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Area, SM-Bank 018

Airships are a form of aircraft.

An airship has a cabin in which the pilots and passengers travel, and cargo is carried. This is shown in the simplified diagram below.

The floor of the cabin is a rectangle, with a length of 9 m and a width of 2.5 m.

The cockpit occupies an area 1.5 m by 2.5 m at the front of the cabin. This is shown shaded in the diagram below.

The remainder of the floor space is available for passengers and cargo.

Calculate the area available for passengers and cargo, in square metres. (2 marks)

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\(18.75\ \text{m}^2\)

Show Worked Solution

\(\text{Area}\)	\(=9\times 2.5-1.5\times 2.5\)
	\(=22.5-3.75\)
	\(=18.75\ \text{m}^2\)

Area, SM-Bank 017

Sequoia owns a farm with a rectangular paddock.

She increases the area of the paddock by adding land that changes it into the shape of a trapezium.

What is the area of Sequoia's new paddock, in square metres? (2 marks)

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\(2599\ \text{m}^2\)

Show Worked Solution

\(\text{Area trapezium}\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{46}{2}(71+42)\)
	\(=2599\ \text{m}^2\)

Area, SM-Bank 015

The floor of a chicken coop is in the shape of a trapezium.

The floor, \(ABCD\), and the chicken coop are shown below.

\(AB = 3\ \text{m}, BC = 2\ \text{m and}\ \ CD = 5\ \text{m.}\)

What is the area of the floor of the chicken coop?

Write your answer in square metres. (2 marks)

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What is the perimeter of the floor of the chicken coop?

Write your answer in metres, correct to one decimal place. (2 marks)

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a. \(8\ \text{m}^2\)

b. \(12.8\ \text{m}\)

Show Worked Solution

a.	\(A\)	\(=\dfrac{h}{2}(a+b)\)
		\(=\dfrac{2}{2}\times (3 + 5)\)
		\(=8\ \text{m}^2\)

\(\text{Using Pythagoras,}\)

\(AD^2\)	\(=2^2+2^2\)
\(AD^2\)	\(=8\)
\(AD\)	\(=\sqrt{8}\)
	\(=2.82\dots\ \text{m}\)

\(\therefore\ \text{Perimeter}\)	\(=3+2+5+2.82\dots\)
	\(=12.8\ \text{m (1 d.p.)}\)

Area, SM-Bank 013

Lucy designs an outdoor table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.

What is the area of Lucy's table top? (2 marks)

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\(2600\ \text{cm}^2\)

Show Worked Solution

\(\text{Method 1: Composite}\)

\(\text{Area}\)	\(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
	\(=(50\times 40) + 2\times\Bigg(\dfrac{1}{2}\times 15\times 40\Bigg)\)
	\(=2000 + 600\)
	\(=2600\ \text{cm}^2\)

\(\text{Method 2: Trapezium}\)

\(\text{Area}\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{40}{2}(80+50)\)
	\(=2600\ \text{cm}^2\)

Area, SM-Bank 012

Luke designs a table that is in the shape of a trapezium.

The dimensions of the table top are shown in the picture below.

What is the area of Luke's table top? (2 marks)

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\(880\ \text{cm}^2\)

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\(\text{Method 1: Composite}\)

\(\text{Area}\)	\(=\text{Area of rectangle}+2\times \text{Area of triangle}\)
	\(=(38\times 20) + 2\times\Bigg(\dfrac{1}{2}\times 6\times 20\Bigg)\)
	\(=760 + 120\)
	\(=880\ \text{cm}^2\)

\(\text{Method 2: Trapezium}\)

\(\text{Area}\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{20}{2}(38+50)\)
	\(=880\ \text{cm}^2\)

Area, SM-Bank 011

This triangle was made by cutting a square in half.

The perimeter of the triangle is 51.21 cm.

