Question

0-10, 11-20, 21-30, 31-40. The
Real Lower Boundary Of The
Class 21-30 Is​

Answer 1

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } GiveN:-}}}}}$

Ranbir bought a rectangular field of area 24000 square meters.

Widhth of the field is 120 m .

He wants to fence it with two rounds of wire.

$\large{\underline{\underline{\red{\tt{\purple{\leadsto } To\:FinD:-}}}}}$

Length of wire required to fence the field .

$\Large{\underline{\underline{\red{\tt{\purple{\leadsto } AnsweR:-}}}}}$

$\sf \underline{\green{Figure :-}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,12)\linethickness{0.4mm}\put(0,0){\line(1,0){5}}</p><p>\put(0,-0.3){ \bf A }\put(5,-0.3){ \bf B } \put(0</p><p>,3.3){ \bf D } \put(5,3.3){ \bf C }\put(5,0){\line(0,1){3}}\put(5,3){\line(-1,0){5}}\put(0,3){\line(0,-1){3}}\put(2.5, -0.3){ \bf 200 m } \put(2.5,3.3){ \bf 200 m } \put(5.2,1.5){ \bf 120 m } \put(-1, 1.5){ \bf 120 m } \end{picture}$

Given that area of the field is 24000 m² .

$\sf \underline{\green{We\:know\:Area\:of\: rectangle \:as. }}$

$\large{\pink{\boxed{\bf Area _{rectangle}=lenght\times Breadth. }}}$

And , the Breadth here is 120 m.

$\tt:\implies Area_{rec.}=(lenght)(breadth)$

$\tt:\implies 24000 m^2 = 120m \times l$

$\tt:\implies l = \dfrac{24000m^2}{120m}$

$\underline{\boxed{\red{\tt:\longmapsto length = 200m}}}$

$\rule{200}2$

$\sf \underline{\green{We\:know\: Perimeter\:of\: rectangle \:as. }}$

$\large{\blue{\boxed{\bf Perimeter _{rectangle}=2(lenght + Breadth) }}}$

$\tt:\implies Perimeter_{rec.}=2(l+b)$

$\tt:\implies Perimeter = 2(200m + 120m)$

$\tt:\implies Perimeter = 2\times 320m$

$\underline{\boxed{\red{\tt:\longmapsto Perimeter = 640m }}}$

Now , since he wants to fence two rounds of wire , hence total required lenght will be

= 2 × 640 m = 1280 m .

Answer 2

Answer:

Answer:21

 Step-by-step explanation:

Given class is 21 to 30.

therefore the lower limit is 21

the upper limit is 30.

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