num-title-ct-core

Right-angled Triangles, SM-Bank 045

Use Pythagoras' Theorem to calculate the height of the building below. (2 marks)

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$80\ \text{metres}$

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$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }a=h,\ b=150\ \text{and }c=170$

$h^2+150^2$	$=170^2$
$h^2$	$=170^2-150^2$
$h^2$	$=6400$
$h$	$=\sqrt{6400}$
$h$	$=80$

$\text{The height of the building is }80\ \text{metres}.$

Right-angled Triangles, SM-Bank 044

Calculate the perpendicular height of the isosceles triangle below, giving your answer correct to one decimal place. (2 marks)

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$\approx 6.9\ (1\text{ d.p.})$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=3\ \text{and }c=7.5$

$a^2+3^2$	$=7.5^2$
$a^2$	$=7.5^2-3^2$
$a^2$	$=47.25$
$a$	$=\sqrt{46.01}$
$a$	$=6.873\dots\approx 6.9\ (1\text{ d.p.})$

Right-angled Triangles, SM-Bank 042

Find the value of the pronumeral in the right-angled triangle below, giving your answer in exact surd form. (2 marks)

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$c=\sqrt{277}$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=9,\ \text{and }b=14$

$\text{Then}\ \ c^2$	$=9^2+14^2$
$c^2$	$=277$
$c$	$=\sqrt{277}$

Right-angled Triangles, SM-Bank 041

Find the value of the pronumeral in the right-angled triangle below, giving your answer in exact surd form. (2 marks)

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$c=\sqrt{269}$

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$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=10,\ \text{and }b=13$

$\text{Then}\ \ c^2$	$=10^2+13^2$
$c^2$	$=269$
$c$	$=\sqrt{269}$

Right-angled Triangles, SM-Bank 040

Find the value of the pronumeral in the right-angled triangle below, giving your answer in exact surd form. (2 marks)

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$c=\sqrt{98}$

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$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=7,\ \text{and }b=7$

$\text{Then}\ \ c^2$	$=7^2+7^2$
$c^2$	$=98$
$c$	$=\sqrt{98}$

Right-angled Triangles, SM-Bank 039

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place. (2 marks)

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$c\approx 6.5\ (1\ \text{d.p.})$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=4.8,\ \text{and }b=4.4$

$\text{Then}\ \ c^2$	$=4.8^2+4.4^2$
$c^2$	$=42.4$
$c$	$=\sqrt{42.4}$
$c$	$=6.511\dots$
$c$	$\approx 6.5\ (1\ \text{d.p.})$

Right-angled Triangles, SM-Bank 038

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place. (2 marks)

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$c\approx 16.2\ (1\ \text{d.p.})$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=15.4,\ \text{and }b=5.1$

$\text{Then}\ \ c^2$	$=15.4^2+5.1^2$
$c^2$	$=263.17$
$c$	$=\sqrt{263.17}$
$c$	$=16.222\dots$
$c$	$\approx 16.2\ (1\ \text{d.p.})$

Right-angled Triangles, SM-Bank 037

Find the value of the pronumeral in the right-angled triangle below, giving your answer correct to 1 decimal place. (2 marks)

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$c\approx 12.5\ (1\ \text{d.p.})$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=10.4,\ \text{and }b=6.9$

$\text{Then}\ \ c^2$	$=10.4^2+6.9^2$
$c^2$	$=155.77$
$c$	$=\sqrt{155.77}$
$c$	$=12.480\dots$
$c$	$\approx 12.5\ (1\ \text{d.p.})$

Right-angled Triangles, SM-Bank 036

Find the value of the pronumeral in the right-angled triangle below. (2 marks)

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$c=70$

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$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=42,\ \text{and }b=56$

$\text{Then}\ \ c^2$	$=42^2+56^2$
$c^2$	$=4900$
$c$	$=\sqrt{4900}$
$c$	$=70$

Right-angled Triangles, SM-Bank 035

Find the value of the pronumeral in the right-angled triangle below. (2 marks)

