Question

3. The shape of a garden is rectangular in the middle and semicircular
at the ends as shown in the diagram. Find the area and the perimeter
T of this garden (Length of rectangle is
7m 20-(35 +3.5) metres ).
​

Answer 1

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Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles

$l.b + 2 \times \frac{1}{2}\pi {r}^{2}$

$= (13 \times 7) {m}^{2} +$

$(2 \times \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5) {m}^{2}$

$= (91 + 38.5) {m}^{2} = 129.5 {m}^{2}$

Perimeter of the garden =2× length of rectangular portion + circumference of the circle

$= (2 \times 13 + 2 \times \frac{22}{7} \times 3.5)m$

$= (26 + 22)m = 48m$

Answer 2

${\large{\bold{Question:-}}}$

The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden.

[Length of rectangle is 20 – (3.5 + 3.5) metres]

${\large{\bold{given:-}}}$

length of rectangle = 20 - (3.5+3.5)

                                  = 20 - 7

                                    = 13m

Area of rectangle = l × b

                                = 13 × 7

                           = 91${m}^{2}$

Area of two circular ends = ${2( \dfrac{1}{2}\:π\:{r}^{2}})$

                                           = ${π\:{r}^{2}}$

                                            = ${ \dfrac{22}{7}\:×\: \dfrac{7}{2}\:×\: \dfrac{7}{2}}$

                                            = ${ \dfrac{77}{2}\:{m}^{2}}$

__________

Total area

                   = Area of rec. + area of two ends

                    = ${91\:{m}^{2}}$ + ${38.5\:{m}^{2}}$

                      = ${129.5\:{m}^{2}}$

__________

Total perimeter

                    = perimeter of rec. + perimeter of two ends

                    = ${2(l+b)+2×(\pi\:r)-2(2\:\pi)}$

                     = ${2×20+22-14}$

                     = ${40+22-14}$

             = ${48m}$