The length of a rectangle is 10 m more than its breadth. If the perimeter of rectangle is 80 m, how do you find the dimensions of the rectangle?
side 1 = 15m, s side 2 = 15m, side 3 = 25m, side 4 = 25m.
As an object's perimeter is equal to the sum of its lengths, in this case, 80m = side1 + side2 + side3 + side4.
There are now two sets of equal-length sides in a rectangle.
Thus, 2xSide1 + 2xSide2 = 80m.
Furthermore, it is stated that the length exceeds the breadth by 10 meters.
Thus, 2xSide1+(10+10) + 2xSide2 = 80m.
Thus, 2xS1+20 + 2S2 = 80m.
80 = 2x + 2y + 20
X + Y would equal 0 if it were a square.
thus
60 = 4 times side 1.
side 1 = 60/4 = 15m, thus
side 1 = 15 m, side 2 = 15 m, side 3 = 15 m + 10 m, and side 4 = 15 + 10 m.
Thus, s1 = 15 m, s2 = 15 m, s3 = 25 m, and s4 = 25 m.
Perimiter is 80 meters, and the rectangle's length is 10 meters greater than its width.
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Let the breadth of the rectangle be ( x ) meters. Then, the length of the rectangle is ( x + 10 ) meters. The perimeter of a rectangle is given by the formula ( 2(length + breadth) ), so we have the equation ( 2(x + x + 10) = 80 ). Solving this equation will give us the value of ( x ), which we can then use to find the length and breadth of the rectangle.
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