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$c=15$

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$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=9,\ \text{and }b=12$

$\text{Then}\ \ c^2$	$=9^2+12^2$
$c^2$	$=225$
$c$	$=\sqrt{225}$
$c$	$=15$

Right-angled Triangles, SM-Bank 034

Find the value of the pronumeral in the right-angled triangle below. (2 marks)

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$x=25$

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$\text{Pythagoras’ Theorem states: }c^2=a^2+b^2$

$\text{Let }a=24,\ b=7\ \text{and }c=x$

$\text{Then}\ \ x^2$	$=24^2+7^2$
$x^2$	$=625$
$x$	$=\sqrt{625}$
$x$	$=25$

Right-angled Triangles, SM-Bank 043

Arthur observes a plane through his binoculars at a distance of 7500 metres. If Ben is directly under the plane and it is flying at an altitude of 5000 metres, how far apart are Arthur and Ben? Give your answer correct to the nearest metre. (2 Marks)

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$5590\ \text{m}$

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$\text{Using Pythagoras’ Theorem: }\ \ a^2+b^2=c^2$

$\text{Let }\ a=d,\ b=5000,\ c=7500$

$d^2+5000^2$	$=7500^2$
$d^2$	$=7500^2-5000^2$
$d^2$	$=31\ 250\ 000$
$d$	$=5590\dots$
$d$	$\approx 5590$

$\therefore\ \text{Arthur and Ben are approximately }5590\ \text{metres apart}.$

Right-angled Triangles, SM-Bank 033

Find the value of the pronumeral in the right-angled triangle below giving your answer in exact surd form. (2 marks)

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$a=\sqrt{45}$

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$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=6\ \text{and }c=9$

$a^2+6^2$	$=9^2$
$a^2+36$	$=81$
$a^2$	$=81-36$
$a^2$	$=45$
$\therefore a$	$=\sqrt{45}$

Right-angled Triangles, SM-Bank 032

Find the value of the pronumeral in the right-angled triangle below giving your answer in exact surd form. (2 marks)

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$a=\sqrt{176}$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=20\ \text{and }c=24$

$a^2+20^2$	$=24^2$
$a^2+400$	$=576$
$a^2$	$=576-400$
$a^2$	$=176$
$\therefore a$	$=\sqrt{176}$

Right-angled Triangles, SM-Bank 031

Find the value of the pronumeral in the right-angled triangle below giving your answer in exact surd form. (2 marks)

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$a=\sqrt{115}$

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$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=9\ \text{and }c=14$

$a^2+9^2$	$=14^2$
$a^2+81$	$=196$
$a^2$	$=196-81$
$a^2$	$=115$
$\therefore a$	$=\sqrt{115}$

Right-angled Triangles, SM-Bank 030

Find the value of the pronumeral in the right-angled triangle below correct to one decimal place. (2 marks)

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$a=28.3\ \text{(1 d.p.)}$

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$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=98\ \text{and }c=102$

$a^2+98^2$	$=102^2$
$a^2+9604$	$=10404$
$a^2$	$=10404-9604$
$a^2$	$=800$
$\therefore a$	$=\sqrt{800}$
	$=28.284\dots\approx 28.3\ \text{(1 d.p.)}$

Right-angled Triangles, SM-Bank 029

Find the value of the pronumeral in the right-angled triangle below correct to one decimal place. (2 marks)

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$a=15.7\ \text{(1 d.p.)}$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }b=14\ \text{and }c=21$

$a^2+14^2$	$=21^2$
$a^2+196$	$=441$
$a^2$	$=441-196$
$a^2$	$=245$
$\therefore a$	$=\sqrt{245}$
	$=15.652\dots\approx 15.7\ \text{(1 d.p.)}$

Right-angled Triangles, SM-Bank 028

Find the value of the pronumeral in the right-angled triangle below correct to one decimal place. (2 marks)

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$x=5.7\ \text{(1 d.p.)}$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }a=x ,\ b=4\ \text{and }c=7$

