sum of the areas of two squares is 157 m square the sum of their perimeters is 68 M find the sides of the two squares
Answers
Answer: Let us say that the sides of the two squares be 'a' and 'b'
Sum of their areas = a^2 + b^2 = 157
Sum of their perimeters = 4a + 4b = 68
=> 4(a+b)=68
=>a+b = 68/4
=>a+b=17
=> a=b-17 ....... (1)
Now, (17-b)^2 + b^2=157
=> 17^2 + b^2 - 2*17*b + b^2 = 157
=> 289 + b^2 - 34b + b^2 = 157
=> 289 + 2b^2 - 34b = 157
=> 289-157 + 2b^2 - 34b = 0
=> 132 + 2b^2 -34b = 0
=> 2b^2 - 34b + 132 = 0
=> b^2 - 17b + 66 = 0 ( Divided by 2 )
=> b^2 - ( 11+6)b + 66 = 0
=> b^2 - 11b - 6b + 66 = 0
=> b(b-11) + 6(b-11) = 0
=> (b-11) (b+6) = 0
=> b = -6 or b = 11
Since, b = -6 is not accepted because the side of a square can't be negative , therefore b = 11 is accepted .
Substituting the value of b in equation (1) , we get,
a = b-17
=> a = 11 - 17
=> a = 6
Therefore answer is a = 6 and b = 11
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Hope this answer helps you !!!!!
Answer:
hope it helps!!!!....
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