Home Practice
For learners and parents For teachers and schools
Textbooks
Full catalogue
Leaderboards
Learners Leaderboard Classes/Grades Leaderboard Schools Leaderboard
Pricing Support
Help centre Contact us
Log in

We think you are located in United States. Is this correct?

8.6 Pythagorean theorem

8.6 Pythagorean theorem (EMCJH)

temp text

Many different methods of proving the theorem of Pythagoras have been formulated over the years. Similarity of triangles is one method that provides a neat proof of this important theorem.

Chapter 8: Theorem: Pythagorean theorem

The square on the hypotenuse of a right-angled triangle is equal to the sum of the squares on the other two sides.

(Reason: Pythagoras or \(\text{right-angled } \triangle \text{s}\))

f825020344de3325f2b2736b18a3e635.png

\(\triangle ABC\) with \(\hat{A}=\text{90}°\)

\(B{C}^{2}=A{B}^{2}+A{C}^{2}\)

Construction: draw \(AD \perp BC\).

\[\begin{array}{rll} \hat{C} + \hat{A}_{2} & = \text{90}° & \text{(}\angle \text{s}\text{ of }\triangle CAD \text{)} \\ \hat{A}_{1} + \hat{A}_{2} & = \text{90}° & \text{(given)} \\ \therefore \hat{A}_{1} & = \hat{C} & \\ & & \\ \hat{A}_{1} + \hat{B} & = \text{90}° & \text{(}\angle \text{s}\text{ of }\triangle ABD \text{)} \\ \therefore \hat{B} & = \hat{A}_{2} & \\ & & \\ \hat{D}_{1} & = \hat{D}_{2}= \hat{A} = \text{90}° & \text{(construction)} \\ & & \\ \therefore \triangle ABD \enspace & ||| \enspace \triangle CBA \enspace ||| \enspace \triangle CAD &\text{(AAA)} \\ & & \\ \therefore \frac{AB}{BC}&=\frac{BD}{AB} & (\triangle ABD \enspace ||| \enspace \triangle CBA ) \\ & & \\ AB^{2} &= BD \times BC & \\ & & \\ \text{Similarly } \frac{AC}{CB}&=\frac{DC}{AC} & (\triangle ABD \enspace ||| \enspace \triangle CBA ) \\ & & \\ AC^{2} &= CB \times DC &\\ & & \\ \therefore AC^{2} + AB^{2} &= (BD \times BC) + (CB \times DC ) &\\ &= BC(BD + DC ) & \\ &= BC(BC ) & \\ &= BC^{2}& \\ & & \\ \therefore BC^{2} &= AC^{2} + AB^{2} & \\ \end{array}\]

Converse: theorem of Pythagoras

If the square of one side of a triangle is equal to the sum of the squares of the other two sides of the triangle, then the angle included by these two sides is a right angle.

Worked example 8: Theorem of Pythagoras

In \(\triangle PQR\), \(P\hat{Q}R = \text{90}°\) and \(QT \perp PR\). If \(PQ = \text{7}\text{ cm}\) and \(QT = 3\sqrt{5} \enspace \text{cm}\), determine \(PR\) and \(QR\) (correct to the nearest integer).

c25a71b9d31ca6175d627e403bf45b7b.png

Use the theorem of Pythagoras to determine \(PT\)

\[\begin{array}{rll} \text{In } \triangle PTQ, \quad PT^{2} &= PQ^{2} - QT^{2} & (\text{Pythagoras}) \\ &= 7^{2} - \left( 3\sqrt{5} \right)^{2} & \\ &= 49 - 45 & \\ \therefore PT &= \sqrt{4} & \\ &= \text{2}\text{ cm} & \end{array}\]

Use proportionality to determine \(PR\) and \(QR\)

\[\begin{array}{rll} P\hat{Q}R &=\text{90}° & \text{(given)} \\ QT &\perp PR & \text{(given)} \\ \therefore \enspace \triangle PQT \enspace & ||| \enspace \triangle QRT \enspace ||| \enspace \triangle PRQ & (\text{right-angled } \triangle \text{s}) \\ & & \\ \therefore \dfrac{QT}{TR} &= \dfrac{PT}{QT} & (\triangle QRT \enspace ||| \enspace \triangle PQT ) \\ & & \\ \therefore QT^{2} &= TR \cdot PT & \\ & & \\ \left( 3\sqrt{5} \right)^{2} &= TR \cdot 2 & \\ & & \\ \frac{45}{2} &= TR & \\ & & \\ \text{And } PR &= PT + TR & \\ &= \dfrac{45}{2} + 2 & \\ &= \text{25}\text{ cm} & \text{(to nearest integer)}\\ & & \\ \text{In } \triangle PQR, \quad QR^{2} &= PR^{2} - PQ^{2} & (\text{Pythagoras}) \\ &= 25^{2} - 7^{2} & \\ \therefore QR &= \sqrt{576} & \\ &= \text{24}\text{ cm} & \end{array}\]

Write the final answer

\(PR = \text{25}\text{ cm}\) and \(QR = \text{24}\text{ cm}\)

For any right-angled \(\triangle MNP\), if \(MQ\) is drawn perpendicular to \(NP\), then:

8f42a94eec0757f48007636cf412f98b.png

Theorem of Pythagoras

Textbook Exercise 8.9

\(B\) is a point on circle with centre \(O\). \(BD \perp AC\) and \(D\) is the midpoint of radius \(OC\).

If the diameter of the circle is \(\text{24}\text{ cm}\), find \(BD\).

Leave answer in simplified surd form.