$x^2+4^2$	$=7^2$
$x^2+16$	$=49$
$x^2$	$=49-16$
$x^2$	$=33$
$\therefore x$	$=\sqrt{33}$
	$=5.744\dots\approx 5.7\ \text{(1 d.p.)}$

Right-angled Triangles, SM-Bank 027

Find the value of the pronumeral in the right-angled triangle below. (2 marks)

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$x=12$

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$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }a=x ,\ b=7\ \text{and }c=25$

$x^2+9^2$	$=15^2$
$x^2+81$	$=225$
$x^2$	$=225-81$
$x^2$	$=144$
$\therefore x$	$=\sqrt{144}$
	$=12$

Right-angled Triangles, SM-Bank 026

Find the value of the pronumeral in the right-angled triangle below. (2 marks)

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$x=24$

Show Worked Solution

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{Let }a=x ,\ b=7\ \text{and }c=25$

$x^2+7^2$	$=25^2$
$x^2+49$	$=625$
$x^2$	$=625-49$
$x^2$	$=576$
$\therefore x$	$=\sqrt{576}$
	$=24$

Right-angled Triangles, SM-Bank 025

Find the height of the power pole below correct to one decimal place. (2 marks)

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$4.8\ \text{m}$

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$\text{Let }a=x ,\ b=3.6\ \text{and }c=6$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$x^2+3.6^2$	$=6^2$
$x^2+12.96$	$=36$
$x^2$	$=36-12.96$
$x^2$	$=23.04$
$\therefore x$	$=\sqrt{23.04}$
	$=4.8$

$\therefore\ \text{The height of the power pole is }4.8\ \text{m}$

Right-angled Triangles, SM-Bank 024

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=19 ,\ b=23\ \text{and }c=24$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=19^2+23^2$
		$=361+529$
		$=890$
		$\ne 24^2$
	$\therefore\ \text{LHS}$	$\ne \text{RHS}$

$\therefore\ \text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 023

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=7 ,\ b=15\ \text{and }c=17$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=7^2+15^2$
		$=49+225$
		$=274$
		$\ne 17^2$
	$\therefore\ \text{LHS}$	$\ne \text{RHS}$

$\therefore\ \text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 022

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=6 ,\ b=7\ \text{and }c=8$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=6^2+7^2$
		$=36+49$
		$=85$
		$\ne 8^2$
	$\therefore\ \text{LHS}$	$\ne \text{RHS}$

$\therefore\ \text{The triangle is not right-angled as the sides do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 021

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=13 ,\ b=84\ \text{and }c=85$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=13^2+84^2$
		$=169+7056$
		$=7225$
		$=85^2$
		$=\text{RHS}$

$\therefore\ \text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 020

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=1.5 ,\ b=2\ \text{and }c=2.5$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=1.5^2+2^2$
		$=2.25+4$
		$=6.25$
		$=2.5^2$
		$=\text{RHS}$

$\therefore\ \text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 019

Use Pythagoras' Theorem to decide if the triangle below is right-angled. (2 marks)

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$\text{See worked solution}$

$\text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=24 ,\ b=7\ \text{and }c=25$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=24^2+7^2$
		$=576+49$
		$=625$
		$=25^2$
		$=\text{RHS}$

$\therefore\ \text{The triangle is right-angled as the sides form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 018

Use Pythagoras' Theorem to decide if the numbers $7, 8$ and $11$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

$7,\ 8\ \text{and }11\ \text{do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=7 ,\ b=8\ \text{and }c=11$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=7^2+8^2$
		$=49+64$
		$=113$
		$\ne 11^2$
	$\therefore\ \ $	$\ne\text{RHS}$

$\therefore\ 7,\ 8\ \text{and }11\ \text{do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 017

Use Pythagoras' Theorem to decide if the numbers $1.4, 4.8$ and $5.2$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

$1.4,\ 4.8\ \text{and }5.2\ \text{do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=1.4 ,\ b=4.8\ \text{and }c=5.2$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\rightarrow\ $	$a^2+b^2$	$=1.4^2+4.8^2$
		$=1.96+23.04$
		$=25$
		$=5^2\ne\text{RHS}$

