# 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 = □ .....[R - Geometry Mathematics 2

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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)

#### 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)

Is there an error in this question or solution?
Chapter 2: Pythagoras Theorem - Q.2 (A)

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