What is the area of the triangle? (2 marks)

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\(112.5\ \text{cm}^2\)

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\(\text{Triangle is isosceles with perimeter}=51.21\ \text{cm.}\)

\(\Rightarrow\ \text{Length of triangle side}\)

\(=\dfrac{1}{2}\times (51.21 – 21.21)\)

\(=\dfrac{1}{2}\times 30\)

\(=15\ \text{cm}\)

\(\therefore\ \text{Area}\)	\(=\dfrac{1}{2}\times bh\)
	\(=\dfrac{1}{2}\times 15\times 15\)
	\(=112.5\ \text{cm}^2\)

Area, SM-Bank 010

A square has an area of 169 square centimetres.

What is the perimeter? (2 marks)

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\(52\ \text{cm}\)

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\(\text{Let}\ \ s=\text{side length of the square}\)

\(\text{Area:}\rightarrow\ \ s^2\)	\(=169\)
\(s\)	\(=\sqrt{169}\)
	\(=13\ \text{cm}\)

\(\therefore\ \text{Perimeter}\)	\(=4\times 13\)
	\(=52\ \text{cm}\)

Area, SM-Bank 009

Jill is playing in her parents' rectangular courtyard.

The courtyard is measured at 6 metres by 10 metres.

Jill draws a triangle on the courtyard with chalk, pictured below.

What is the area of this triangle? (2 marks)

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\(14\ \text{m}^2\)

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\(\text{Base}=7\ \text{m}, \text{ Height}=4\ \text{m}\)

\(\text{Area of}\ \Delta\)	\(=\dfrac{1}{2}\times bh\)
	\(=\dfrac{1}{2}\times 7\times 4\)
	\(=14\ \text{m}^2\)

Area, SM-Bank 008

The length of this rectangle is one and a half times its width.

The perimeter of the rectangle is 50 centimetres.

What is the area of the rectangle? (2 marks)

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\(150\ \text{cm}^2\)

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\(\text{Let}\ x=\text{width, then}\ \ 1.5x=\text{length}\)

\(\text{Perimeter:}\ \rightarrow\ \)	\(\ \ 2\times x+2\times 1.5 x\)	\(=50\)
	\(5x\)	\(=50\)
	\(x\)	\(=10\)

\(\therefore\ \text{width}=10\ \text{cm, length}=15\ \text{cm}\)

\(\therefore\ \text{Area}\)	\(=10\times 15\)
	\(=150\ \text{cm}^2\)

Area, SM-Bank 005

The area of the shaded rectangle below is \(84\ \text{cm}^2\).

What is the length of the shaded rectangle? (2 marks)

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\(14\ \text{cm}\)

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\(\text{Let}\ \ l =\text{length of rectangle}\)

\(l\times 6\)	\(=84\)
\(\therefore\ l\)	\(=\dfrac{84}{6}\)
	\(=14\ \text{cm}\)

Area, SM-Bank 004 MC

A neighbourhood soccer oval is marked out with the dimensions shown below.

What is the area of the field?

\(298\ \text{m}^2\)
\(2842.75\ \text{m}^2\)
\(5261.25\ \text{m}^2\)
\(8123.25\ \text{m}^2\)

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\(C\)

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\(\text{Area of the soccer field}\)

\(=57.5\times 91.5\)

\(= 5261.25\ \text{m}^2\)

\(\Rightarrow C\)

Area, SM-Bank 003 MC

A resort has 4 pools.

Which pool has the largest surface area?

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\(D\)

Show Worked Solution

\(\text{Consider the surface area of each pool:}\)

\(\text{Pool A}=6\times 19=114\ \text{m}^2\)

\(\text{Pool B}=7\times 18=126\ \text{m}^2\)

\(\text{Pool C}=10\times 15=150\ \text{m}^2\)

\(\text{Pool D}=12.5\times 12.5=156.25\ \text{m}^2\)

\(\therefore\ \text{Pool D,}\ 12.5\times 12.5,\text{ has the largest}\)

\(\text{surface area.}\)

\(\Rightarrow D\)

Right-angled Triangles, SM-Bank 053

Use Pythagoras' Theorem to calculate the length of the hypotenuse in the isosceles triangle below. Give your answer correct in exact surd form. (2 marks)

