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From given figure, in ∆MNK, if ∠MNK = 90°, ∠M = 45°, MK = 6, then for finding value of MN and KN, complete the following activity.

Activity:

In ∆MNK, ∠MNK = 90°, ∠M = 45° …...[Given]

∴ ∠K = `square` .....[Remaining angle of ∆MNK]

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

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

∴ MN = `1/sqrt(2) xx square` and KN = `1/sqrt(2) xx 6`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

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#### Solution

In ∆MNK, ∠MNK = 90°, ∠M = 45° …...[Given]

∴ ∠K = **45°** .....[Remaining angle of ∆MNK]

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

∴ **MN** = `1/sqrt(2)` MK and **KN** = `1/sqrt(2)` MK

∴ MN = `1/sqrt(2) xx 6` and KN = `1/sqrt(2) xx 6`

∴ MN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))` and KN = `(1 xx 6 xx sqrt(2))/(sqrt(2) xx sqrt(2))` .......[Multiply numerator and denominator by `sqrt(2)`]

∴ MN = `(6 xx sqrt(2))/2` and KN = `(6 xx sqrt(2))/2`

∴ MN = `3sqrt(2)` and KN = `3sqrt(2)`

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