Question

As shown in figure, LK = ${\displaystyle 6\sqrt{2}}$ then

1. MK = ?
2. ML = ?
3. 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 = ${\displaystyle \frac{\sqrt{3}}{2}}$ MK    ......[Side opposite to 60°]

∴ ${\displaystyle 6\sqrt{2}=\frac{\sqrt{3}}{2}}$ MK   ......[Given]

∴ MK = ${\displaystyle 6\sqrt{2}\times \frac{2}{\sqrt{3}}}$

= ${\displaystyle \frac{12\sqrt{2}}{\sqrt{3}}}$

= ${\displaystyle \frac{12\times \sqrt{2}\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}}$   ......[Multiply numerator and denominator by ${\displaystyle \sqrt{3}}$]

= ${\displaystyle \frac{12\times \sqrt{2\times 3}}{3}}$

MK = ${\displaystyle 4\sqrt{6}}$ units

 

(ii) In ∆MLK,

∠MLK = 90°    ......[Given]

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

∴ ${\displaystyle {\left(4\sqrt{6}\right)}^2={\text{ML}}^2+{\left(6\sqrt{2}\right)}^2}$   ......[From (i) and given]

∴ (16 × 6) = ML² + (36 × 2)

∴ 96 = ML² + 72

∴ ML² = 24

∴ ML = ${\displaystyle \sqrt{24}}$     .......[Taking square root of both sides]

∴ ML = ${\displaystyle \sqrt{4\times 6}}$

∴ ML = ${\displaystyle 2\sqrt{6}}$ units

 

(iii) In ∆NKM,

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

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

∴ ∆NKM is 45° – 45° – 90° triangle.

∴ MK = ${\displaystyle \frac{1}{\sqrt{2}}}$ MN    ......[Theorem of 45° – 45° – 90° triangle]

∴ ${\displaystyle 4\sqrt{6}=\frac{1}{\sqrt{2}}}$ MN     ......[From (i)]

∴ MN = ${\displaystyle 4\sqrt{6}\times \sqrt{2}}$

= ${\displaystyle 4\sqrt{6\times 2}}$

= ${\displaystyle 4\sqrt{4\times 3}}$

= ${\displaystyle 4\times 2\times \sqrt{3}}$

∴ MN = ${\displaystyle 8\sqrt{3}}$ units