Given that the area of S 1 is 1 0 cm 2 and the magnitude of the perimeter of S 2 is double the magnitude of the area of S 1 , find the perimeter of T in meters
Side note: Sorry that my MS Paint skills are not too great. If anyone knows any software that makes images with the L A T E X and shapes, feel free to post that as a solution because it will solve one of my "problems"
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First, I made both equations look nicer. Area of S 1 = 4 ( 2 x − y ) = 8 x − 4 y = set 1 0 cm 2 Perimeter of S 2 = 2 ( 3 x + 2 y + 2 x − y ) = 2 ( 5 x + y ) = 1 0 x + 2 y = set 2 × Area of S 1 = 2 0 cm
I rearranged the top equation for y , but there are other ways of solving these: 8 x − 4 y = 1 0 add 4 y and subtract 1 0 8 x − 1 0 = 4 y divide by 4 y = 2 x − 2 5 cm
Now we can substitute this into our second equation and solve for x : 1 0 x + 2 y = 2 0 sub y 1 0 x + 2 ( 2 x − 2 5 ) = 2 0 ⟹ 1 4 x − 5 = 2 0 add 5 1 4 x = 2 5 divide by 1 4 x = 1 4 2 5 cm
We can use this to find y : y = 2 x − 2 5 sub x y = 2 ( 1 4 2 5 ) − 2 5 = 7 2 5 − 2 5 = 1 4 1 5 cm
Now that we have both x and y , we can start to find the perimeter of T . Firstly, we must know that the sides are of length x cm and 2 x − y cm . Secondly, we need to find the hypotenuse of T ; for this, we can use the Pythagorean theorem since it is a right angled triangle: Hypotenuse of T = x 2 + ( 2 x − y ) 2 = ( 1 4 2 5 ) 2 + ( 2 5 ) 2 = 9 8 9 2 5 = 1 4 5 7 4 cm
Finally, we can calculate the perimeter of T : Perimeter of T = 1 4 5 7 4 + 1 4 2 5 + 2 ( 1 4 2 5 ) − 1 4 1 5 = 1 4 6 0 + 5 7 4 cm
However, the answer has to be in meters so the actual answer is 1 4 6 0 + 5 7 4 cm ÷ 1 0 0 = 2 8 0 1 2 + 7 4 meters = 0 . 0 7 3 5 7 9 7 3 3 1 … meters
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Area of S 1 is 4 ( 2 x − y ) = 1 0 ⟹ 8 x − 4 y = 1 0 . . . ( 1 ) also ⟹ 2 x − y = 4 1 0 = 2 5 .
Perimeter of S 2 is 2 ( 2 x − y ) + 2 ( 3 x + 2 y ) = 2 0 ⟹ 5 + 6 x + 4 y = 2 0 ⟹ 6 x + 4 y = 1 5 . . . ( 2 ) .
( 1 ) + ( 2 ) : 1 4 x = 2 5 ⟹ x = 1 4 2 5 .
Perimeter of T is given by:
x + 2 x − y + x 2 + ( 2 x − y ) 2 = 1 4 2 5 + 2 5 + ( 1 4 2 5 ) 2 + ( 3 5 ) 2 ≈ 7 . 3 5 7 9 7 3 3 cm ≈ 0 . 0 7 3 6 m