Question

The sides of a pentagon are produced in order and the exterior angles so obtained are of measure
xº, (2x), (3x + 10), (4x + 5) and (5x)". Find the value of "x" and also find the
measure of each exterior angle of a pentagon​

Answer 1

Given :

• First angle = x°
• Second angle = 2x°
• Third angle = (3x + 10)°
• Fourth angle = (4x + 5)°
• Fifth angle = 5x°

To find :

• Value of x
• Measure of each exterior angle

Solution :

First let us find the value of sum of all the interior angles of a Pentagon .

We know the identity find the formula i.e,

$\boxed{\bf{A = (n - 2) \times 180^{\circ}}}$

Where :

• A = Sum of angles of a Polygon
• n = No. of sides of the polygon

Using the above relation and substituting the values in it, we get :

$:\implies \bf{A = (n - 2) \times 180^{\circ}}$

$:\implies \bf{A = (5 - 2) \times 180^{\circ}}$

$:\implies \bf{A = 3 \times 180^{\circ}}$

$:\implies \bf{A = 3 \times 180^{\circ}}$

$:\implies \bf{A = 540^{\circ}}$

$\boxed{\therefore \bf{A = 540^{\circ}}}$

Hence, the sum of all the interior angles of a Polygon is 540°.

To find the value of x :

Since , we know that the sum of angles and we are provided with the angles (in terms of x).

So by using the property that sum of all the interior angles sum up to 540° , we get the Equation as :

$:\implies \bf{x^{\circ} + 2x^{\circ} + (3x + 10)^{\circ} + (4x + 5)^{\circ} + 5x^{\circ} = 540^{\circ}}$

Now , by solving it we get :

$:\implies \bf{x^{\circ} + 2x^{\circ} + 3x^{\circ} + 10^{\circ} + 4x^{\circ} + 5^{\circ} + 5x^{\circ} = 540^{\circ}}$

$:\implies \bf{15x^{\circ} + 15^{\circ} = 540^{\circ}}$

$:\implies \bf{15x^{\circ} = - 15^{\circ} + 540^{\circ}}$

$:\implies \bf{15x^{\circ} = 525^{\circ}}$

$:\implies \bf{x = \dfrac{525^{\circ}}{15^{\circ}}}$

$:\implies \bf{x = 35^{\circ}}$

$\boxed{\therefore \bf{x = 35^{\circ}}}$

Hence , the value of x is 35°.

Each interior angle :

By putting the value of x in the angles , we can find the required value.

• First angle = x° = 35°

• Second angle = 2x° = 70°

• Third angle = (3x + 10)° = 115°

• Fourth angle = (4x + 5)° = 145°

• Fifth angle = 5x° = 175°

Each exterior angle :

• First angle = (180° - 35°) = 145°

• Second angle = (180° - 70°) = 110°

• Third angle = (180° - 115°) = 65°

• Fourth angle = (180° - 145°) = 35°

• Fifth angle = (180° - 175°) = 5°

Answer 2

$\frac{8 + 3 \sqrt{2} + 7 - 52 - (3 - 4 \sqrt{2}) }{6 - 2 \sqrt{2} } = a + b \sqrt{2}$

$\frac{8 + 3 \sqrt{2} + 7 - \sqrt{2} - 3 + 4 \sqrt{2} }{6 - 2 \sqrt{2} } a + b \sqrt{2}$

$\frac{8 + 70 - 3 + 3 \sqrt{2} - \sqrt{2} + 4 \sqrt{2} }{6 - 2 \sqrt{2} } = a + b \sqrt{2}$

$\frac{12 + 6 \sqrt{2} }{6 - 2 \sqrt{2} } = a + b \sqrt{2}$

$\frac{2(6 + 3 \sqrt{2}) }{2(3 \sqrt{2} )} = a + b \sqrt{2}$

$\frac{6 + 3 \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}$

$\frac{6 + 3 \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 - \sqrt{2} } = a + b \sqrt{2}$

$\frac{18 + 6 \sqrt{2} + 9 \sqrt{2} + 6}{a - 42} = a + b \sqrt{2}$

$\frac{24 + 15 \sqrt{2} }{7} = a + b \sqrt{2}$

$\frac{24}{7} + \frac{15}{17} \sqrt{2} = a + b \sqrt{2}$

$(a , b) = ( \frac{24}{7} 15 \: /7)$