# Shikha and Neetu are friends. They both are very health conscious. They jog daily in the morning. Shikha jogs around a rectangular park of dimensions (ab +bc) m and (ab-bc)m. Neetu jogs around a square park of side (2x+3y)m. i) How much distance will Shikha cover in 5 rounds? ii) How much distance will Neetu cover in 1 round? iii) Find the area of each park.

Dear Student,

Please find below the solution to the asked query:

Given :  Shikha jogs around a rectangular park of dimensions ( ab + bc ) m and ( ab - bc ) m .

And

Neetu jogs around a square park of side ( 2x + 3y ) m .

i ) We know perimeter of rectangle  =  2 ( Length + Width ) , So

Distance travel in one round of that rectangular park = 2 ( ab + bc +  ab - bc ) = 2 ( 2 a b ) = 4 ab  m

Therefore ,

Distance travel in 5 round of that rectangular park = 5 ( 4 ab ) =  20 ab  m                                            ( Ans )

ii ) We know perimeter of square  = 4 ( Side ) , So

Distance travel in one round of that square park = 4 ( 2x + 3y )  =  ( 8 x + 12 y ) m                                ( Ans )

iii )  We know area of rectangle  =  Length $×$ Width , So

Area of rectangular park  = ( ab + bc ) ( ab - bc ) =  ( a2 b2 - b2 c2 ) m2                                                    ( Ans )

And

We know area of square =  ( Side )2  , So

Area of rectangular park  = ( 2x + 3y )2=  ( 4 x2 + 9 y2 + 12 x y ) m2                                                          ( Ans )

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• 2
i) Distance Shika will cover in 5 rounds

5*Perimeter
=5* 2(ab+bc+ab-bc)
=5* 2(2ab)
=5* 4ab
=20ab

Therefore Shika will cover in (20ab)m in 5 rounds

ii) Distance Neetu will cover in 1 round
Perimeter of a Square = 4*Side

=4*(2x+3y)
​=8x+12y

Therefore Neetu will cover (8x+12y)m in 1 round

iii) Area of 1st park = Length*Breadth

=(ab+bc)(ab-bc)
=ab(ab-bc) +bc(ab-bc)
=a2b2 - ab2c + ab2c - b2c2
=a2b2 - b2c2

Area of 2nd park = Side*Side

=(2x+3y)(2x+3y)
=(2x+3y)2
=(2x)2  + (2*2x*3y) + (3y)​2
=4x​2 + 12xy + 9y​2

Hope that helps

• 8
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