If the interior angles of a pentagon are (3x + 4)°, (x + 20)°, (3x - 5)°, (2x + 10)° and (4x - 9)°, then find the value of x.

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Solution

Sum of the interior angles of a pentagon = $540^{\circ}$ (1 mark)

$\Rightarrow (3 x + 4) ° + (x + 20) ° + (3 x - 5) ° + (2 x + 10) ° + (4 x - 9) ° = 540^{\circ}$

$\Rightarrow 13 x + 20^{\circ} = 540^{\circ}$

$\Rightarrow 13 x = 540^{\circ} - 20^{\circ}$

$\Rightarrow x = \frac{520}{13}$

$\Rightarrow x = 40^{\circ}$ (1 mark)

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The angles of a hexagon are $x + 10^{o}, 2 x + 20^{o}, 2 x - 20^{o}, 3 x - 50^{o}, x + 40^{o}, a n d x + 20^{o}$ . Find x.

If the interior angles of a pentagon are (3x + 4)°, (x + 20)°, (3x - 5)°, (2x + 10)° and (4x - 9)°, then x is equal to

(3 x + 4)^{0}, (x + 20)^{0}, (3 x - 5)^{0}, (2 x + 10)^{0}

(4 x - 9)^{0}

x

3 x + 4^{o}, x + 20^{o}, 3 x - 5^{o}, 2 x + 10^{o}

4 x - 9^{o}

^{'} x^{'}

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