$\therefore\ 1.4,\ 4.8\ \text{and }5.2\ \text{do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 016

Use Pythagoras' Theorem to decide if the numbers $36, 48$ and $63$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

$36,\ 48\ \text{and }63\ \text{do not form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=36 ,\ b=48\ \text{and }c=63$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\longrightarrow\ $	$a^2+b^2$	$=36^2+48^2$
		$=1296+2304$
		$=3600$
		$=60^2\ne\text{RHS}$

$\therefore\ 36,\ 48\ \text{and }63\ \text{do not form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 015

Use Pythagoras' Theorem to decide if the numbers $1.1 ,\ 6$ and $6.1$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

$1.1,\ 6\ \text{and }6.1\ \text{form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=1.1 ,\ b=6\ \text{and }c=6.1$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\longrightarrow\ $	$a^2+b^2$	$=1.1^2+6^2$
		$=1.21+36$
		$=37.21$
		$=6.1^2=\text{RHS}$

$\therefore\ 1.1,\ 6\ \text{and }6.1\ \text{form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 014

Use Pythagoras' Theorem to decide if the numbers $12 , 35$ and $37$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

$12,\ 35\ \text{and }37\ \text{form a Pythagorean triad.}$

Show Worked Solution

$\text{Let }a=12 ,\ b=35\ \text{and }c=37$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\longrightarrow\ $	$a^2+b^2$	$=12^2+35^2$
		$=144+1225$
		$=1369$
		$=37^2=\text{RHS}$

$\therefore\ 12,\ 35\ \text{and }37\ \text{form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 013

Use Pythagoras' Theorem to decide if the numbers $5 , 12$ and $13$ form a Pythagorean triad. (2 marks)

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$\text{See worked solution}$

Show Worked Solution

$\text{Let }a=5 ,\ b=12\ \text{and }c=13$

$\text{Pythagoras’ Theorem states: }a^2+b^2=c^2$

$\text{LHS: }\longrightarrow\ $	$a^2+b^2$	$=5^2+12^2$
		$=25+144$
		$=169$
		$=13^2=\text{RHS}$

$\therefore\ 5,\ 12\ \text{and }13\ \text{form a Pythagorean triad.}$

Right-angled Triangles, SM-Bank 012

Which side is the hypotenuse in these right angled triangles? (3 marks)

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a. $w$

b. $q$

c. $e$

Show Worked Solution

a. $\text{Side: }w$

b. $\text{Side: }q$

c. $\text{Side: }e$

Right-angled Triangles, SM-Bank 011

Which side is the hypotenuse in these right angled triangles? (3 marks)

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a. $x$

b. $m$

c. $a$

Show Worked Solution

a. $\text{Side: }x$

b. $\text{Side: }m$

c. $\text{Side: }a$

Right-angled Triangles, SM-Bank 010 MC

Which of the following is true of the hypotenuse in a right-angled triangle?

The hypotenuse is the side opposite to the right angle.
The hypotenuse is the side adjacent to the right angle.
The hypotenuse is one of the shorter sides in a right angled triangle.
The hypotenuse is equal to the sum of the square of the other two sides.

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$A$

Show Worked Solution

$\text{The hypotenuse is the side opposite to the right angle.}$

$\Rightarrow A$

Right-angled Triangles, SM-Bank 009 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$b^2=c^2-a^2$
$c^2=b^2+c^2$
$a^2=b^2-c^2$
$c^2=a^2-b^2$

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$C$

Show Worked Solution

$a^2=b^2-c^2$

$\Rightarrow C$

Right-angled Triangles, SM-Bank 008 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$w^2=x^2+z^2$
$z^2=w^2-x^2$
$x^2=z^2-w^2$
$w^2=x^2-z^2$

Show Answers Only

$D$

Show Worked Solution

$w^2=x^2-z^2$

$\Rightarrow D$

Right-angled Triangles, SM-Bank 007 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$c^2=a^2-b^2$
$b^2=c^2-a^2$
$a^2=b^2-c^2$
$c^2=a^2+b^2$

