The length and breadth of Mihir’s rectangular garden is 50m. and 30m. respectively. There is a road in the middle of the garden of width 4m. parallel to the length of the garden. This road divides the garden into two rectangular region of equal area. Let’s draw a picture and find out the area of the road and note it down.

A. If the road of width 4 m passing through the middle of the garden is parallel to the breadth and this road divides the garden into two equal parts, then let’s draw a picture and find the area of the road and note it down.

B. If there are two roads of width 4m parallel to the length and breadth of Mihir’s garden passing through the middle and divides the garden into four equal parts, then let’s draw a picture to find the area of the road and note it down.

According to the question,

Let Mihir’s garden be ABCD.

Then, AB = 50 m

BC = 30 m

A road EFGH is in the middle of this garden, of width 4 m.

⇒ EF = AB = 50 m

And FG = HE = 4 m

The road is of dimension (50 × 4) m².

We need to find the area of the road.

Area of EFGH = EF × FG [∵, Area of rectangle = length × breadth]

⇒ Area of EFGH = 50 × 4

⇒ Area of EFGH = 200 m²

Hence, area of the road is 200 m².

(A). We have,

Given that, the road EFGH is parallel to the breadth of ABCD.

Then, HE = 4 m [as given]

EF = AD = 30 m

Then, area of the road EFGH is given as,

Area of EFGH = EF × HE [∵, Area of rectangle = length × breadth]

⇒ Area of EFGH = 30 × 4

⇒ Area of EFGH = 120 m²

Hence, area of the road is 120 m².

(B). We have

According to the question,

There are two roads of width 4 m, parallel to each length and breadth.

Since, these roads divide the rectangle ABCD into four equal parts.

We can find QE and MQ.

If DA = 30 m and PM = 4 m. Then,

AM + PD = DA – PM

⇒ AM + PD = 30 – 4

⇒ AM + PD = 26

⇒ AM + AM = 26 [∵, AM = PD as these roads divide ABCD into 4 equal parts]

⇒ 2 AM = 26

⇒ AM = 13 m

Thus, AM = PD = 13 m

Also, AM = QE = 13 m

And, PD = TF = 13 m

Similarly, if AB = 50 m and EH = 4 m. Then,

AE + HB = AB – EH

⇒ AE + HB = 50 – 4

⇒ AE + HB = 46

⇒ AE + AE = 46 [∵, AE = HB as these roads divide ABCD into 4 equal parts]

⇒ 2 AE = 46

⇒ AE = 23 m

Thus, AE = HB = 23 m

Also, AE = MQ = 23 m

And, HB = RN = 23 m

Area of QRST is given by,

Area₁ = 4 × 4 [∵, QRST is a square with dimensions 4 m each]

⇒ Area₁ = 16 m² …(i)

Area of EHRQ is given by,

Area₂ = 4 × 13 [∵, EH = 4 m & QE = 13 m]

⇒ Area₂ = 52 m²

So,

Area of EHRQ = Area of FGST = 52 m² …(ii)

Similarly, area of ONRS is given by,

Area₃ = 4 × 23 [∵, ON = 4 m and RN = 23 m]

⇒ Area₃ = 92 m²

So,

Area of ONRS = Area of PMQT = 92 m² …(iii)

From equation (i), (ii) and (iii), we get

Area of road = Area of QRST + Area of EHRQ + Area of FGST + Area of ONRS + Area of PMQT

⇒ Area of the road = 16 + 52 + 52 + 92 + 92

⇒ Area of the road = 304 m²

Hence, area of the road is 304 m².