Question

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A rectangular park is to be designed whose breadth is 3m less than its length.Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m.Find it's length and breadth



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class 10 question ​

Answer 1

$\underline{\underline{\red{\bold{\: \: SOLUTION\: \: }}}}$➡

$\bold{\underline{For \: \: the \: \: reactangular \: \: park}} \\ \\ \bold{Let,} \\ \\ \bold{Breadth = x \: \: metre} \\ \\ \bold{So, } \\ \\ \: \bold{ Length= (x + 3) \: \: metre} \\ \\ \bold{Hence,} \\ \\ \bold{Are a= Length \times Breadth} \\ \\ = \bold{((x + 3) \times x) \: \: sq.m} \\ \\ \bold{ = (x {}^{2} + 3x) \: \: sq.m }$

$\bold{\underline{For \: \: the \: \: park \: \: which \: \: is \: \: in \: \:the \: \: shape \: \: }}\\ \bold{\underline{of \: \: isosceles \: \: triangle} }\\ \\ \bold{Given,} \\ \\ \bold{Base \: \: of \: \: the \: \: triangular \: \: park = breadth }\: \\ \bold{of \: \: the \: \: reactangular \: \: park} \\ \\ \bold{ = > Base = x \: \: metre} \\ \\ \bold{Altitude = 12 \: \: metres \: \: \: \: \: } \\ \\ \bold{Area = \frac{1}{2} \times Base \times Altitude} \\ \\ \bold{ =( \frac{1}{2} \times x \times 12) \: \: \: sq.m} \\ \\ \bold{ = 6x \: \: \: \: sq.m}$

$\bold{\underline{According \: \: to \: \: question,}}$

$\bold{ x {}^{2} + 3x = 6x + 4 } \\ \\ \bold{ = > x {}^{2} + 3x - 6x - 4 = 0} \\ \\ = > \bold{x {}^{2} - 3x - 4 = 0} \\ \\ \\ \bold{ Here,} \\ \\ \bold{a = 1} \\ \bold{ b= - 3} \\ \bold{ c= - 4}$

$\bold{\underline{Using \: \: quadratic \: \: formula}}$

$\bold{x = \frac{ - b \pm \sqrt{b {}^{2} - 4ac \: \: } }{2a} } \\ \\ \bold{ = \frac{ - ( - 3) \pm \sqrt{( - 3) {}^{2} - 4 \times 1 \times ( - 4) \: } }{2 \times 1} } \\ \\ \bold{ = \frac{ 3 \pm \sqrt{9 + 16} }{2} } \\ \\ \bold{ = \frac{ 3 \pm \sqrt{25} }{2} } \\ \\ \bold{ = \frac{ 3 \pm 5 }{2} } \\ \\ \\ \bold{Now ,} \\ \\ \bold{x = \frac{3 + 5}{2} } \\ \\ = > \bold{x = \frac{8}{2} = 4} \\ \\ \bold{Or,} \\ \\ \bold{x = \frac{3 - 5}{2} } \\ \\ \bold{ = > x = \frac{ - 2}{2} = - 1}$

$\bold{Here, \: \: x \: \: is \: \: not \: \: equal \: \: to \: \: ( - 1)} \\ \\ \bold{So,} \\ \\ \bold{x =\boxed{ 4}} \\ \\ \bold{And,} \\ \\ \bold{x + 3 = 4 + 3 =\boxed{ 7}} \\ \\ \\ \bold{Hence,} \\ \\ \bold{Length \: \: of \: \: the \: \: reactangular \: \: park = 7 \: \: metres} \\ \\ \bold{and} \\ \\ \bold{Breadth \: \: of \: \: the \: \: reactangular \: \: park = 4 \: \: metres.}$

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$\underline{\underline{\bold{\red{\: \: VERIFICATION\: \: }}}}$➡

$\bold{\underline{For \: \: the \: \: reactangular \: \: park}} \\ \\\bold{Length = 7 \: \: m\: \: and \: \: Breadth = 4 \: \: m} \\ \\ \bold{Area = Length \times Breadth} \\ \\ \bold{ =( 7 \times 4) \: \: sq.m = 28 \: \: sq.m} \\ \\ \\ \bold{\underline{For \: \: the \: \: triangular \: \: park}} \\ \\ \bold{Altitude = 12 \: \: m \: \: and \: \: Base = 4\: \: m} \\ \\ \bold{Area = \frac{1}{2} \times Base \times Altitude} \\ \\ \bold{ = (\frac{1}{2} \times 4 \times 12) \: \: sq.m} \\ \\ \bold{ = 24 \: \: sq.m} \\ \\ \bold{So,} \\ \\ \bold{Area \: \: of \: \: triangular \: \: park + 4 \: \: sq.m} \\ \\\bold{ =(24 + 4) \: \: sq.m} \\ \\ \bold{ = 28 \: \: sq .m} \\ \\ \bold{ = Area \: \: of \: \: the \: \: reactangular \: \: park}$

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✝$\mathfrak{\red{THANKS..}}$✝

Answer 2

$\huge\red{hi \: here \: your \: answer \: dear}$

Let the length be l

therefore breadth=l-3

area=l(l-3)

the area of isosceles triangle = 1/2×(l-3)12

6(l-3)

given:l(l-3)-4=6(l-3)

l²-3l-4=6l-18

l²-9l+14=0

l²-2l-7l+14l=0

l(l-7)-2(l-7)=0

(l-2)(l-7)=0

l=7or 2

and the breadth is 4

since breath cannnot be negative.