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Hint: Try to get an equation in x, using the area of the path outside the park by dividing the path in 4 small rectangles and finding the area individually and then at last adding them all to get \[305{m^2}\] . In this way we can get the value of x and then putting it in \[5x\] & \[ 2x\] we can get the dimensions of the park.

Complete step-by-step answer:

According to the question it is given that there's a 2.5m path running all along the perimeter of the rectangular park

Therefore, \[TD = DU = VC = WC = XA = AO = BN = BM = 2.5m\]

Again from the Figure it is clear that \[SR = UW = XM = PQ\] and also

\[SP = TO = VN = RQ\]

As \[CB = AD = 2x\] & \[ DC = AB = 5x\]

Then, \[UW = XM = 5x + 5\]

Again it can be observed that Area of rectangle UWRS = Area of rectangle XMQP

Also, Area of rectangle UDAX = Area of rectangle CWMB

\[\therefore \] Area of Path = Area of UDAX + Area of CWMB + Area of XMQP + Area of UWRS

\[\therefore \] Area of Path = 2(Area of UDAX + Area of XMQP)

Now Area of UDAX = \[UD \times AX = 2x \times 2.5 = 5x\]

Area of XMQP = \[XM \times MQ = 2.5(5x + 5)\]

\[\therefore \] Area of Path = \[305{m^2}\]

\[\begin{array}{l}

\Rightarrow 305 = 2\{ 2.5(5x + 5) + 5x\} \\

\Rightarrow 305 = 5(5x + 5) + 10x\\

\Rightarrow 305 = 25x + 25 + 10x\\

\Rightarrow 305 = 35x + 25\\

\Rightarrow 280 = 35x\\

\Rightarrow x = 8

\end{array}\]

Now as we have the value of \[x = 8\] Then it is given that dimension are 5x and 2x

\[\therefore 5x = 40\] & \[ 2x = 10\]

Therefore the dimensions of the rectangular park are 10m and 40m.

Note: One must be careful while observing the diagram because that is the key to the whole question and also while adding the length of the path in both sides of the rectangular sides. For the side CD(rectangular side) the lengths CW and DU must be added separately, a lot of students missed that and found the wrong answer. Also you can extend the sides of CB and AD as TO and VN which will be \[2x + 5\] and then by proceeding in the same manner we can get the correct answer.

Complete step-by-step answer:

According to the question it is given that there's a 2.5m path running all along the perimeter of the rectangular park

Therefore, \[TD = DU = VC = WC = XA = AO = BN = BM = 2.5m\]

Again from the Figure it is clear that \[SR = UW = XM = PQ\] and also

\[SP = TO = VN = RQ\]

As \[CB = AD = 2x\] & \[ DC = AB = 5x\]

Then, \[UW = XM = 5x + 5\]

Again it can be observed that Area of rectangle UWRS = Area of rectangle XMQP

Also, Area of rectangle UDAX = Area of rectangle CWMB

\[\therefore \] Area of Path = Area of UDAX + Area of CWMB + Area of XMQP + Area of UWRS

\[\therefore \] Area of Path = 2(Area of UDAX + Area of XMQP)

Now Area of UDAX = \[UD \times AX = 2x \times 2.5 = 5x\]

Area of XMQP = \[XM \times MQ = 2.5(5x + 5)\]

\[\therefore \] Area of Path = \[305{m^2}\]

\[\begin{array}{l}

\Rightarrow 305 = 2\{ 2.5(5x + 5) + 5x\} \\

\Rightarrow 305 = 5(5x + 5) + 10x\\

\Rightarrow 305 = 25x + 25 + 10x\\

\Rightarrow 305 = 35x + 25\\

\Rightarrow 280 = 35x\\

\Rightarrow x = 8

\end{array}\]

Now as we have the value of \[x = 8\] Then it is given that dimension are 5x and 2x

\[\therefore 5x = 40\] & \[ 2x = 10\]

Therefore the dimensions of the rectangular park are 10m and 40m.

Note: One must be careful while observing the diagram because that is the key to the whole question and also while adding the length of the path in both sides of the rectangular sides. For the side CD(rectangular side) the lengths CW and DU must be added separately, a lot of students missed that and found the wrong answer. Also you can extend the sides of CB and AD as TO and VN which will be \[2x + 5\] and then by proceeding in the same manner we can get the correct answer.

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