Question

The sides of a triangular field are 51 m , 37 m and 20 m . find the number of flower beds that can be prepared if each bed is to contain 2×3 sq m space.​

Answer 1

$\large\boxed{\textsf{\textbf{\blue{S0LUTI0N\::-}}}}$

$\underline{\bf{\dag} \:\mathfrak{step\;by\;step\; explanation\: :}}$⠀

s = a+b+c/2

$51 +37+20/2$

$\mapsto$54

$area = √s(s-a)(s-b)(s-c)$

$= √54 (54-51)(54-37)(54-20)$

$=√54(3)(17)(34) {simplest form}$

$= √(3*2*3*3)*(3)*(17)*(17*2)$

$= 3*2*3*17$

$= 306m²$

area needed for flower beds

$\mapsto$ 306/9

$\mapsto$ 34

${\underline{\textsf{ \textbf{∴34 flower beds are required to occupy the space}}}}$

Answer 2

${\large{\bf{\red{\clubs{\underline{Given :}}}}}}$

The sides of a triangular field are 51 m , 37 m and 20 m .

￣￣￣￣￣￣￣￣￣￣￣￣

${\large{\bf{\blue{\clubs{\underline{To Find :}}}}}}$

find the number of flower beds that can be prepared if each bed is to contain 2×3 sq m space.

￣￣￣￣￣￣￣￣￣￣￣￣

${\large{\bf{\orange{\clubs{\underline{Solution :}}}}}}$

${\blue{\bigstar{\mathfrak{\pink{\boxed{S = \frac{a + b + c}{2} }}}}}}$

${\sf{\green{= \frac{51 + 37 + 20}{2} }}}$

${\sf{\green{= {\cancel\frac{108}{2} }}}}$

${\sf{\red{s= 54}}}$

Than,

${\blue{\bigstar{\pink{\boxed{area=√s(s−a)(s−b)(s−c)}}}}}$

${\sf{\green{= \sqrt{54 (54-51)(54-37)(54-20} )}}}$

${\sf{\green{= \sqrt{54 \times (3) \times (17) \times (34} )}}}</u></p><p><u>[tex]{\sf{\green{= \sqrt{54 \times (3) \times (17) \times (34} )}}}$

${\sf{\green{= \sqrt{93636} }}}$

${\sf{\red{= 306 {m}^{2} }}}$

Square beds can be contained :

${\sf{\green{Area \: of \: 1 \: bed = 2 \times 3 = 6 {m}^{2} }}}$

${\sf{\green{Area \: of \: triangular \: field = 306 {m}^{2} }}}$

${\sf{\green{Beds \: can \: be \: contained = {\cancel \frac{306}{6} }}}}$

${\underbrace{\bf{\red{51 }}}}$

So,

51 beds could be contained inside this triangular field.

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${\bf{\underline{\fcolorbox{green}{red}{\blue{XitzNobitaxX}}}}}$