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This is the second of our many quizzes on Types of angles. Click on this to access more such competitive quizzes on Geometry.

Types of angles Quiz 2

mathematics

 In a polygon, two of the external angles are $60{}^\circ \text{ and }40{}^\circ $ and the others are equal, each equal to $20{}^\circ $. Find the total measure of all the internal angles.

 Consider the formula $\text{Sum of internal angles + Sum of exterior angles = 180}{}^\circ \text{ }\times \text{ Number of sides}$ then equate the sum of all the interior angles to $180{}^\circ $. 

  We know that, $ \\ $ Sum of internal angles $ + $ Sum of exterior angles $ = 180{}^\circ \times $ Number of sides $  \\ \Rightarrow \left( n-2 \right)\times 180{}^\circ +60{}^\circ +40{}^\circ +\left( n-2 \right)\times 20{}^\circ =180{}^\circ \times n  \\ \Rightarrow ~180{}^\circ \times n-360{}^\circ +100{}^\circ +20\times n-40{}^\circ =180{}^\circ \times n  \\ \Rightarrow n~=\dfrac{300}{20}=15  \\ $ Hence, sum of internal angles $ =\left( n-2 \right)\times 180{}^\circ =2340{}^\circ $ 

 A square carpet is to be woven such that the two triangles have the same base and one of them is an equilateral triangle as given below. Find the value of angle $x.$ $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/f3570549-68da-4010-a2f4-61dfacffbc4d13834277728527809.png"/>

 First take the equilateral triangle then find $\angle EBA$ , from there find $\angle BAC\text{ and }\angle ECB$, then $x$ can be found out considering the sum of the interior angles of the smallest triangle.

  BEC is an equilateral triangle, so all angles $ =60{}^\circ $. $ \\ \angle ABE=\angle DCE=90{}^\circ -60{}^\circ =30{}^\circ  \\ $ Now, AC is a diagonal of the square, then $ \angle ABC=45{}^\circ .\\ $ Then $ \angle ACB=180{}^\circ -45{}^\circ -90{}^\circ =45{}^\circ \\ $ Hence, $ x{}^\circ =180{}^\circ -60{}^\circ -45{}^\circ =75{}^\circ $ 

 The interior angles, ${{a}_{1}},{{a}_{2}}...{{a}_{n}}$  of a polygon such that ${{a}_{2}}-{{a}_{1}}={{a}_{3}}-{{a}_{2}}$. Find the value of n if ${{a}_{1}}=72{}^\circ $ the difference between two consecutive angles is $12{}^\circ $, ${{a}_{n}}<180{}^\circ $.

 First consider the formula, The sum of the series = sum of the interior angles of the polygon, then by the summation formula, $S=\dfrac{n}{2}\left[ 2a+(n-1)d \right]$ of the arithmetic series find the value of $n$.

Let the number of sides of the polygon be $n$. $ \\ $ The angles of the polygon are in A.P. with the first term as $ 72{}^\circ$ and difference as $ 12{}^\circ $ $ \\ $ The sum of the series $=$ sum of the interior angles of the polygon $ \\ \Rightarrow 72{}^\circ +84{}^\circ +...=(n-2)180{}^\circ  \\ \Rightarrow \dfrac{n}{2}\left[ 144{}^\circ +(n-1)12{}^\circ  \right]=(n-2)180{}^\circ \Rightarrow {{n}^{2}}-19n+60=0 \\ \Rightarrow n=4\text{ or }15 \\ $ Since 15 exceeds the angle above $ 180{}^\circ$ which is not possible so, the polygon has 4 vertices.

 The area of a sector of a circle is $2.88\pi c{{m}^{2}}$. The area of a sector of another circle with the angle being $5{}^\circ $ less than the first sector is $5.76\pi c{{m}^{2}}$. If the radii of the circles are in the ratio $2:3$ respectively, then what will be the area of the equilateral triangle inscribed by the second circle?

 By the formula of area of a sector, $\dfrac{\pi 4{{r}^{2}}d}{360{}^\circ }$, for the two expressions, find out the radius then find out the length of the side of the triangle for the area.

 For the first sector, area $ =\dfrac{\pi 4{{r}^{2}}d}{360{}^\circ }=\dfrac{\pi {{r}^{2}}d}{90{}^\circ }\Rightarrow \dfrac{{{r}^{2}}d}{90{}^\circ }=2.88\Rightarrow {{r}^{2}}d=259.2\\ $ For the second sector, area $ =\dfrac{\pi 9{{r}^{2}}(d-5)}{360{}^\circ }=\dfrac{{{r}^{2}}(d-5)}{40{}^\circ }\\  \Rightarrow \dfrac{{{r}^{2}}(d-5)}{40{}^\circ }=5.76\Rightarrow 5{{r}^{2}}=259.2-230.4=28.8,r=2.4\\ $ So, radius of the second circle $ =7.2cm \\ $ Length of each side of the equilateral triangle $ =2\sqrt{3}\times 7.2=14.4\sqrt{3}cm$  $ \\ $ Area $ =\dfrac{\sqrt{3}}{4}{{\left( 14.4\sqrt{3} \right)}^{2}}=155.52\sqrt{3}c{{m}^{2}}$

 A playground has two swimming pools I and II, which have a shape of polygons, with $x$ and $y$ as sides. Each interior angle of pool II are 2 times of the interior angles of pool I. The interior angles with $\left( x+y \right)$sides is$140{}^\circ $. What is the value of $y$?

 First use the formula for the interior angle of a regular polygon having $n$ sides $=\left[ \dfrac{\left( n-2 \right)}{2} \right]\times 180{}^\circ $ then consider the given relation of two angles, solve for the unknown variables.

 Interior angle of a regular polygon having n sides $ =\left[ \dfrac{\left( n-2 \right)}{2} \right]\times 180{}^\circ$ $ \\ $ Given that $ \angle y=2\times \angle x \\ \Rightarrow \left[ \dfrac{\left( y-2 \right)}{y} \right]\times 180{}^\circ \text{ = }2\times \left[ \dfrac{\left( x-2 \right)}{x} \right]\times 180{}^\circ \\ $ Solving, we get $ 4y-2x=xy $ $\\ $ Each interior angle of a regular polygon having $ \left( x+y \right)\text{ sides } = \left[ \dfrac{\left( x+y-2 \right)}{\left( x+y \right)} \right]\times 180{}^\circ =140{}^\circ \\ $ Solving, we get $ x+y=9$ $ \\ $ Solving $(I)$ and $(II)$ we get $ x=3,12$ $ \\ $ If $ x=3, y=9-3=6 \\ $ If $ x=12, y=9-12=-3 \text{ (which is not possible as } y > 0)\\ \text{Therefore } x=3,\text{ }y=6.$ 