In the figure, if ∠A+∠B+∠C+∠D+∠E+∠F=k× right angle, then k is

It is given that in the given figure,

∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = k \times

right angle.

Angle sum property of a triangle states that the sum of interior angles of a triangle is 180°.
Therefore, in

△ ABC

, we have,

∠ A + ∠ B + ∠ C = 180^{\circ} (1)

Similarly in

△ DEF,

we have,

∠ D + ∠ E + ∠ F = 180^{\circ} (2)

By adding equation (1) and (2), we obtain,

∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = 180^{\circ} + 180^{\circ}

\Rightarrow

∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = 360^{\circ} (3)

We have been given,

∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = k \times

right angle (4)
Comparing equations (3) and (4),

\Rightarrow 360^{\circ} = k \times

right angle

\Rightarrow

k = \frac{360^{o}}{90^{o}}

\Rightarrow

k = 4

Therefore, the value of k is 4.
Hence, option (3) is correct.

In the figure, if $∠ A + ∠ B + ∠ C + ∠ D + ∠ E + ∠ F = k \times$ right angle, then $k$ is