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What is the size of the diagonal of a rectangle whose length is twice its width plus 3.5 cm and has a perimeter of 59.5 cm showing all work?
Wiki User
∙ 8y ago
Let the shorter side be 'a'. Then the longer side is 2a + 3.5
To find the perimeter we add the 4 sides: a+a+(2a+3.5) + (2a+3.5) = 6a+7
Now we know the perimeter is 59.5cm
So 6a+7=59.5
==> 6a = 52.5
==> a = 52.5 ÷ 6 = 8.75
So the shorter side is 8.75 and the longer side is (2 * 8.75) + 3.5 = 21. (where * means multiply)
Now to find the diagonal, we use Pythagoras a^2 + b^2 = c^2 (where ^2 means to the power of 2 or squared)
So substituting the two sides of the rectangle,
c^2 (the diagonal) = 21^2 + (8.75)^2 = 441 + 76.5625 = 517.5625
==> c = sq rt (517.5625) = 27.75cm
Additional Information:-
All of the above is correct except for the fact that the square root of 517.5625 is 22.75cm which is the length of the diagonal
Wiki User
∙ 8y ago
Wiki User
∙ 8y agoYou first need to find out the length and width. Write one equation for the relation between length and width (l = 2w + 3.5), one for the perimeter (stating that the perimeter is the sum of the four sides), and calculate "l" and "w". Then use Pythagoras's Law to find the diagonal.
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