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Question-83369




Question Number 83369 by Power last updated on 01/Mar/20
Commented by mr W last updated on 01/Mar/20
∠A=90°  ⇒AD=BD=DC=5
$$\angle{A}=\mathrm{90}° \\ $$$$\Rightarrow{AD}={BD}={DC}=\mathrm{5} \\ $$
Commented by mathmax by abdo last updated on 01/Mar/20
we use the formulae AB^2  +AC^2 =2AD^2  +((BC^2 )/2) ⇒  8^2  +6^2 =2AD^2  +((100^2 )/2) ⇒2AD^2 =100−50 =50 ⇒AD^2 =25 ⇒  AD=5
$${we}\:{use}\:{the}\:{formulae}\:{AB}^{\mathrm{2}} \:+{AC}^{\mathrm{2}} =\mathrm{2}{AD}^{\mathrm{2}} \:+\frac{{BC}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow \\ $$$$\mathrm{8}^{\mathrm{2}} \:+\mathrm{6}^{\mathrm{2}} =\mathrm{2}{AD}^{\mathrm{2}} \:+\frac{\mathrm{100}^{\mathrm{2}} }{\mathrm{2}}\:\Rightarrow\mathrm{2}{AD}^{\mathrm{2}} =\mathrm{100}−\mathrm{50}\:=\mathrm{50}\:\Rightarrow{AD}^{\mathrm{2}} =\mathrm{25}\:\Rightarrow \\ $$$${AD}=\mathrm{5} \\ $$
Answered by mind is power last updated on 01/Mar/20
cos(B)=((10^2 +8^2 −6^2 )/(2.10.8)) =((128)/(160))=(4/5)  AD^2 =8^2 +5^2 −2.8.5cos(B)=64+25−64=25  AD=5
$${cos}\left({B}\right)=\frac{\mathrm{10}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} −\mathrm{6}^{\mathrm{2}} }{\mathrm{2}.\mathrm{10}.\mathrm{8}}\:=\frac{\mathrm{128}}{\mathrm{160}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${AD}^{\mathrm{2}} =\mathrm{8}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{2}.\mathrm{8}.\mathrm{5}{cos}\left({B}\right)=\mathrm{64}+\mathrm{25}−\mathrm{64}=\mathrm{25} \\ $$$${AD}=\mathrm{5} \\ $$

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