0f19acb4e699fb5c73b70849d1cd16c7.png
\[\begin{array}{rll} ABC&= \text{90}° & (\angle \text{ in semi-circle}) \\ BD&\perp AC & \\ \therefore BD^{2}&= AC.DC & \\ AO &= OC & (\text{equal radii}) \\ &= \frac{1}{2} AC & \\ &= \text{12}\text{ cm} & \\ \therefore DC &= \text{6}\text{ cm} & \\ \therefore AD&= \text{18}\text{ cm} & \\ BD^{2}&= AD \cdot DC & (\text{right-angled } \triangle \text{s}) \\ BD^{2}&= 18 \times 6 & \\ BD&= \sqrt{108} & \\ &= \sqrt{36 \times 3} & \\ &= 6 \sqrt{3} \text{ cm} & \end{array}\]

In \(\triangle PQR\), \(RQ \perp QP\) and \(QT \perp RP\). \(PQ = 2\) units, \(QR = b\) units, \(RT = 3\) units and \(TP = a\) units. Determine \(a\) and \(b\), giving reasons.

771ed6f2d52292bce21b16f7a7840500.png
\[\begin{array}{rll} QP^2 &= PT \cdot PR & (\text{right-angled } \triangle \text{s}) \\ 2^{2} &= a(a + 3) & \\ 4 &= a^{2} + 3a & \\ 0 &= a^{2} + 3a -4 & \\ 0 &= (a - 1)(a + 4) & \\ \therefore a &= 1 \text{ or } a = -4 & \\ \text{Length must be positive } & \therefore a = 1 \text{ unit} & \end{array}\] \[\begin{array}{rll} QR^2 &= RT \cdot RP & (\text{right-angled } \triangle \text{s}) \\ b^{2} &= 3(3 + 1) & \\ &= 3(4) & \\ &= 12 & \\ \therefore b &= \pm \sqrt{12} & \\ \text{Length must be positive } & \therefore b = 2\sqrt{3} \text{ units} & \end{array}\]

Chord \(AQ\) of circle with centre \(O\) cuts \(BC\) at right angles at point \(P\).

a4fc911810006bea7aa2daa8db6da475.png

Why is \(\triangle ABP \enspace ||| \enspace \triangle CBA\)?

\[\begin{array}{rll} B\hat{A}C&= \text{90}° & (BC \text{ is diameter of circle }O) \\ \end{array}\]

In \(\triangle ABP\) and \(\triangle CBA\):

\[\begin{array}{rll} B\hat{P}A&= B\hat{A}C =\text{90}° & (\text{given}) \\ \hat{B}&= \hat{B} & (\text{common } \angle) \\ \therefore \triangle ABP& \enspace ||| \enspace \triangle CBA & (\text{AAA}) \end{array}\]

If \(AB = \sqrt{6}\) units and \(PO = 2\) units, calculate the radius of the circle.

In \(\triangle ABP\):

\[\begin{array}{rll} AP^{2} &= BA^{2} - BP^{2} & (\text{Pythagoras}) \\ BP &= BO - PO & \\ &= r - 2 & (BO = r, \text{ given } PO = 2) \\ AP^{2} &= \left( \sqrt{6} \right)^{2} - (r-2)^{2} & \\ &= 6 - (r-2)^{2} & \\ & & \\ AP^{2} &= BP.PC & (B\hat{A}C = \text{90}°, AP \perp BC) \\ &= (r-2).PC & \\ \text{And } PC &= PO + OC & \\ &= 2 + r & \\ \therefore AP^{2} &= (r-2)(2 + r) & \end{array}\] \[\begin{array}{rll} 6 - (r-2)^{2} &= (r-2)(2 + r) & \\ 6 - r^{2} + 4r - 4 &= r^{2} - 4 & \\ 2r^{2} - 4r -6 &= 0 & \\ r^{2} - 2r - 3 &= 0 & \\ (r-3)(r+1) &= 0 & \\ r = 3 \text{ or } r &= -1 & \\ \therefore r &= \text{3}\text{ units} & \\ \end{array}\]

In the diagram below, \(XZ\) and \(WZ\) are tangents to the circle with centre \(O\) and \(X\hat{Y}Z = \text{90}°\).

e428b5acfa1294833a3a5ef498bbd04a.png

Show that \(XY^{2} = OY \cdot YZ\).

\[\begin{array}{rll} \text{In } \triangle OXZ: & & \\ X\hat{Y}Z & = \text{90}° & (\text{given}) \\ O\hat{X}Z & = \text{90}° & (\text{tangent } \perp \text{ radius}) \\ \therefore XY^{2} & = OY \cdot YZ & (\text{right-angled } \triangle \text{s}) \end{array}\]

Prove that \(\frac{OY}{YZ} = \frac{OW^{2}}{WZ^{2}}\).

\[\begin{array}{rll} \text{In } \triangle OWZ: & & \\ W\hat{Y}Z & = \text{90}° & (\text{given}) \\ O\hat{W}Z & = \text{90}° & (\text{tangent } \perp \text{ radius}) \\ \therefore WZ^{2} & = ZY \cdot ZO & (\text{right-angled } \triangle \text{s}) \\ \text{And } WO^{2} & = OY \cdot ZO & (\text{right-angled } \triangle \text{s}) \\ \therefore \frac{WO^{2}}{WZ^{2}} & = \frac{OY \cdot ZO}{ZY \cdot ZO} & \\ \frac{WO^{2}}{WZ^{2}} & = \frac{OY}{ZY} & \\ \therefore \frac{OY}{YZ} &= \frac{OW^{2}}{WZ^{2}} & \end{array}\]