Question

6: The length of a rectangle shaped park exceeds its breadth by 17 meters. If the perimeter of the park is 178 meters find the dimensions of the park?​

Answer 1

Step-by-step explanation:

Let the breadth of the park be =x meters

Then the length of the park =x+17 meters

∴ perimeter of the park =2(length+breadth)

=2(x+17+x) meters

=2(2x+17) meters

But it is given that the perimeter of the rectangle is 178 meters

∴ 2(2x+17)=178

4x+34=178

4x=178−34

4x=144

x=

4

144

=36

Therefore the breadth of the park =36 meters

Length of the park =36+17=53 meters.

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Answer 2

Answer:  Length = 53 cm and Breadth = 36 cm.

Step-by-step explanation:

We know that the length of the rectangle exceeds its Breadth by 17 m.

So let us take breadth as y m.

Then, ultimately the length would be y + 17 m.

Now lets come to the main part of the question.

We know the perimeter is 178 m.

We also know that the perimeter of a rectangle = 2 * Length + 2 * Breadth or in short form 2(l + b).

Then we can say that 2(l + b) = 178 m.

Now, lets replace the values of Length and Breadth.

We get : 2(y + y + 17)m = 178m.

Then, 2y + 2y + 34m = 178m.

Then, 4y = 178m - 34m

4y = 144m.

Lets cancel 4 on both sides.

We get : y = 144/4 m

y = 36 m.

y = Breadth.

Hence Breadth = 36m.

Now, y + 17m = Length.

Then, Length = 36m + 17m

= 53m.

Therfore we can conclude that Breadth -= 36m and Length = 53m.

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