Mixed Problems
Every question with a fully worked solution.
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Theory
Every Pythagoras question starts with one decision: is the side you want the hypotenuse (add the squares) or a shorter side (subtract from the hypotenuse squared)? This Year 8 maths page mixes both, with exact, surd and rounded answers, and right-angled triangles hidden inside other shapes.
Find the right angle. The hypotenuse is the side opposite it, and it is the longest side.
Hypotenuse wanted: add. \(c^2 = a^2 + b^2\). Shorter side wanted: subtract. \(a^2 = c^2 - b^2\), with the hypotenuse squared first. Take the square root last and keep only the positive answer.
No figure? Sketch one. \(\angle ABC = 90^\circ\) puts the right angle at \(B\), the middle letter, so the hypotenuse is \(AC\), the side that does not touch \(B\).
Exact or rounded? An exact answer keeps the square root, such as \(\sqrt{29}\), a surd. A decimal answer is rounded as the question asks. Change every length to the same unit before you square.
Simplest form of a surd. Split the number under the root into a square number times another number, then take the root of the square: \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\). Use the largest square factor you can find.
Inside a shape, find a right-angled triangle: the height of an isosceles triangle cuts the base in half; a line parallel to a side splits a right trapezium into a rectangle and a right-angled triangle; the diagonals of a kite or rhombus cross at right angles. The two shorter sides meet at a right angle, so they are a base and height for area.
Comparing two hypotenuses? Compare \(a^2 + b^2\) for each. The larger total gives the longer hypotenuse.
Decide, then add or subtract:
How to solve a mixed problem
- Sketch the triangle and mark the right angle.
- Decide: is the unknown the hypotenuse (add) or a shorter side (subtract)?
- Change to the same unit, substitute and work out.
- Take the positive square root; give it exact, as a simplest surd or rounded.
How to simplify a surd
- Find the largest square number that divides the number under the root.
- Split: \(\sqrt{72} = \sqrt{36} \times \sqrt{2}\).
- Take the root of the square: \(6\sqrt{2}\).
| \(x^2\) | \(=\) | \(113^2 - 112^2\) |
| \(\) | \(=\) | \(225\) |
| \(x\) | \(=\) | \(\sqrt{225} = 15\) |
\(x = 15\)
| \(y^2\) | \(=\) | \(24^2 + 143^2\) |
| \(\) | \(=\) | \(21\,025\) |
| \(y\) | \(=\) | \(\sqrt{21\,025} = 145\) |
\(y = 145\)
| \(h^2\) | \(=\) | \(9.4^2 - 6.5^2\) |
| \(\) | \(=\) | \(88.36 - 42.25 = 46.11\) |
| \(h\) | \(=\) | \(\sqrt{46.11} = 6.7904\ldots\) |
\(h \approx 6.8\) m
The right angle is at \(R\), so \(PQ\), which does not touch \(R\), is the hypotenuse.
| \(PQ^2\) | \(=\) | \(6^2 + 9^2 = 36 + 81 = 117\) |
\(117\) is not a square number: \(PQ = \sqrt{117}\) cm (about \(10.8\) cm).
The right angle is at \(B\), so \(AC\) is the hypotenuse and \(AB\) is a shorter side.
| \(AB^2\) | \(=\) | \(9^2 - 3^2 = 81 - 9 = 72\) |
| \(AB\) | \(=\) | \(\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}\) |
\(AB = 6\sqrt{2}\) (about \(8.5\))
(i) Find its perimeter.
(ii) Find its area.
The height cuts the base in half: \(30 \div 2 = 15\). In one half, \(s\) is the hypotenuse.
| \(s^2\) | \(=\) | \(15^2 + 20^2 = 225 + 400 = 625\) |
| \(s\) | \(=\) | \(\sqrt{625} = 25\) |
(i) \(30 + 25 + 25 = 80\) cm
(ii) \(\dfrac{1}{2} \times 30 \times 20 = 300\) cm\(^2\)
(i) Find \(EB\).
(ii) Find \(BC\).
(iii) Find the perimeter of \(ABCD\).
| \(EB\) | \(=\) | \(27 - 15 = 12\) |
| \(BC^2\) | \(=\) | \(12^2 + 16^2\) |
| \(\) | \(=\) | \(144 + 256 = 400\) |
| \(BC\) | \(=\) | \(\sqrt{400} = 20\) |
(i) \(12\) cm
(ii) \(20\) cm
(iii) \(27 + 20 + 15 + 16 = 78\) cm
(ii) The diagonals of kite \(ABCD\) cross at right angles at \(E\), with \(AE = EC = 15\) cm, \(BE = 36\) cm and \(ED = 20\) cm. Find the perimeter.
(iii) Which hypotenuse is longer: shorter sides \(4\) and \(8\), or \(6\) and \(6\)?
(i) Same units: \(2.4\) m \(= 240\) cm.
| \(\sqrt{240^2 + 70^2}\) | \(=\) | \(\sqrt{62\,500} = 250\) |
| \(AB^2\) | \(=\) | \(15^2 + 36^2 = 1521\) |
| \(AD^2\) | \(=\) | \(15^2 + 20^2 = 625\) |
(i) \(250\) cm \(= 2.5\) m
(ii) \(AB = BC = 39\) cm and \(AD = DC = 25\) cm, so the perimeter is \(2 \times 39 + 2 \times 25 = 128\) cm
(iii) \(4^2 + 8^2 = 80 > 72 = 6^2 + 6^2\), so the first one
Common pitfalls
Frequently asked questions
How do I know whether to add or subtract?
Add if the side you want is the hypotenuse; subtract if it is one of the two shorter sides.
Which side is the hypotenuse if the angle is named?
In \(\angle ABC = 90^\circ\) the right angle is at \(B\), so the hypotenuse is \(AC\).
How do you write a surd in simplest form?
Take out the largest square factor: \(\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}\).
How do you find the height of an isosceles triangle?
Halve the base, then subtract: height squared equals the equal side squared minus half the base squared.
Why can x not be negative?
Both \(14^2\) and \((-14)^2\) are \(196\), but a side length must be positive.