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

Classifying triangles Quiz 2

mathematics

 In $\Delta PQR$, each side is $15cm$. Find the area of the triangle. $ \\ $<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/50b81b37-2616-45e0-a1a6-f86d1202abdb8524945051961083855.png"/>

 The area of an equilateral triangle is $\dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}$. 

  All sides are equal. Thus it is an equilateral triangle. $ \\  \therefore {\text{ area of }}\Delta POR = \dfrac{{\sqrt 3 }}{4} \times {\left( {side} \right)^2}  \\   \Rightarrow \dfrac{{\sqrt 3 }}{4} \times 15 \times 15  \\   \Rightarrow \dfrac{{225\sqrt 3 }}{4}c{m^2}  \\ $

 If $\Delta ABC$ has $a = b + 2c$ then find the value of $\sec A,cos\dfrac{A}{2}$.

 Apply $cos$ law $\operatorname{cosA}  = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}$ and $\cos \dfrac{\theta }{2} = \sqrt {\dfrac{{1 + \cos \theta }}{2}} $.

  $  {\text{Given }}a = b + 2c  \\   \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}  \\   \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {{\left( {b + 2c} \right)}^2}}}{{2bc}}  \\   \Rightarrow \cos A = \dfrac{{{b^2} + {c^2} - {b^2} - 4{c^2} - 4bc}}{{2bc}}  \\  \cos  A =\dfrac{-4bc-3c^2}{2bc} \\ \cos A =-\dfrac{c\left(4b+3c\right)}{2bc} \\ \cos A =-\dfrac{4b+3c}{2b} \\ \Rightarrow  \sec A =-\dfrac{2b}{4b+3c} \\ \sec A = -\dfrac{2(a-2c)}{4b+3c} \\ \sec A = -\dfrac{2a-4c}{4b+3c} \\  \Rightarrow \cos \dfrac{A}{2} = \sqrt {\dfrac{{1 + \cos A}}{2}} \\ \cos \dfrac{A}{2} = \sqrt {\dfrac{{1 - \dfrac{4b+3c}{2b}}}{2}} \\ \cos \dfrac{A}{2} = \sqrt {\dfrac{{2b-(4b+3c)}}{4b}} \\ \cos \dfrac{A}{2} = \sqrt {\dfrac{{2(a-2c)-(4b+3c)}}{4b}} $

 Ratio of sides of a triangle is $6:8:10$. The triangle is

 Use Pythagoras’ Theorem to check if the triangle is right-angled.

Given ratio of sides is $6:8:10$ $ \\ $ Let sides be $6x,\;8x,\;10x$ \\ $  {\left( {10x} \right)^2} = 100{x^2}  \\  {\left( {6x} \right)^2} + {\left( {8x} \right)^2} = 36{x^2} + 64{x^2} = 100{x^2}  \\  $ Pythagoras’ Theorem holds true. Thus the triangle is right-angled.

 In $\Delta PQR$, If ${B^2} + {A^2} = 144{h^2}$ then find the value of $\dfrac{{\tan R}}{{\tan P + \tan R}}$. Also classify the triangle.

 Substitute values of $\tan P$ and $\tan R$ in the expression and simplify. 

 <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/351827dd-f5d1-408e-a9ad-ad6f5366fcdb313851030671060141.png"/>$ \\ $ $  {B^2} + {A^2} = 144{h^2}  \\   \Rightarrow {h^2} = {\left( {\dfrac{B}{{12}}} \right)^2} + {\left( {\dfrac{A}{{12}}} \right)^2}  \\  {\text{It is a right angled triangle}}  \\   \Rightarrow \tan P = \dfrac{B}{{12}} \times \dfrac{{12}}{A} = \dfrac{B}{A}  \\   \Rightarrow \tan R = \dfrac{A}{B}  \\  \therefore \dfrac{{\operatorname{tanR} }}{{\tan P + \tan R}}  \\   \Rightarrow \dfrac{{\dfrac{A}{B}}}{{\dfrac{B}{A} + \dfrac{A}{B}}}  \\   \Rightarrow \dfrac{A}{B} \times \dfrac{{AB}}{{\left( {{A^2} + {B^2}} \right)}}  \\   \Rightarrow \dfrac{{{A^2}}}{{{A^2} + {B^2}}}  \\  $

 In $\Delta ABC$, $a = 15\;cm$, $b = 12\;cm$ and $c = 9\;cm$, and $D$ is the midpoint of $BC$, then

 Establish that the given triangle is a right-angled triangle. For a right-angled triangle the midpoint of the hypotenuse is the centre of the circumcircle.

 ${15^2} = {12^2} + {9^2}$ $ \\ $ Thus it is a right-angled triangle. In a right-angled triangle, the line joining the midpoint of the hypotenuse and the right-angled vertex is the circumradius. Thus $Circumradius = \dfrac{a}{2} = AD$. $ Area = \dfrac{1}{2} \times base \times height = \dfrac{1}{2} \times b \times c  \\  Inradius = \dfrac{{a + b + c}}{2} = \dfrac{{15 + 12 + 9}}{2} = 18\;cm  \\  $