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A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 meters, assuming the width to be x, form an equation in x and solve it to find the value of x.

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Abhishek Das Posted

A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 meters, assuming the width to be x, form an equation in x and solve it to find the value of x.

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Candace N

To visualize, first draw a rectangle with dimensions 10x16m. Then draw another rectangle outside this one (showing the walkway) that’s 16+2x on one side and 10+2x on the other side. X is the width of the walkway. The equation to solve is (10+2x)(16+2x)-160=120. The answer is x=2.

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Naren Kartik

A rectangle has dimensions 10x16m.

the walkway has dimensions 16+2x on one side and 10+2x on the other side

So the equation becomes (10+2x)(16+2x)-160=120.

And the final answer is x=2

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Mohan Naik

 The rectangle has dimensions 10x16m. and is surrounded by a walk of uniform width (of say x) and total area of  the walk  is 120 mts.

So the equation is : 2(10x)+2(16x)+4(x X x)=120

So,

4(x X x)+ 52x-120=0 When  we solve this, we get a positive solution as x=2 

So width is 2 mts

By. monaik

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ANKUR MUNDRA

total area of the walk = 2*(10*x+16*x)+4*(x*x)=120

i.e. 4x^2+52x=120, or x^2+13x=30, i.e., x^2+13x-30=0, i.e. x^2+15x-2x-30=0

i.e. x(x+15)-2(x+15)=0. i.e., (x+15)(x-2)=0, as x>0, x=2

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Neha Kajla

First let us see what the question is demanding and then provide as per the demand of the question we will proceed step wise to get the exact answer.

Given –

The length and width of rectangle = 10 and 16

from which we can find the area of rectangle

formula for area of rectangle = Length x width

As per the question

Area of the rectangle = 16 x 10 = 160 m ( we will call this outer rectangle).

Area of other rectangle after the construction of wall = 120 (given in question. we will call this inner rectangle).

Demand of question is to form and equation in “x” and then solve for “x”

area of inner rectangle = area of outer rectangle (walk+ wall) – area of inner rectangle(walk only)

120 (given in question) = 10 *16 – (16+2x) * (10+2x)

120 = 160 – [10(16+2x)+2x(16+2x)]

120=160 – [10*16+10*2x + 2x*16+2x*2x]

120=160 – [160 + 20x + 32x + 4 x²  ]

120 = 160 – [160 + 52x +4 x²  ]

120 = 160 – 160 -52x – 4 x²

120 = -52x -4 x²

4 x²  +52x -120 = 0

Now solve for x

4 x²  + 52x – 120 = 0

Dividing the whole equation by 4, we get

x²  + 13 x – 30 = 0

now, using trial and error method we will find the best combination,

we have to find such  nos.which on addition or subtraction produces 13 and the same nos. if multiplied produce 30

nos. will be 15-23 = 13 and 15*2=30

putting these in above equation we get,

x²  + 15x -2x – 30 = 0

x(x+15) – 2 (x + 15) = 0

(x+15) and (x-2) = 0

x=-15 and x = 2

-15 is not acceptable hence x= 2 is the answer.

 

 

 

 

 

 

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