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Pythagoras Theorem | Pythagoras Theorem Exercise 12

Question 7

(a) In figure (i) given below, BC = 5 cm,

∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.

(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove

that \mathbf{A B}^{2}=4 \mathbf{A} \mathbf{D}^{2}-3 \mathbf{A} \mathbf{C}^{2}

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(a)Given BC = 5 cm,

\begin{aligned} &\angle B=90^{\circ}, A B=5 \mathrm{AE}\\ &\mathrm{CD}=2 \mathrm{AE} \text { and } \mathrm{AC}=\mathrm{ED}\\ &\triangle \mathrm{ABC} \text { is a right triangle. } \end{aligned}

\begin{aligned} &\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \quad \ldots \text { (i) } \quad[\text { Pythagoras theorem }]\\ &\triangle \mathrm{BED} \text { is a right triangle. }\\ &\therefore \mathrm{ED}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \quad \text { [Pythagoras theorem] }\\ &\therefore \mathrm{AC}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \quad \ldots \text { (ii) } \quad[\because \mathrm{AC}=\mathrm{ED}] \end{aligned}

Comparing (i) and (ii)

\begin{array}{l} \mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \\ (5 \mathrm{AE})^{2}+5^{2}=(4 \mathrm{AE})^{2}+(\mathrm{BC}+\mathrm{CD})^{2} \quad[\because \mathrm{BE}=\mathrm{AB}-\mathrm{AE}=5 \mathrm{AE}-\mathrm{AE}=4 \mathrm{AE}] \\ (5 \mathrm{AE})^{2}+25=(4 \mathrm{AE})^{2}+(5+2 \mathrm{AE})^{2} \ldots(\text { iii }) \\ \text { Let } \mathrm{AE}=\mathrm{x} . \text { So (iii) becomes, } \\ (5 \mathrm{x})^{2}+25=(4 \mathrm{x})^{2}+(5+2 \mathrm{x})^{2} \\ 25 \mathrm{x}^{2}+25=16 \mathrm{x}^{2}+25+20 \mathrm{x}+4 \mathrm{x}^{2} \\ 25 \mathrm{x}^{2}=20 \mathrm{x}^{2}+20 \mathrm{x} \\ 5 \mathrm{x}^{2}=20 \mathrm{x} \end{array}

\begin{aligned} &\therefore x=20 / 5=4\\ &\therefore A E=4 \mathrm{cm}\\ &\therefore \mathrm{CD}=2 \mathrm{AE}=2 \times 4=8 \mathrm{cm}\\ &\therefore \mathrm{AB}=5 \mathrm{AE}\\ &\therefore A B=5 \times 4=20 \mathrm{cm}\\ &\triangle \mathrm{ABC} \text { is a right triangle. }\\ &\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \quad[\text { Pythagoras theorem }]\\ &\therefore \mathrm{AC}^{2}=20^{2}+5^{2}\\ &\therefore \mathrm{AC}^{2}=400+25\\ &\therefore \mathrm{AC}^{2}=425 \end{aligned}

Taking square root on both sides,

AC = √425 = √(25×17)

AC = 5√17 cm

Hence EA = 4 cm, CD = 8 cm, AB = 20 cm and AC = 5√17 cm.

(b)Given D is the midpoint of BC.

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Question 7

(a) In figure (i) given below, BC = 5 cm,

∠B =90°, AB = 5AE, CD = 2AE and AC = ED. Calculate the lengths of EA, CD, AB and AC.

(b) In the figure (ii) given below, ABC is a right triangle right angled at C. If D is mid-point of BC, prove

that \mathbf{A B}^{2}=4 \mathbf{A} \mathbf{D}^{2}-3 \mathbf{A} \mathbf{C}^{2}

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(a)Given BC = 5 cm,

\begin{aligned} &\angle B=90^{\circ}, A B=5 \mathrm{AE}\\ &\mathrm{CD}=2 \mathrm{AE} \text { and } \mathrm{AC}=\mathrm{ED}\\ &\triangle \mathrm{ABC} \text { is a right triangle. } \end{aligned}

\begin{aligned} &\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \quad \ldots \text { (i) } \quad[\text { Pythagoras theorem }]\\ &\triangle \mathrm{BED} \text { is a right triangle. }\\ &\therefore \mathrm{ED}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \quad \text { [Pythagoras theorem] }\\ &\therefore \mathrm{AC}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \quad \ldots \text { (ii) } \quad[\because \mathrm{AC}=\mathrm{ED}] \end{aligned}

Comparing (i) and (ii)

\begin{array}{l} \mathrm{AB}^{2}+\mathrm{BC}^{2}=\mathrm{BE}^{2}+\mathrm{BD}^{2} \\ (5 \mathrm{AE})^{2}+5^{2}=(4 \mathrm{AE})^{2}+(\mathrm{BC}+\mathrm{CD})^{2} \quad[\because \mathrm{BE}=\mathrm{AB}-\mathrm{AE}=5 \mathrm{AE}-\mathrm{AE}=4 \mathrm{AE}] \\ (5 \mathrm{AE})^{2}+25=(4 \mathrm{AE})^{2}+(5+2 \mathrm{AE})^{2} \ldots(\text { iii }) \\ \text { Let } \mathrm{AE}=\mathrm{x} . \text { So (iii) becomes, } \\ (5 \mathrm{x})^{2}+25=(4 \mathrm{x})^{2}+(5+2 \mathrm{x})^{2} \\ 25 \mathrm{x}^{2}+25=16 \mathrm{x}^{2}+25+20 \mathrm{x}+4 \mathrm{x}^{2} \\ 25 \mathrm{x}^{2}=20 \mathrm{x}^{2}+20 \mathrm{x} \\ 5 \mathrm{x}^{2}=20 \mathrm{x} \end{array}

\begin{aligned} &\therefore x=20 / 5=4\\ &\therefore A E=4 \mathrm{cm}\\ &\therefore \mathrm{CD}=2 \mathrm{AE}=2 \times 4=8 \mathrm{cm}\\ &\therefore \mathrm{AB}=5 \mathrm{AE}\\ &\therefore A B=5 \times 4=20 \mathrm{cm}\\ &\triangle \mathrm{ABC} \text { is a right triangle. }\\ &\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2} \quad[\text { Pythagoras theorem }]\\ &\therefore \mathrm{AC}^{2}=20^{2}+5^{2}\\ &\therefore \mathrm{AC}^{2}=400+25\\ &\therefore \mathrm{AC}^{2}=425 \end{aligned}

Taking square root on both sides,

AC = √425 = √(25×17)

AC = 5√17 cm

Hence EA = 4 cm, CD = 8 cm, AB = 20 cm and AC = 5√17 cm.

(b)Given D is the midpoint of BC.

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