-->

Statistics

Question
CBSEENMA9003279

Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.


Solution

For Triangular Area I
a = 5 cm b = 5 cm c = 1 cm
therefore space space space straight s equals fraction numerator straight a plus straight b plus straight c over denominator 2 end fraction equals fraction numerator 5 plus 5 plus 1 over denominator 2 end fraction equals 11 over 2 equals 5.5 space cm

therefore space space Area space straight I space equals space square root of straight s left parenthesis straight s minus straight a right parenthesis left parenthesis straight s minus straight b right parenthesis left parenthesis straight s minus straight c right parenthesis end root
space space space space space space space space space space space space space space equals square root of 5.5 left parenthesis 5.5 minus 5 right parenthesis left parenthesis 5.5 minus 5 right parenthesis left parenthesis 5.5 minus 1 right parenthesis end root
space space space space space space space space space space space space space space equals square root of 5.5 left parenthesis.5 right parenthesis left parenthesis.5 right parenthesis left parenthesis 4.5 right parenthesis end root
space space space space space space space space space space space space space space equals left parenthesis.5 right parenthesis square root of left parenthesis 5.5 right parenthesis left parenthesis 4.5 right parenthesis end root
space space space space space space space space space space space space space space equals left parenthesis.5 right parenthesis square root of left parenthesis.5 right parenthesis left parenthesis 11 right parenthesis left parenthesis.5 right parenthesis left parenthesis 9 right parenthesis end root
space space space space space space space space space space space space space space equals left parenthesis.5 right parenthesis left parenthesis.5 right parenthesis left parenthesis 3 right parenthesis square root of 11
space space space space space space space space space space space space space space equals 0.75 square root of 11 equals 0.75 left parenthesis 3.3 right parenthesis space left parenthesis approx right parenthesis
space space space space space space space space space space space space space space equals space 2.5 space cm squared space left parenthesis approx right parenthesis space
Area II = 6.5 x 1 = 6.5 cm2
For Area III
Area space III space equals space 1 half cross times 2 cross times fraction numerator square root of 3 over denominator 2 end fraction plus 1 half cross times 1 cross times fraction numerator square root of 3 over denominator 2 end fraction
space space space space space space space space space space space space equals fraction numerator square root of 3 over denominator 2 end fraction plus fraction numerator square root of 3 over denominator 4 end fraction
space space space space space space space space space space space space equals space fraction numerator 3 square root of 3 over denominator 4 end fraction equals fraction numerator 3 cross times 1.732 over denominator 4 end fraction space left parenthesis approx right parenthesis
space space space space space space space space space space space space equals space fraction numerator 5.196 over denominator 4 end fraction equals 1.3 space cm squared space left parenthesis approx right parenthesis

Area space IV equals fraction numerator 6 cross times 1.5 over denominator 2 end fraction equals 4.5 space cm squared
Area space straight V equals fraction numerator 6 cross times 1.5 over denominator 2 end fraction equals 4.5 space cm squared

Total area of the paper used

= Area I + Area II + Area III + Area IV + Area V
= 2.5 cm2 + 6.5 cm2 + 1.3 cm2 + 4.5 cm+ 4.5 cm2
= 19.3 cm2.

 

 

Some More Questions From Statistics Chapter