# ML Aggarwal Solutions Class 9 Mathematics Solutions for Pythagoras Theorem Exercise 12 in Chapter 12 - Pythagoras Theorem

Question 6 Pythagoras Theorem Exercise 12

(a) In figure (i) given below, AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm. Calculate the length

of BD.

(b) In figure (ii) given below, ∠PSR = 90°, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of

PR.

(c) In figure (iii) given below, ∠ D = 90°, AB = 16 cm, BC = 12 cm and CA = 6 cm. Find CD.

(a)Given AB = 12 cm, AC = 13 cm, CE = 10 cm and DE = 6 cm

\begin{aligned} &\triangle \mathrm{ABC} \text { is a right triangle. }\\ &\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \quad[\text { Pythagoras theorem }]\\ &\therefore 13^{2}=12^{2}+\mathrm{BC}^{2}\\ &\therefore \mathrm{BC}^{2}=13^{2}-12^{2}\\ &\therefore \mathrm{BC}^{2}=169-144 \end{aligned}

Taking square root on both sides,

BC = 5 cm

\begin{aligned} &\triangle \mathrm{CDE} \text { is a right triangle. }\\ &\therefore \mathrm{CE}^{2}=\mathrm{CD}^{2}+\mathrm{DE}^{2} \quad[\text { Pythagoras theorem] }\\ &\therefore 10^{2}=\mathrm{CD}^{2}+6^{2}\\ &\therefore 100=\mathrm{CD}^{2}+36\\ &\therefore C D^{2}=100-36\\ &\therefore C D^{2}=64 \end{aligned}

Taking square root on both sides,

CD = 8 cm

\begin{aligned} &\therefore B D=B C+C D\\ &\therefore B D=5+8\\ &\therefore \mathrm{BD}=13 \mathrm{cm}\\ &\text { Hence the length of BD is } 13 \mathrm{cm} \text { . } \end{aligned}

(b) Given

\angle \mathrm{PSR}=90^{\circ}, \mathrm{PQ}=10 \mathrm{cm}, \mathrm{QS}=6 \mathrm{cm} \text { and } \mathrm{RQ}=9 \mathrm{cm}

\begin{aligned} &\triangle \mathrm{PSQ} \text { is a right triangle. }\\ &\therefore \mathrm{PQ}^{2}=\mathrm{PS}^{2}+\mathrm{QS}^{2} \quad[\text { Pythagoras theorem] }\\ &10^{2}=\mathrm{PS}^{2}+6^{2}\\ &100=\mathrm{PS}^{2}+36\\ &\therefore P S^{2}=100-36\\ &\therefore P S^{2}=64 \end{aligned}

Taking square root on both sides,

PS = 8 cm

\begin{aligned} &\triangle \text { PSR is a right triangle. }\\ &\mathrm{RS}=\mathrm{RQ}+\mathrm{QS}\\ &R S=9+6\\ &\mathrm{RS}=15 \mathrm{cm}\\ &\therefore P R^{2}=P S^{2}+R S^{2} \quad[\text { Pythagoras theorem }]\\ &\mathrm{PR}^{2}=8^{2}+15^{2}\\ &\mathrm{PR}^{2}=64+225\\ &\mathrm{PR}^{2}=289 \end{aligned}

Taking square root on both sides,

PR = 17 cm

Hence the length of PR is 17 cm.

\begin{aligned} &\text { (c) } \angle \mathrm{D}=90^{\circ}, \mathrm{AB}=16 \mathrm{cm}, \mathrm{BC}=12 \mathrm{cm} \text { and } \mathrm{CA}=6 \mathrm{cm}\\ &\triangle \mathrm{ADC} \text { is a right triangle. }\\ &\therefore \mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{CD}^{2} \quad[\text { Pythagoras theorem }] \end{aligned}

\begin{aligned} &256=\mathrm{AD}^{2}+144+24 \mathrm{CD}+\mathrm{CD}^{2}\\ &256-144=A D^{2}+C D^{2}+24 C D\\ &\mathrm{AD}^{2}+\mathrm{CD}^{2}=112-24 \mathrm{CD}\\ &6^{2}=112-24 \mathrm{CD} \end{aligned}

36 = 112-24CD

24CD = 112-36

24CD = 76

\begin{aligned} &\therefore \mathrm{CD}=7624=\\ &\therefore C D=3 \frac{1}{6}\\ &\text { Hence the length of } \mathrm{CD} \text { is } 3 \frac{1}{6} \mathrm{cm} \end{aligned}

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