Pythagoras

Find the side opposite the small square.

The unknown is the longest side. Add the squares.

The right angle is between the two known sides.

Subtract the square of the known shorter side from the hypotenuse squared.

First identify the hypotenuse.

Add the squares, take the square root, then round.

1. c^2=7^2+9^2=49+81=130.
2. c=\sqrt{130}=11.401754\ldots cm, so the answer is 11.4 cm.

Use the right-angle mark to identify the longest side. Write the equation before rearranging.

1. The 14 m side is opposite the right angle. Therefore x^2+9^2=14^2.
2. x^2=196-81=115, so x=\sqrt{115}=10.723805\ldots m.
3. To 3 significant figures, x=10.7 m.

Is the missing side the hypotenuse or a shorter side?

1. The missing side squared is 17^2-8^2. Take a square root to turn this into a length.

Which operation reverses squaring?

1. 169 is the square of the hypotenuse, not its length. \sqrt{169}=13, so the hypotenuse is 13 cm.

Calculate in centimetres first, then convert.

1. c^2=90^2+120^2=8100+14400=22500.
2. c=\sqrt{22500}=150 cm. Convert to metres: 150\div100=1.5 m.

Draw the right triangle made by the diagonal.

1. The diagonal joins opposite corners, so it is opposite the right angle in a triangle with sides 90 cm and 120 cm.
2. d^2=90^2+120^2=8100+14400=22500.
3. d=\sqrt{22500}=150 cm.

The wall height is a shorter side.

1. The ladder is the hypotenuse, so h^2=4.5^2-1.2^2=20.25-1.44=18.81.
2. h=\sqrt{18.81}=4.337049\ldots m, so it reaches 4.3 m.

Find the sloping length before adding the extra cable.

1. The straight part has length \sqrt{2.4^2+3.2^2}=\sqrt{5.76+10.24}=\sqrt{16}=4 m.
2. Add the fastening allowance: 4+0.6=4.6 m.

Find the third side, then add all three sides.

1. The missing side is \sqrt{25^2-7^2}=\sqrt{625-49}=\sqrt{576}=24 cm.
2. The perimeter is 25+7+24=56 cm.

The perpendicular height splits the base into two equal halves.

1. Draw a perpendicular from the top vertex to the base. In this isosceles triangle it divides the base into two equal lengths, each 10\div2=5 cm.
2. Use either right-angled triangle: h^2+5^2=13^2.
3. h^2=169-25=144, so h=\sqrt{144}=12 cm.

Use the area to find the missing side of the rectangle.

1. First find its length: 60\div5=12 cm.
2. The diagonal is \sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13 cm.

Calculate both routes, then find their difference.

1. The edge route is 10+8=18 m.
2. The diagonal is \sqrt{10^2+8^2}=\sqrt{164}=12.806248\ldots m.
3. The difference is 18-\sqrt{164}=5.193751\ldots m, or 5.2 m. Keep the diagonal unrounded until subtracting.

Use A=\frac12(a+b)h to find the other parallel side. Then consider a right-angled triangle beside the rectangular part.

1. Let DC=t. The trapezium area gives \frac12(10+t)\times6=84.
2. 10+t=84\div3=28, so DC=18 cm.
3. Imagine drawing a vertical line from B to DC. The right-angled triangle on the right has horizontal side 18-10=8 cm and vertical side 6 cm.
4. BC=\sqrt{8^2+6^2}=\sqrt{100}=10 cm.
5. The perimeter is 10+10+18+6=44 cm. The imagined line is inside the shape, so it is not included.

Find the original distance from the wall before moving the foot.

1. Initially the foot is \sqrt{5^2-4^2}=\sqrt9=3 m from the wall.
2. After moving, the foot is 3+1=4 m from the wall. The new height is \sqrt{5^2-4^2}=3 m.
3. The top moves down 4-3=1 m.

First use the perimeter to find the two equal sides. A perpendicular from the top vertex divides the base in half.

1. The two equal sides total 46-12=34 cm, so each is 34\div2=17 cm.
2. Draw the perpendicular height to the base. It bisects the base, giving two right-angled triangles with a base of 12\div2=6 cm.
3. h^2=17^2-6^2=289-36=253, so h=\sqrt{253}=15.905973\ldots cm.
4. Area =\frac12\times12\times\sqrt{253}=95.435842\ldots\text{ cm}^2.
5. To 3 significant figures, the area is 95.4\text{ cm}^2. Keep the height unrounded until calculating the area.