4.8 - Pythagoras in two dimensions

4.8 - Pythagoras in two dimensions

You will use a right angle to connect three side lengths, find a missing length and solve problems in a flat diagram. The same method helps you calculate a diagonal or a sloping support. You will also decide whether a final answer needs ordinary rounding or rounding up.

The hypotenuse and Pythagoras

Two supports each span 2.4 m2.4\,\mathrm{m} horizontally and 1.8 m1.8\,\mathrm{m} vertically. How much timber must you buy for both, with 10%10\% spare, if timber is sold in whole metres? The sloping lengths are hidden for now. A right triangle will reveal them.

You already know

Squaring multiplies a number by itself; a square root reverses that operation for a non-negative number. Use the same length unit before adding or subtracting squared lengths.

Start with a smaller question. In a right triangle, perpendicular sides of 9 cm9\,\mathrm{cm} and 12 cm12\,\mathrm{cm} have squared lengths 81 cm281\,\mathrm{cm}^2 and 144 cm2144\,\mathrm{cm}^2. Their sum is 225 cm2225\,\mathrm{cm}^2, the square of 15 cm15\,\mathrm{cm}. Pythagoras explains why this finds the third side.

Hypotenuse

The hypotenuse is the side opposite the right angle in a right-angled triangle. It is the longest side. Rotating the triangle does not change which side is the hypotenuse.

A marked right angle, a rectangle's corner, or stated horizontal and vertical directions can establish the right angle. A triangle that merely looks right-angled is not enough.

[DIAGRAM: asset_slug: m069_pythagoras; caption: A right-angled triangle with legs a and b and hypotenuse c; description: The marked right angle lies between a and b and the opposite side c is the hypotenuse. Labels are symbolic and the figure is schematic.; file: m069_pythagoras.png; recommended_method: matplotlib]
A right-angled triangle with legs a and b and hypotenuse c

The squares built on the two shorter sides together have the same area as the square built on the hypotenuse. This is a relationship between squared lengths, not a rule for adding side lengths.

Pythagoras' theorem

For a right-angled triangle with perpendicular sides a,ba,b and hypotenuse cc,

a2+b2=c2a^2+b^2=c^2

Absent from both the Foundation and Higher sheets. Not in the formulae sheet: learn it.

Use the equation first, then calculate its positive square root because a side length is positive.

Worked example 1: the longest side

Question. Find the hypotenuse of a right triangle with perpendicular sides 9 cm9\,\mathrm{cm} and 12 cm12\,\mathrm{cm}.

Plan. Add the squared perpendicular sides, then take the positive square root.

c2=92+122Pythagoras=81+144=225square, then addc=225=15positive length\begin{aligned} c^2&=9^2+12^2 &&\text{Pythagoras}\\ &=81+144=225 &&\text{square, then add}\\ c&=\sqrt{225}=15 &&\text{positive length} \end{aligned}

Why add? Both known sides meet at the right angle; the unknown is opposite it.

Answer. 15 cm15\,\mathrm{cm}.

Check. 15>12>915>12>9, and 152=92+12215^2=9^2+12^2.

Change only the scale: double both given lengths. The right angle and side roles stay the same.

Worked example 2: double the triangle

Question. Find the hypotenuse when the perpendicular sides are 18 cm18\,\mathrm{cm} and 24 cm24\,\mathrm{cm}.

Plan. Apply the same equation to the enlarged triangle.

c2=182+242Pythagoras=324+576=900square, then addc=900=30positive square root\begin{aligned} c^2&=18^2+24^2 &&\text{Pythagoras}\\ &=324+576=900 &&\text{square, then add}\\ c&=\sqrt{900}=30 &&\text{positive square root} \end{aligned}

Why is the answer doubled rather than quadrupled? The square is quadrupled, but taking its square root doubles the length.

Answer. 30 cm30\,\mathrm{cm}.

Check. 30=2×1530=2\times15, matching the scale factor.

A useful surprising fact follows: one right triangle creates infinitely many more by enlargement. For any k>0k>0, (3k)2+(4k)2=9k2+16k2=(5k)2(3k)^2+(4k)^2=9k^2+16k^2=(5k)^2. The familiar 3,4,53,4,5 relationship survives every positive scale factor.

Watch out

The line c=9+12=21c=9+12=21 adds lengths, so it does not use Pythagoras. Write c2=92+122c^2=9^2+12^2, then take the square root to obtain c=15c=15.

Try a different pair of perpendicular sides.

Finding a shorter side

When the hypotenuse is supplied, keep its square alone on one side of the equation. Rearranging then tells you to subtract the known shorter-side square.

If cc is the hypotenuse and aa is the known shorter side, b2=c2−a2b^2=c^2-a^2. The missing length is b=c2−a2b=\sqrt{c^2-a^2}. This rearrangement is also absent from both tier sheets. A negative value inside the root indicates impossible lengths or a mistaken hypotenuse; it does not mean a negative side length.

Worked example 3: subtract a square

Question. A right triangle has hypotenuse 13 m13\,\mathrm{m} and one shorter side 5 m5\,\mathrm{m}. Find the other shorter side.

Plan. Write Pythagoras with 13213^2 alone, then rearrange.

x2+52=132identify the hypotenusex2=169−25=144subtract 52x=144=12positive square root\begin{aligned} x^2+5^2&=13^2 &&\text{identify the hypotenuse}\\ x^2&=169-25=144 &&\text{subtract }5^2\\ x&=\sqrt{144}=12 &&\text{positive square root} \end{aligned}

Why subtract? The unknown square and 525^2 must add to the hypotenuse square.

Answer. 12 m12\,\mathrm{m}.

Check. 52+122=169=1325^2+12^2=169=13^2, and 12<1312<13.

Change only the scale again: double both known lengths. This checks whether the side roles survive a change in size.

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