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

Triangles Quiz 2

mathematics

 Calculate the ratio of area of two similar triangles whose perimeter ratio is  $8:7$.

 Apply the formula ${\text{Ratio of Area}} = {\left( {{\text{Ratio of perimeter}}} \right)^2}$. 

  $  {\text{Ratio of perimeter }} = \dfrac{8}{7}   \\  {\text{Ratio of Area  = }}{\left( {{\text{Ratio of perimeter}}} \right)^{\text{2}}}   \\   \Rightarrow {\left( {\dfrac{8}{7}} \right)^2}   \\   \Rightarrow \dfrac{{64}}{{49}}   \\  {\text{Hence ratio of area }} = 64:49   \\  $

 Evaluate the value of $x$ in the given figure, if $A{C^2} - B{C^2} = {x^2} $ $ \\ $ <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/5fd24d87-ca19-4b98-a580-dde896dcca914646541633557241100.png"/>

 Apply the Pythagoras theorem ${H^2} = {P^2} + {B^2}$ on both the triangles and then subtract both the equations to calculate the value of “x”. 

  $  {\text{In }}\Delta ACD,{\text{ by Pythagoras theorem}}   \\   \Rightarrow A{D^2} = A{C^2} + C{D^2}...........(i)   \\  {\text{In }}\Delta BCD,{\text{by Pythagoras theorem}}   \\   \Rightarrow C{D^2} + B{C^2} = B{D^2}............(ii)   \\  {\text{on  subtracting equation }}\left( {{\text{ii}}} \right){\text{  form equation }}\left( {\text{i}} \right){\text{ we get}}   \\   \Rightarrow A{C^2} - B{C^2} = A{D^2} - B{D^2}   \\   \Rightarrow A{C^2} - B{C^2} = A{B^2}\left[ {\because A{D^2} - B{D^2} = A{B^2}} \right]   \\  {\text{Hence, }}x = AB   \\  $

 If a boy stands in front of a building then find the distance between boy and building if the distance between foot of the building and boy is $17$ more than the height, the distance between the top of the building and boy  is ${\text{25m}}$.

 Find the relation between the height of the building and distance between the boy and foot of the building. After that apply PGT to find the height of the building. 

<img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/3c5dd347-93f8-4540-8819-83d63bc09fff6414695397411462717.png"/>$ \\  {\text{In }}\vartriangle ABC,{\text{ by using pythagoras theorem}}   \\  {\left( {AC} \right)^2} = {\left( {BC} \right)^2} + {\left( {AB} \right)^2}   \\   \Rightarrow {\left( {25} \right)^2} = {x^2} + {\left( {17 + x} \right)^2}   \\   \Rightarrow 625 = 2{x^2} + 289 + 34x   \\  \therefore \left( {x + 24} \right)\left( {x - 7} \right) = 0   \\  x + 24 = 0   \\   \Rightarrow x =  - 24{\text{ rejected}}   \\  x - 7 = 0   \\   \Rightarrow x = 7   \\  \therefore {\text{Distance between boy and top of the building}} = 7 + 17   \\   = 24m   \\  $

 Find the base of the right angle triangle if the area of the right angle triangle is $3$ more than the twice of the equilateral triangle, if the side of the equilateral triangle is $15cm$ and height of the right angle triangle is $16cm$. 

 Apply the formula of equilateral and right angle triangles. 

 $  {\text{Area of right angle triangle}} = 2 \times {\text{area of equilateral triangle}} + 3   \\   \Rightarrow \dfrac{1}{2} \times base \times height = 2 \times \dfrac{{\sqrt 3 }}{4}{\left( {side} \right)^2} + 3   \\   \Rightarrow \dfrac{1}{2} \times b \times 16 = 2 \times \dfrac{{1.73}}{2} \times 15 \times 15 + 3   \\   \Rightarrow b = 25cm   \\  \therefore base{\text{ is }}25cm   \\  $

 The three sides of a triangle are 8m, 10m and 14m respectively, and then what is the area of the given triangle

 By using $s = \dfrac{{a + b + c}}{2}$ and ${\text{Area of triangle}} = \sqrt {s(s - a)(s - b)(s - c)} $  

 $  {\text{As }}s = \dfrac{{a + b + c}}{2} = \dfrac{{8 + 10 + 14}}{2} = 16   \\  {\text{Area of triangle}} = \sqrt {s(s - a)(s - b)(s - c)}    \\   = \sqrt {16(16 - 8)(16 - 10)(16 - 14)}    \\   = \sqrt {16 \times 8 \times 6 \times 2}  = 39.19{m^2}   \\ $ 