# As shown in figure, LK = 62 then (i) MK = ? (ii) ML = ? (iii) MN = ? - Geometry

Sum

As shown in figure, LK = 6sqrt(2) then

(i) MK = ?

(ii) ML = ?

(iii) MN = ?

#### Solution

(i) In ∆MLK,

∠MLK = 90° and ∠LKM = 30°    ......[Given]

∴ ∠LMK = 60°     ......[Remaining angle of a triangle]

∴ ∆MLK is a 30° – 60° – 90° triangle.

∴ LK = sqrt(3)/2 MK    ......[Side opposite to 60°]

∴ 6sqrt(2) = sqrt(3)/2 MK   ......[Given]

∴ MK = 6sqrt(2) xx 2/sqrt(3)

= (12sqrt(2))/sqrt(3)

= (12 xx sqrt(2) xx sqrt(3))/(sqrt(3) xx sqrt(3))   ......[Multiply numerator and denominator by sqrt(3)]

= (12 xx sqrt(2 xx 3))/3

MK = 4sqrt(6) units

(ii) In ∆MLK,

∠MLK = 90°    ......[Given]

∴ MK2 = ML2 + LK2     ......[Pythagoras theorem]

∴ (4sqrt(6))^2 = "ML"^2 + (6sqrt(2))^2   ......[From (i) and given]

∴ (16 × 6) = ML2 + (36 × 2)

∴ 96 = ML2 + 72

∴ ML2 = 24

∴ ML = sqrt(24)     .......[Taking square root of both sides]

∴ ML = sqrt(4 xx 6)

∴ ML = 2sqrt(6) units

(iii) In ∆NKM,

∠NKM = 90° and ∠MNK = 45°    ......[Given]

∴ ∠KMN = 45°      ......[Remaining angle of a triangle]

∴ ∆NKM is 45° – 45° – 90° triangle.

∴ MK = 1/sqrt(2) MN    ......[Theorem of 45° – 45° – 90° triangle]

∴ 4sqrt(6) = 1/sqrt(2) MN     ......[From (i)]

∴ MN = 4sqrt(6) xx sqrt(2)

= 4sqrt(6 xx 2)

= 4sqrt(4 xx 3)

= 4 xx2 xx sqrt(3)

∴ MN = 8sqrt(3) units

Concept: Property of 30°- 60°- 90° Triangle Theorem
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