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Using your result from part (a) above, calculate the perimeter of the shape below, correct to 1 decimal place. (2 marks)

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a. \(\sqrt{50}\ \text{cm (exact surd form)}\)

b. \(30.6\ \text{cm (1 d.p.)}\)

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

\(\text{Let }a=5\ \text{and }b=5\)

\(\text{Then}\ \ c^2\)	\(=5^2+5^2\)
\(c^2\)	\(=50\)
\(c\)	\(=\sqrt{50}\ \text{cm (exact surd form)}\)

b.	\(\text{Perimeter}\)	\(=\text{chord (a)}\ +\dfrac{3}{4}\times\text{circumference}\)
		\(=\sqrt{50}+\dfrac{3}{4}\times 2\pi r\)
		\(=\sqrt{50}+\dfrac{3}{4}\times 2\pi\times 5\)
		\(=30.633\dots\)
		\(\approx 30.6\ \text{cm (1 d.p.)}\)

Right-angled Triangles, SM-Bank 052

Use Pythagoras' Theorem to calculate the perimeter of the isosceles triangle below, correct to the nearest centimetre. (3 marks)

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\(29\ \text{cm}\ (\text{nearest cm})\)

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\(\text{Find length of the equal sides of the triangle.}\)

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

\(\text{Let }a=5\ \text{and }b=8\)

\(\text{Then}\ \ c^2\)	\(=5^2+8^2\)
\(c^2\)	\(=89\)
\(c\)	\(=\sqrt{89}\)
\(c\)	\(=9.433\dots\)

\(\text{Perimeter}\)

\(=2\times 9.433\dots+10\)

\(=28.867\dots\approx 29\ \text{cm}\ (\text{nearest cm})\)

Right-angled Triangles, SM-Bank 051

Use Pythagoras' Theorem to calculate the perimeter of the trapezium below, correct to 1 decimal place. (3 marks)

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\(31.1\ \text{mm}\ (1\ \text{d.p.})\)

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\(\text{Find length of sloped side of trapezium.}\)

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

\(\text{Let }a=4\ \text{and }b=7)

\(\text{Then}\ \ c^2\)	\(=4^2+7^2\)
\(c^2\)	\(=65\)
\(c\)	\(=\sqrt{65}\)
\(c\)	\(=8.062\dots\)

\(\text{Perimeter}\)

\(=6+7+10+8.062\dots\)

\(=31.062\dots\approx 31.1\ \text{mm}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 050

Use Pythagoras' Theorem to calculate the perimeter of the rectangle below, correct to 1 decimal place. (3 marks)

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\(13.0\ \text{m}\ (1\ \text{d.p.})\)

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\(\text{Find side length of rectangle.}\)

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

\(\text{Let }b=3\ \text{and }c=4.6\)

\(\text{Then}\ \ a^2+3^2\)	\(=4.6^2\)
\(a^2\)	\(=4.6^2-3^2\)
\(a\)	\(=\sqrt{12.16}\)
\(a\)	\(=3.487\dots\)

\(\text{Perimeter}\)

\(=2\times 3.487\dots+ 2\times 3\)

\(=12.974\dots\approx 13.0\ \text{m}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 049

Calculate the length of the hypotenuse in the following right-angled triangle. Give your answer correct to the nearest metre. (2 marks)
Find the perimeter of the triangle. (1 mark)

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a. \(7\ \text{m}\)

b. \(18\ \text{m}\)

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

\(\text{Let }a=5\ \text{and }b=6\)

\(\text{Then}\ \ c^2\)	\(=5^2+6^2\)
\(c^2\)	\(=61\)
\(c\)	\(=\sqrt{61}\)
\(c\)	\(=7.141\dots\approx 7\)

\(\text{The hypotenuse is }7\ \text{metres (nearest metre})\)

b. \(\text{Perimeter}\)

\(=7+5+6\)

\(=18\ \text{m}\)

Right-angled Triangles, SM-Bank 048

The size of a computer monitor is determined by the length of the diagonal of the screen. Use Pythagoras' Theorem to calculate the size of the monitor in the diagram below. (2 marks)

