Question

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

(i) MK = ?

(ii) ML = ?

(iii) MN = ?

Answer 1

(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