Question

In the given figure, find the length of AC
AB=10cm,BC=17cm,BD=8cm
Step by Step

Answer 1

consider triangle ABD

given :- BD=8cm

BA=10cm

using Pythagoras thereom

AD^2=AB^2-BD^2

AD^2= 10^2-8^2

AD^2=100-64

AD^2=36

AD=√36

AD=6cm

consider triangle BDC

given :- BC =17cm

BD= 8cm

using Pythagoras thereom

DC^2=BC^2-BD^2

DC^2=17^2-8^2

DC^2=289 - 64

DC^2= 225

DC=√225

DC=15cm

therefore AC = AD+DC

=6+15

= 21

Answer 2

AnsweR :

$\purple { \boxed{ \sf AC = 21cm}}$

Step by Step Explanation :

$\sf \:by \: using \: pythagoras \: theorem.$

$\green{\boxed{ \sf \pink{ {a}^{2} + {b}^{2} = {c}^{2}} }}$

Find AD :

$\sf \: {BD}^{2} + {AD}^{2} = {AB}^{2} \\ \\ \sf \: AB = 10cm \\ \sf \: BD = 8cm \\ \\ \sf {(8)}^{2} + {AD}^{2} = {(10)}^{2} \\ \sf = 64 + {AD}^{2} = 100 \\ \sf = {AD}^{2} = 100 - 64 \\ \sf {AD}^{2} = 36 \\ \sf \: AD = \sqrt{36} \\ \sf \: AD = 6cm$

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Find DC :

$\sf \: {BD}^{2} + {DC}^{2} = {BC}^{2} \\ \\ \sf \: BD = 8cm \\ \sf \: BC = 17cm \\ \\ \sf \: {(8)}^{2} + {DC}^{2} = {(17)}^{2} \\ \sf = 64 +{DC}^{2} = 289 \\ \sf {DC}^{2} = 289 - 64 \\ \sf {DC}^{2} = 225 \\ \sf \: DC = \sqrt{225} \\ \sf \: DC = 15cm$

________________________

Find AC :

$\sf \: AD + DC = AC \\ \sf6cm + 15cm = AC \\ \sf \: AC = 21cm$