Question

From the given figure, in ∆ABC, if AD ⊥ BC, ∠C = 45°, AC = `8sqrt(2)` , BD = 5, then for finding value of AD and BC, complete the following activity.

Activity: In ∆ADC, if ∠ADC = 90°, ∠C = 45°    ......[Given]

∴ ∠DAC = `square`   .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ `square = 1/sqrt(2)` AC and `square = 1/sqrt(2)` AC

∴ AD =`1/sqrt(2) xx square` and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD +DC

= 5 + 8

= 13

Answer 1

In ∆ADC, if ∠ADC = 90°, ∠C = 45°    ......[Given]

∴ ∠DAC = 45°    .....[Remaining angle of ∆ADC]

By theorem of 45° – 45° – 90° triangle,

∴ AD = `1/sqrt(2)` AC and DC = `1/sqrt(2)` AC

∴ AD =`1/sqrt(2) xx 8sqrt(2)` and DC = `1/sqrt(2) xx 8sqrt(2)`

∴ AD = 8 and DC = 8

∴ BC = BD +DC

= 5 + 8

= 13