Free Study Material

NCERT Solutions

NCERT Books

Revision Notes

Practice Problems

This is one of our many quizzes on Special right triangle. Click on this to access more such competitive quizzes on Geometry.

Special right triangle Quiz 2

mathematics

 Which of the following given below is a special right triangle?

 Use the property of triangles in which a certain angle is ${90^ \circ }$.

  $45 - 45 - 90$ and $30 - 60 - 90$ are special right triangles. In triangle $45 - 45 - 90$ sides are in the special ratio of $1:1:2$

 What is the length of the square whose diagonals is $3\sqrt 2 cm$. $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/fd5b2272-2f23-4d58-8720-5456ba605a414165575952713398026.png"/>

 Find the length of one side using the diagonal length. Also use the property that all the internal angles and sides are equal in a square.

  <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/dcb5c617-8718-48d2-9301-def4b695379e4885972767527270427.png"/> $ \\  {\text{length of diagonal}} = 3\sqrt 2 cm  \\  \text{In }\Delta ABC  \\\text{By using Pythagora's theorem. We get,}  \\  {\left( H \right)^2} = {\left( P \right)^2} + {\left( B \right)^2}  \\  {\left( {3\sqrt 2 } \right)^2} = {\left( x \right)^2} + {\left( x \right)^2}  \\  3\sqrt 2  \times 3\sqrt 2  = {n^2} + {x^2}  \\  9 \times 2 = 2{x^2}  \\  \dfrac{{18}}{2} = {x^2}  \\  {x^2} = 9  \\  x = \sqrt 9   \\  x = 3, - 3  \\ \text{length can never be negative}  \\  x = 3  \\  $

 Calculate the hypotenuse of the right angle triangle in which the height of the triangle is $2$ more than of base and area of the right angle triangle is $144c{m^2}$.

 Use the area of the triangle to find the value of base and height. 

$  {\text{Let height of triangle}} = x  \\ \text{Base }= x - 2  \\  \text{Area }= 144  \\  \therefore {\text{Area of triangle}} = \dfrac{1}{2} \times B \times h  \\   \Rightarrow 144 = \dfrac{1}{2} \times \left( {x - 2} \right) \times x  \\   \Rightarrow 288 = {x^2} - 2x  \\   \Rightarrow {x^2} - 2x - 288 = 0  \\   \Rightarrow \left( {x - 18} \right)\left( {x + 16} \right) = 0  \\  \therefore x - 18 = 18  \\   \Rightarrow x + 16 = 0  \\   \Rightarrow x =  - 16{\text{ rejected length cannot }}\left(  -  \right)ve.  \\  \therefore \text{Base }= 16  \\  \therefore {h^2} = {B^2} + {A^2}  \\   \Rightarrow h = \sqrt {{{\left( {18} \right)}^2} + {{\left( {16} \right)}^2}}   \\   \Rightarrow h = \sqrt {580}   \\   \Rightarrow h = 2\sqrt {145} cm  \\ $

 The length of a hypotenuse and other side of a right triangle are  $24cm$ and $20cm$. Find the length of the other side?

 Use Pythagora's theorem to solve this problem.

 $ \text{Hypotenuse }= 24cm  \\  {\text{One side of triangle}} = 20cm  \\  {\text{By using pythagoras theorem;}}  \\  {(hypotenuse)^2} = {l^2} + {b^2}  \\   \Rightarrow {24^2} = {20^2} + {b^2}  \\   \Rightarrow 576 = 400 + {b^2}  \\   \Rightarrow 576 - 400 = {b^2}  \\   \Rightarrow b = 13.267  \\ \text{Length of other side }\left( x \right) = 13.267cm  \\  $ 

 If the diagonal of a right triangle is $6cm$. Find the length of the other two sides of the triangle given that one of its angles is $60$ degree?

 Use the ratio of  $30 - 60 - 90$  in the right triangle to find the other side.

 $30 - 60 - 90\text{ right triangle use the ratio}.\\  x : x\sqrt{3} : 2x  \\  \text{Diagonal }\left( \text{hypotenuse} \right)= 6cm  \\  2x = 6  \\  x = \dfrac{6}{2}  \\  x = 3  \\  \text{Put } x \text{ in } x\sqrt 3   \\  x\sqrt 3  = 3\sqrt 3   \\  \text{Short side of right triangle is }  3cm \\ \text{Longer side of right triangle is } 3\sqrt 3cm $