Show Answers Only

$A$

Show Worked Solution

$c^2=a^2-b^2$

$\Rightarrow A$

Right-angled Triangles, SM-Bank 006 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$m^2=n^2-p^2$
$p^2=m^2-n^2$
$n^2=p^2-m^2$
$n^2=m^2+p^2$

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$B$

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$p^2=m^2-n^2$

$\Rightarrow B$

Right-angled Triangles, SM-Bank 005 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$m^2=n^2+p^2$
$p^2=m^2+n^2$
$n^2=p^2+m^2$
$n^2=m^2-p^2$

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$A$

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$m^2=n^2+p^2$

$\Rightarrow A$

Right-angled Triangles, SM-Bank 004 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$w^2=x^2+z^2$
$x^2=w^2+z^2$
$z^2=x^2+w^2$
$x^2=w^2-z^2$

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$B$

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$x^2=w^2+z^2$

$\Rightarrow B$

Right-angled Triangles, SM-Bank 003 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$c^2=a^2+b^2$
$b^2=c^2-a^2$
$a^2=b^2+c^2$
$b^2=a^2+c^2$

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$C$

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$a^2=b^2+c^2$

$\Rightarrow C$

Right-angled Triangles, SM-Bank 002 MC

Which of the following correctly states Pythagoras' Theorem for this triangle?

$c^2=a^2+b^2$
$b^2=c^2-a^2$
$a^2=b^2+c^2$
$b^2=a^2+c^2$

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$D$

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$b^2=a^2+c^2$

$\Rightarrow D$

Right-angled Triangles, SM-Bank 001 MC

Henry flies a kite attached to a long string, as shown in the diagram below.

The horizontal distance of the kite to Henry’s hand is 8 m.

The vertical distance of the kite above Henry’s hand is 15 m.

The length of the string, in metres, is

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$B$

Show Worked Solution

$\text{Using Pythagoras}$

$s^2$	$= 8^2 + 15^2$
	$= 289$
$\therefore\ s$	$=\sqrt{289}$
	$= 17\ \text{metres}$

$\Rightarrow B$

Composite Figures, SM-Bank 033

The design below is made up of one sector with an angle of $\theta^\circ$ and one equilateral triangle.

Calculate the the value of $\large \theta$ if the perimeter of the shape in terms of $\large \pi$ is $(27+3\pi)$ metres. (3 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$60^\circ$

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$\text{Radius of sector}\ r=9\ \text{m, Sector angle}=\theta^\circ$

$\text{Perimeter}=27+3\pi$

$\text{Perimeter}$	$=3\ \text{straight edges}+1\ \text{arc length}$
$27+3\pi$	$=3\times 9+\Bigg(\dfrac{\theta}{360}\times 2\pi \times 9\Bigg)$
$27+3\pi$	$=27+\Bigg(\dfrac{\theta \times 18\pi}{360}\Bigg)$
$\dfrac{\theta\times \pi}{20}$	$=3\pi$
$\theta$	$=3\times 20=60^\circ$

$\therefore\ \theta=60^\circ$

Composite Figures, SM-Bank 032

The design below is made up of one $106^\circ$ sector arc and two right angled triangles.

Calculate the perimeter of the design, correct to two decimal place. (2 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$23.25\ \text{mm (2 d.p.)}$

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$\text{Radius of sector}\ r=5\ \text{m, Sector angle }\theta=106^\circ$

$\text{Perimeter}$	$=3\ \text{straight edges}+1\ \text{arc length}$
	$=2\times 3+8+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)$
	$=14+\Bigg(\dfrac{106}{360}\times 2\pi\times 5\Bigg)$
	$=14+\dfrac{53}{18}\times \pi$
	$=23.2502\dots$
	$\approx 23.25\ \text{mm (2 d.p.)}$

Composite Figures, SM-Bank 031

The design below is made up of one $40^\circ$ sector arc and a right angled triangle.

