Question

In the adjoining figure,find the value of :-

i) ED
ii) AE​

Answer 1

here is your answer

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Answer 2

i) ED
In ∆ECD,
as it is right angled triangle,
${ ed}^{2} = \: {ec}^{2} + {cd}^{2} \\ \: {ed}^{2} = \: {4}^{2} + {3}^{2} \\ \: ed = \sqrt{16 + 9} \\ \: ed = \sqrt{25} \\ ed = 5 \\ \\ \\ \\ \:$
ii)AE
In ∆ ABE
${ae}^{2} = \: {be}^{2} + {ab}^{2} \\ \: {ae}^{2} = {5}^{2} + {9}^{2} \\ \: ae = \sqrt{25 + 81} \\ ae = \sqrt{106}$