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This is one of our many quizzes on Central, inscribed, and circumscribed angles. Click on this to access more such competitive quizzes on Geometry.

Central, inscribed, and circumscribed angles Quiz 2

mathematics

 A Right angle is made by Prakash such that all its sides have integer values, and one side is 12 cm. What is the radius of the largest possible circle that can be made, such that it touches all sides of the triangle? 

 Incircle of the triangle is to be found, for right angle triangle the incircle is given as\[\text{radius of incircle=}\dfrac{a+b-c}{2}\].

The figure is as shown <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/439ed599-4b21-4b0c-b649-4193414a070f5897558056158275229.png"/> Incircle of the triangle is to be found, for right angle triangle the incircle is given as \[\text{radius of incircle=}\dfrac{a+b-c}{2}\] As one side is 12 cm. the other side will be $x$ and the hypotenuse will be $\sqrt{144+{{x}^{2}}}$ Substituting the values \[\text{radius of incircle=}\dfrac{12+x-\sqrt{144+{{x}^{2}}}}{2}=6+\dfrac{1}{2}\left( x-\sqrt{144+{{x}^{2}}} \right)\] Given all sides are integer so $144={{h}^{2}}-{{x}^{2}}=(h+x)(h-x)$ The possible solutions are $2\times 72,4\times 36,6\times 24,8\times 18,12\times 12$ This gives us the value of radius as $\{5,4,3,2\}$ Hence the largest value is 5.

Three people are standing on the circumference of a circular track as shown. Here the angle inscribed between them is ${{24}^{\circ }}$. What is the measure between the inscribed angle? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fd83c314-1573-4668-b965-861be977ea24585351328756411523.png"/>

 A conjecture about inscribed angles states the inscribed angle is one half the arc it cuts. The converse is also true; an arc is double the measure of the angle by which is cut.

A conjecture about inscribed angles states the inscribed angle is one half the arc it cuts. The converse is also true; an arc is double the measure of the angle by which is cut. With this conjecture we know that$arcAC=2.m\angle AC=2\times 24=48$ A circle must be ${{360}^{\circ }}$Since \[arcABC+arcAC={{360}^{\circ }}\] Therefore &nbsp;$\begin{align} &nbsp;&nbsp;& \Rightarrow arcABC+48={{360}^{\circ }} \\ &nbsp;& \Rightarrow arcABC={{312}^{\circ }} \\ \end{align}$

A metallic plate is kept on a circular barrel such that the line forms a tangent with the circle. The angle between the points on the tangent makes ${{24}^{\circ }}$ with the centre as shown, what is the value of $\angle ABC$? <img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/27f8a698-5638-42e0-bf99-ee088338d4dd7395804594056904335.png"/>

 The angle made by the tangent and the centre of the circle are right angles, the triangle inscribed by a point on the tangent with the radius is a right angled triangle.

Given that $\overleftrightarrow{AC}$ is the tangent to the circle and B is the centre of the circle, $\Rightarrow \angle CAB={{90}^{\circ }}$ Given that $\angle ACB={{24}^{\circ }}$ Angle sum property of triangle states that the sum of interior angles of a triangle is 180 so in $\vartriangle ABC$ $\begin{align} &nbsp;&nbsp;& \Rightarrow \angle ABC=180-\angle CAB-\angle ACB \\ &nbsp;& \Rightarrow \angle ABC=180-90-24=66 \\ \end{align}$

 Parul has a right angle triangle such that two sides inscribing the right angle are 5 units and 12 units. She wants to make the largest possible circle, what is the radius of such a circle? 

 The largest possible circle will be the incircle of the triangle, which touches all its sides.

Consider a triangle as $\vartriangle ABC$ right angled at B So, $AB=P=5units$ and $BC=B=12units$ The hypotenuse of a right angle triangle by Pythagoras theorem is given by, $h=\sqrt{{{p}^{2}}+{{b}^{2}}}$ Substituting values, $\Rightarrow h=\sqrt{{{5}^{2}}+{{12}^{2}}}=\sqrt{25+144}=\sqrt{169}=13$ In a right angled triangle, the radius of incircle is given as \[r=\dfrac{P+B-H}{2}=\dfrac{5+12-13}{2}=2units\]

 Suppose I draw a right angled triangle $\vartriangle ABC$ where the value of perpendicular is 10 cm and base is 6 cm, one of the altitude is 4 cm. A point on this altitude is so formed, that it is the diameter of the circumcircle, then find the length of circumradius in cm. 

 The area of the triangle is given as $area=\dfrac{1}{2}\times b\times altitude$. Area of a triangle is given in terms of circumcircle and is given as $area=\dfrac{abc}{4R}$. Finally find R by equating the area.

In $\vartriangle ABC$ given that $AB=P=10cm$ and $AC=H=6cm$, it is inferred that the altitude $AE=4$ Let the new point on the altitude be D such that $AD$ is the circum diameter. Using Pythagoras theorem finding the base $\begin{align} &nbsp;&nbsp;& b=\sqrt{{{h}^{2}}-{{p}^{2}}} \\ &nbsp;& \Rightarrow b=\sqrt{100-36} \\ &nbsp;& \Rightarrow b=BC=8 \\ \end{align}$ The area of triangle is given as $\begin{align} &nbsp;&nbsp;& area=\dfrac{1}{2}\times b\times altitude \\ &nbsp;& \Rightarrow area=\dfrac{1}{2}\times 8\times 4=16c{{m}^{2}} \\ \end{align}$ Area of a triangle in terms of circumcircle is given as $\begin{align} &nbsp;&nbsp;& area=\dfrac{abc}{4R} \\ &nbsp;& \Rightarrow 16=\dfrac{10\times 8\times 6}{4\times R} \\ &nbsp;& \Rightarrow R=7.5cm \\ \end{align}$