Calculate the perimeter of the design, correct to one decimal place. (2 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$29.1\ \text{cm (1 d.p.)}$

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$\text{Radius of sector}\ r=13\ \text{m, Sector angle }\theta=40^\circ$

$\text{Perimeter}$	$=3\ \text{straight edges}+1\ \text{arc length}$
	$=12+5+13+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)$
	$=30+\Bigg(\dfrac{40}{360}\times 2\pi\times 13\Bigg)$
	$=30+\dfrac{26}{9}\times \pi$
	$=29.075\dots$
	$\approx 29.1\ \text{cm (1 d.p.)}$

Composite Figures, SM-Bank 030

The design below is made up of one $120^\circ$ sector arc and three identical rhombuses.

Calculate the perimeter of the design, correct to one decimal place. (2 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$34.0\ \text{cm (1 d.p.)}$

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$\text{Radius of sector}\ r=4.2\ \text{m, Sector angle }\theta=120^\circ$

$\text{Perimeter}$	$=6\ \text{straight edges}+1\ \text{arc length}$
	$=6\times 4.2+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)$
	$=25.2+\Bigg(\dfrac{120}{360}\times 2\pi\times 4.2\Bigg)$
	$=25.2+\dfrac{1}{3}\times 8.4\pi$
	$=33.996\dots$
	$\approx 34.0\ \text{cm (1 d.p.)}$

Composite Figures, SM-Bank 029

Jonti is constructing the stage for the local music festival. The design is made up of one $60^\circ$ sector arc and two equilateral triangles.

Calculate the perimeter of Jonti's stage. Give your answer correct to one decimal place. (2 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$25.2\ \text{m (1 d.p.)}$

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$\text{Radius of sector}\ r=5\ \text{m, Sector angle }\theta=60^\circ$

$\text{Perimeter}$	$=4\ \text{straight edges}+1\ \text{arc length}$
	$=4\times 5+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)$
	$=20+\Bigg(\dfrac{60}{360}\times 2\pi\times 5\Bigg)$
	$=20+\dfrac{5}{3}\pi$
	$=25.235\dots$
	$\approx 25.2\ \text{m (1 d.p.)}$

Composite Figures, SM-Bank 028

Pixie is designing a new company logo. The design is made up of three identical sectors with the radius of each sector being 18 millimitres.

Calculate the perimeter of Pixie's. Give your answer correct to the nearest millimetre. (3 marks)

NOTE: $\text{Arc length: }l=\dfrac{\theta}{360}\times 2\pi r$

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$127\ \text{cm (nearest mm)}$

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$\text{Radius of sector}\ r=18\ \text{mm, Sector angle }\theta=60^\circ$

$\text{Perimeter}$	$=6\ \text{straight edges}+3\times \text{arc length}$
	$=6r+\Bigg(\dfrac{\theta}{360}\times 2\pi r\Bigg)$
	$=6\times 18+\Bigg(\dfrac{60}{360}\times 2\pi\times 18\Bigg)$
	$=108 +\Bigg(\dfrac{60\times 36\pi}{360}\Bigg)$
	$=108+6\pi$
	$=126.849\dots$
	$\approx 127\ \text{cm (nearest mm)}$

Composite Figures, SM-Bank 027

Pedro had one piece of pizza left to eat. The piece of pizza was one-eighth of the whole pizza and had a radius of 12 centimetres.

Calculate the perimeter of Pedro's piece of pizza. Give your answer correct to the nearest centimetre. (2 marks)

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$89\ \text{cm (nearest cm)}$

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$\text{Radius of sector}=12\ \text{cm}$

$\text{Perimeter}$	$=2\ \text{straight edges}+\dfrac{1}{8}\times \text{circumference} $
	$=2r+\Bigg(\dfrac{1}{8}\times 2\pi r\Bigg)$
	$=2\times 12 +\Bigg(\dfrac{1}{8}\times 2\pi \times 12\Bigg)$
	$=24+\dfrac{24\pi}{8}$
	$=80+3\pi$
	$=89.424\dots$
	$\approx 89\ \text{cm (nearest cm)}$

