \frac{2}{\cos \frac{π}{n}}, (x \sin \frac{π}{n} \neq 0)$ (or equivalent) A1

EITHER

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $P = n y$ (M1)

$P = 2 n (\frac{2}{\cos \frac{π}{n}}) (\sin \frac{π}{n})$ A1

Note: Other approaches are possible. For example, award A1 for $P = 2 n x \cos \frac{π}{n} \tan \frac{π}{n}$ and M1 for substituting $x = \frac{2}{\cos \frac{π}{n}}$ into $P$ .

substitutes $x = \frac{2}{\cos \frac{π}{n}}$ (or equivalent) into $A = n A_{T}$ (M1)

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (\sin \frac{2 π}{n})$

$A = \frac{1}{2} n {(\frac{2}{\cos \frac{π}{n}})}^{2} (2 \sin \frac{π}{n} \cos \frac{π}{n})$ A1

THEN

$A = P = 4 n \tan \frac{π}{n}$ AG

[7 marks]

attempts to use the Maclaurin series for $\tan x$ with $x = \frac{π}{n}$ (M1)

$\tan \frac{π}{n} = \frac{π}{n} + \frac{{(\frac{π}{n})}^{3}}{3} + \frac{2 {(\frac{π}{n})}^{5}}{15} (+ \dots)$

$4 n \tan \frac{π}{n} = 4 n (\frac{π}{n} + \frac{π^{3}}{3 n^{3}} + \frac{2 π^{5}}{15 n^{5}} (+ \dots))$ (or equivalent) A1

$= 4 (π + \frac{π^{3}}{3 n^{2}} + \frac{2 π^{5}}{15 n^{4}} + \dots)$

$\Rightarrow \lim_{n \to \infty} (4 n \tan \frac{π}{n}) = 4 π$ A1

Note: Award a maximum of M1A1A0 if $\lim_{n \to \infty}$ is not stated anywhere.

[3 marks]

e.i.

(as $n \to \infty, P \to 4 π$ and $A \to 4 π$ )

the polygon becomes a circle of radius $2$ R1

Note: Award R1 for alternative responses such as:
the polygon becomes a circle of area $4 π$ OR
the polygon becomes a circle of perimeter $4 π$ OR
the polygon becomes a circle with $A = P = 4 π$ .
Award R0 for polygon becomes a circle.

[1 mark]

e.ii.

$A = \frac{1}{2} a b$ and $P = a + b + \sqrt{a^{2} + b^{2}}$ (A1)(A1)

equates their expressions for $A$ and $P$ M1

$A = P \Rightarrow a + b + \sqrt{a^{2} + b^{2}} = \frac{1}{2} a b$

$\sqrt{a^{2} + b^{2}} = \frac{1}{2} a b - (a + b)$ M1

Note: Award M1 for isolating $\sqrt{a^{2} + b^{2}}$ or $\pm 2 \sqrt{a^{2} + b^{2}}$ . This step may be seen later.

$a^{2} + b^{2} = {(\frac{1}{2} a b - (a + b))}^{2}$

$a^{2} + b^{2} = \frac{1}{4} a^{2} b^{2} - 2 (\frac{1}{2} a b) (a + b) + {(a + b)}^{2}$ M1

$(= \frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + a^{2} + 2 a b + b^{2})$

Note: Award M1 for attempting to expand their RHS of either $a^{2} + b^{2} = \dots$ or $4 (a^{2} + b^{2}) = \dots$ .

EITHER

$a b (\frac{1}{4} a b - a - b + 2) = 0 (a b \neq 0)$ A1

$\frac{1}{4} a b - a - b + 2 = 0$

$a b - 4 a = 4 b - 8$

$\frac{1}{4} a^{2} b^{2} - a^{2} b - a b^{2} + 2 a b = 0$

$a (\frac{1}{4} b^{2} - b) + (2 b - b^{2}) = 0 (a (b^{2} - 4 b) + (8 b - 4 b^{2}) = 0)$ A1

$a = \frac{4 b^{2} - 8 b}{b^{2} - 4 b}$

THEN

$\Rightarrow a = \frac{4 b - 8}{b - 4}$ A1

$a = \frac{4 b - 16 + 8}{b - 4}$

$a = \frac{8}{b - 4} + 4$ AG

Note: Award a maximum of A1A1M1M1M0A0A0 for attempting to verify.
For example, verifying that $A = P = \frac{16}{b - 4} + 2 b + 4$ gains $4$ of the $7$ marks.

[7 marks]

using an appropriate method (M1)

eg substituting values for $b$ or using divisibility properties

$(5, 12, 13)$ and $(6, 8, 10)$ A1A1

Note: Award A1A0 for either one set of three correct side lengths or two sets of two correct side lengths.

[3 marks]

g.i.

$A = P = 30$ and $A = P = 24$ A1

Note: Do not award A1FT.

[1 mark]

g.ii.

Examiners report

[N/A]

e.i.

[N/A]

e.ii.

[N/A]

g.i.

[N/A]

g.ii.

Syllabus sections

Topic 5 —Calculus » AHL 5.13—Limits and L’Hopitals

Show 27 related questions

Hide related questions

Topic 5 —Calculus

Question

Markscheme

Examiners report

Syllabus sections

View options