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\(51\ \text{centimetres}\)

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

\(\text{Let }a=24\ \text{and }b=45\)

\(\text{Then}\ \ c^2\)	\(=24^2+45^2\)
\(c^2\)	\(=2601\)
\(c\)	\(=\sqrt{2601}\)
\(c\)	\(=51\)

\(\text{The size of the monitor is }51\ \text{centimetres}\)

Right-angled Triangles, SM-Bank 047

Use Pythagoras' Theorem to calculate the length of the diagonal in the rectangle below, correct to 1 decimal place. (2 marks)

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\(8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

Show Worked Solution

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

\(\text{Let }a=4.2\ \text{and }b=7.3\)

\(\text{Then}\ \ c^2\)	\(=4.2^2+7.3^2\)
\(c^2\)	\(=70.98\)
\(c\)	\(=\sqrt{70.98}\)
\(c\)	\(=8.4219\dots\approx 8.4\ (1\ \text{d.p.})\)

\(\text{The length of the string is }8.4\ \text{centimetres}\ (1\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 046

Lewis is holding a string attached to a kite. Lewis is 8 metres from the kite and the kite is flying 4 metres above his hand. Use Pythagoras' Theorem to calculate the length of the string attached to the kite, correct to 2 decimal places. (2 marks)

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\(8.94\ \text{metres}\ (2\ \text{d.p.})\)

Show Worked Solution

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

\(\text{Let }a=4\ \text{and }b=8\)

\(\text{Then}\ \ c^2\)	\(=4^2+8^2\)
\(c^2\)	\(=80\)
\(c\)	\(=\sqrt{80}\)
\(c\)	\(=8.9442\dots\approx 8.94\ (2\ \text{d.p.})\)

\(\text{The length of the string is }8.94\ \text{metres}\ (2\ \text{d.p.})\)

Right-angled Triangles, SM-Bank 045

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

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\(80\ \text{metres}\)

Show Worked Solution

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

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

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

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

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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\(c=\sqrt{277}\)

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

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

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

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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\(c=\sqrt{98}\)

Show Worked Solution

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

\(\text{Let }a=7,\ \text{and }b=7\)

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

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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\(a=\sqrt{45}\)

Show Worked Solution

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

\(\text{Let }b=6\ \text{and }c=9\)

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

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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\(a=\sqrt{176}\)

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

\(\text{Let }b=20\ \text{and }c=24\)

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

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

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

\(\text{Let }b=14\ \text{and }c=21\)

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

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

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

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

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

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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\(x=24\)

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

Right-angled Triangles, SM-Bank 023

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

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\(\text{See worked solution}\)

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

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

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

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

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

Right-angled Triangles, SM-Bank 022

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

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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

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

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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\(\text{See worked solution}\)

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

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

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

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

\(\therefore\ \text{Total Area}\)	\(=\text{Area 1 rectangle}+2\times\ \text{Area of triangle}\)
	\(=6\times 21+2\times\Bigg(\dfrac{1}{2}\times 3\times 14\Bigg)\)
	\(=126+42\)
	\(=168\ \text{cm}^2\)

\(\text{Triangle A}\)	\(=\dfrac{1}{2}\times 3\times 2=3\ \text{square units}\)
\(\text{Triangle B}\)	\(=\dfrac{1}{2}\times 3\times 3=4.5\ \text{square units}\ \checkmark\)
\(\text{Triangle C}\)	\(=\dfrac{1}{2}\times 2\times 4=4\ \text{square units}\)
\(\text{Triangle D}\)	\(=\dfrac{1}{2}\times 1\times 6=3\ \text{square units}\)

\(\text{Area}\)	\(=\text{Area of large rectangle}-\text{Area of small rectangle}\)
	\(=(120\times 200)-(60\times 40)\)

\(\text{Area}\)	\(=\dfrac{h}{2}(a+b)\)
	\(=\dfrac{60}{2}(80+60)\)
	\(=4200\)