Composite Figures, SM-Bank 026

Find the exact perimeter of the composite shape below. (2 marks)

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$80+10\pi)\ \text{cm}$

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$\text{Radius of quadrant}=29\ \text{mm}$

$\text{Perimeter}$	$=4\ \text{straight edges}+\dfrac{1}{4}\times \text{circumference} $
	$=4\times 20+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)$
	$=80 +\Bigg(\dfrac{1}{4}\times 2\pi \times 20\Bigg)$
	$=80+\dfrac{40\pi}{4}$
	$=(80+10\pi)\ \text{cm}$

Composite Figures, SM-Bank 025

A guitar plectrum was made using a semicircle and a quadrant as shown in the diagram below.

Calculate the perimeter of the plectrum, correct to the one decimal place. (2 marks)

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$8.3\ \text{cm (1 d.p.)}$

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$\text{Radius of quadrant}=2\ \text{cm, Diameter of semicircle}=2\ \text{cm}$

$\text{Perimeter}$	$=1\ \text{straight edge}+\dfrac{1}{4}\times \text{circumference(1)} +\dfrac{1}{2}\times \text{circumference(2)}$
	$=2+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)+\Bigg(\dfrac{1}{2}\times \pi d\Bigg)$
	$=2 +\Bigg(\dfrac{1}{4}\times 2\pi \times 2\Bigg)+\Bigg(\dfrac{1}{2}\times \pi\times 2\Bigg)$
	$=2+\dfrac{4\pi}{4}+\dfrac{2\pi}{2}$
	$=2+2\pi$
	$=8.283\dots$
	$=8.3\ \text{cm (1 d.p.)}$

Composite Figures, SM-Bank 024

James ate one-quarter of his extra large pizza. Calculate the perimeter of the remaining pizza, correct to the nearest centimetre. (2 marks)

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$107\ \text{cm (nearest cm)}$

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$\text{Perimeter}$	$=2\ \text{straight sides} +\dfrac{3}{4}\times \text{circumference}$
	$=2\times 16 +\Bigg(\dfrac{3}{4}\times 2\pi r\Bigg)$
	$=32 +\Bigg(\dfrac{3}{4}\times 2\pi\times 16\Bigg)$
	$=32+24\pi$
	$=107.398\dots$
	$=107\ \text{cm (nearest cm)}$

Composite Figures, SM-Bank 023

Calculate the perimeter of the following shape correct to 1 decimal place. (2 marks)

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$72.8\ \text{cm (1 d.p.)}$

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$\text{Perimeter}$	$=4\ \text{straight sides} +\dfrac{1}{4}\times \text{circumference}$
	$=2\times 15+12+7.5+4.5 +\Bigg(\dfrac{1}{4}\times 2\pi\times 12\Bigg)$
	$=54+6\pi$
	$=72.849\dots$
	$=72.8\ \text{cm (1 d.p.)}$

Composite Figures, SM-Bank 022

The diagram below shows the design for a chicken pen at Gayle's farm.

Gayle wishes to construct a fence around the pen. Calculate the amount of fencing required, correct to the next metre. (2 marks)

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Gayle has chosen fencing that costs $22 per metre, with an additional cost of $240 for the installation of a gate. Calculate the total cost of fencing the pen. (2 marks)

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a. $17\ \text{m (next metre)}$

b. $$614$

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a.	$\text{Perimeter}$	$=2\ \text{radii}+\dfrac{1}{4}\ \text{circumference}$
		$=2\times 4.5+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)$
		$=9+\Bigg(\dfrac{1}{4}\times 2\pi \times 4.5\Bigg)$
		$=9 +\dfrac{1}{2}\times 4.5\pi$
		$=16.068\dots$
		$\approx 17\ \text{m (next metre)}$

b.	$\text{Total cost}$	$=$22\times 17 +$240$
		$=$614$

Composite Figures, SM-Bank 021

Write an expression in terms of $\large \pi$ for the perimeter of the shape below. Give your answer in simplest form. (2 marks)

