Question

⏸️Solve property .........Class 10 (triangle)✔️✔️✔️

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Answer 1

SOLUTION

Here AC= 9cm , AD= 3cm

So,

CD= AC-AD

=) 9cm-3cm= 6cm

Now we apply Pythagoras theorem in ∆ABC & get

${AB}^{2} + {BC}^{2} = {AC}^{2} \\ = > {AB}^{2} + {BC}^{2} = {9}^{2} \\ = > {AB}^{2} + {BC}^{2} = 81.............(1)$

Now, we apply Pythagoras theorem in ∆ABD & get;

${BD}^{2} + {AD}^{2} = {AB}^{2} \\ = > {BD}^{2} + {3}^{2} = {AB}^{2} \\ = > {BD}^{2} = {AB}^{2} - 9..............(2)$

Now, we apply Pythagoras theorem in ∆CBD & gets:

${BD}^{2} + {CD}^{2} = {BC}^{2} \\ = > {BD}^{2} + {6}^{2} = {BC}^{2} \\ = > {BD}^{2} = {BC}^{2} - 36............(3)$

Add equation (2) & (3), we get:

$2 {BD}^{2} = {AB}^{2} + {BC}^{2} - 45 \: \: [substitute \: value \: from \: equation \: (1) ]\\ = > 2 {BD}^{2} = 81 - 45 \\ = > 2 {BD}^{2} = 36 \\ = > {BD}^{2} = 18 \\ = > BD = \sqrt{18} \\ = > BD= 3 \sqrt{2} cm$

Option (ii)✓

hope it helps ☺️