6.02d, 10.05a - Pythagoras' theorem

6.02d, 10.05a - Pythagoras' theorem

A rectangular garden is 6 metres by 8 metres. How long is the straight path between opposite corners? Use squared areas to explain Pythagoras' theorem, then find missing lengths in right-angled triangles and simple two-dimensional shapes.

An area proof of Pythagoras

A diagonal takes a shortcut across the garden, but adding the two edge lengths measures a different route. The connection we need is between the squares of the lengths.

In a right-angled triangle, the two sides meeting at the right angle are perpendicular. The side opposite that angle is the hypotenuse, the longest side. Call the perpendicular side lengths aa and bb, and the hypotenuse cc. All three lengths are positive and measured in the same units.

[DIAGRAM: asset_name: Four triangles prove Pythagoras; asset_slug: pythagoras-area-proof; caption: Four identical right-angled triangles leave a square of side c.; description: An outer square has side a plus b. Four congruent right-angled triangles of perpendicular sides a and b are placed against its four corners, leaving a central tilted square with all four sides c. The two acute angles of a triangle are alpha and beta, adding to 90 degrees.]
Four identical right-angled triangles leave a square of side c.

The outer square has area (a+b)2(a+b)^2. The four triangles together have area 4×12ab=2ab4\times\frac12ab=2ab. The remaining shape has four sides of length cc. Each of its angles is 180∘−(α+β)=90∘180^\circ-(\alpha+\beta)=90^\circ, since the two acute angles of a right-angled triangle add to 90∘90^\circ. So the centre really is a square, with area c2c^2.

Calculate the same remaining area in two ways:

c2=(a+b)2−2abouter area minus triangles=a2+2ab+b2−2abexpand the square=a2+b2cancel equal terms\begin{aligned} c^2&=(a+b)^2-2ab &&\text{outer area minus triangles}\\ &=a^2+2ab+b^2-2ab &&\text{expand the square}\\ &=a^2+b^2 &&\text{cancel equal terms} \end{aligned}

Pythagoras' theorem

For a right-angled triangle with hypotenuse cc,

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

The squares on the two perpendicular sides have the same total area as the square on the hypotenuse. The June and November 2027 OCR formula sheets supply this formula for both tiers; you still need to identify the right angle and the hypotenuse.

Try actual areas before using the rule to calculate lengths.

Worked example 1: the remaining square

Question. In the arrangement above, let a=3 cma=3\,\mathrm{cm} and b=4 cmb=4\,\mathrm{cm}. Find the area of the central square.

Plan. Subtract the four triangle areas from the outer square area.

Outer area=(3+4)2=49Four triangles=4×12×3×4=24c2=49−24=25\begin{aligned} \text{Outer area}&=(3+4)^2=49\\ \text{Four triangles}&=4\times\tfrac12\times3\times4=24\\ c^2&=49-24=25 \end{aligned}

Answer. 25 cm225\,\mathrm{cm}^2.

Check. The two smaller squares have total area 32+42=9+16=25 cm23^2+4^2=9+16=25\,\mathrm{cm}^2, agreeing with the subtraction.

Change only one perpendicular side. The central area need not be a square number.

Worked example 2: a different triangle

Question. Repeat the area calculation when a=3 cma=3\,\mathrm{cm} and b=5 cmb=5\,\mathrm{cm}.

Plan. Use the same arrangement with the new side length.

Outer area=(3+5)2=64Four triangles=4×12×3×5=30c2=64−30=34\begin{aligned} \text{Outer area}&=(3+5)^2=64\\ \text{Four triangles}&=4\times\tfrac12\times3\times5=30\\ c^2&=64-30=34 \end{aligned}

Answer. 34 cm234\,\mathrm{cm}^2.

Check. 32+52=9+25=343^2+5^2=9+25=34. The theorem concerns areas even when cc is not a whole number.

The right angle is essential: the area arrangement used it to make the central square. Do not apply this equation to a triangle merely because its sketch looks nearly right-angled.

Finding the hypotenuse

When both perpendicular sides are known, add their squares and take the positive square root to find the hypotenuse.

The garden's edges meet at a right angle. Its diagonal is opposite that angle, so it is the hypotenuse. The label cc is a convention, not a position on the page: a rotated triangle still has the same hypotenuse.

[DIAGRAM: asset_name: A garden diagonal is a hypotenuse; asset_slug: garden-diagonal; caption: The diagonal is opposite the right angle in triangle ABC.; description: A rectangle ABCD has AB equal to 8 metres and BC equal to 6 metres. Its diagonal AC splits it into two right-angled triangles. In triangle ABC the right angle is at B, so AC is the hypotenuse.]
The diagonal is opposite the right angle in triangle ABC.

Worked example 3: the garden shortcut

Question. A rectangular garden has sides 6 m and 8 m. Find its diagonal length.

Plan. Use the two perpendicular edges as the shorter sides.

c2=62+82add squared lengths=36+64=100c=100=10positive length\begin{aligned} c^2&=6^2+8^2 &&\text{add squared lengths}\\ &=36+64=100\\ c&=\sqrt{100}=10 &&\text{positive length} \end{aligned}

Answer. 10 m10\,\mathrm m.

Check. 102−82=100−64=36=6210^2-8^2=100-64=36=6^2. Also, 10 m is longer than either edge and shorter than the 14 m route round two edges.

Change the 8 m edge to 7 m. The method stays the same, but a calculator is useful for the final root.

Worked example 4: a non-integer diagonal

Question. A rectangle has sides 6 m and 7 m. Find its diagonal to 3 significant figures.

Plan. Keep the exact squared length until the final calculation.

c2=62+72=36+49=85c=85=9.219544…\begin{aligned} c^2&=6^2+7^2=36+49=85\\ c&=\sqrt{85}=9.219544\ldots \end{aligned}

Answer. 9.22 m9.22\,\mathrm m to 3 significant figures.

Check. 92<85<1029^2<85<10^2, so the diagonal lies between 9 m and 10 m. Its rounded value is in that interval.

The mistake 62+72=6+7\sqrt{6^2+7^2}=6+7 would make the diagonal equal the longer edge route. A square root does not distribute over addition.

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