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$2r +\dfrac{3\pi r}{2}$

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$\text{Radius}=r$

$\text{Perimeter}$	$=2\ \text{straight sides}+\dfrac{3}{4}\ \text{circumference}$
	$=2\times r+\Bigg(\dfrac{3}{4}\times 2\pi r\Bigg)$
	$=2r+\Bigg(\dfrac{6}{4}\times \pi \times r\Bigg)$
	$=2r +\dfrac{3\pi r}{2}$

Composite Figures, SM-Bank 020

Calculate the perimeter of the composite shape below correct to the nearest centimetre. (2 marks)

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$39\ \text{cm (nearest centimetre)}$

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$\text{Radius}=8\ \text{cm}$

$\text{Perimeter}$	$=4\ \text{straight sides}+1\ \text{quadrant}$
	$=2\times 5+10+ 6+\Bigg(\dfrac{1}{4}\times 2\pi r\Bigg)$
	$=26+\Bigg(\dfrac{1}{4}\times 2\pi \times 8\Bigg)$
	$=26+4\pi$
	$=38.566\dots$
	$\approx 39\ \text{cm (nearest centimetre)}$

Composite Figures, SM-Bank 019 MC

The courtyard shown below incorporates quadrants and rectangles in its design.

The landscaping company needs to calculate the perimeter of the courtyard so they can provide estimates for fencing.

Which answer represents the approximate length of fencing required, to the nearest metre?

$23\ \text{m}$
$33\ \text{m}$
$37\ \text{m}$
$47\ \text{m}$

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$D$

Show Worked Solution

$\text{Radius}=4+2=6\ \text{m}$

$\text{Perimeter}$	$=6\ \text{straight sides}+2\ \text{quadrants}$
	$=2\times 4+2\times 8+ 2\times 2+\Bigg(2\times \dfrac{1}{4}\times 2\pi r\Bigg)$
	$=28+\pi\times 6$
	$=46.849\dots$
	$\approx 47\ \text{m (nearest metre)}$

$\Rightarrow D$

\(h^2+150^2\)	\(=170^2\)
\(h^2\)	\(=170^2-150^2\)
\(h^2\)	\(=6400\)
\(h\)	\(=\sqrt{6400}\)
\(h\)	\(=80\)

\(a^2+3^2\)	\(=7.5^2\)
\(a^2\)	\(=7.5^2-3^2\)
\(a^2\)	\(=47.25\)
\(a\)	\(=\sqrt{46.01}\)
\(a\)	\(=6.873\dots\approx 6.9\ (1\text{ d.p.})\)

\(\text{Then}\ \ c^2\)	\(=9^2+14^2\)
\(c^2\)	\(=277\)
\(c\)	\(=\sqrt{277}\)

\(\text{Then}\ \ c^2\)	\(=10^2+13^2\)
\(c^2\)	\(=269\)
\(c\)	\(=\sqrt{269}\)

\(\text{Then}\ \ c^2\)	\(=7^2+7^2\)
\(c^2\)	\(=98\)
\(c\)	\(=\sqrt{98}\)

\(\text{Then}\ \ x^2\)	\(=24^2+7^2\)
\(x^2\)	\(=625\)
\(x\)	\(=\sqrt{625}\)
\(x\)	\(=25\)

\(d^2+5000^2\)	\(=7500^2\)
\(d^2\)	\(=7500^2-5000^2\)
\(d^2\)	\(=31\ 250\ 000\)
\(d\)	\(=5590\dots\)
\(d\)	\(\approx 5590\)

\(a^2+6^2\)	\(=9^2\)
\(a^2+36\)	\(=81\)
\(a^2\)	\(=81-36\)
\(a^2\)	\(=45\)
\(\therefore a\)	\(=\sqrt{45}\)

\(a^2+20^2\)	\(=24^2\)
\(a^2+400\)	\(=576\)
\(a^2\)	\(=576-400\)
\(a^2\)	\(=176\)
\(\therefore a\)	\(=\sqrt{176}\)