मराठी

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 - Mathematics

Advertisements
Advertisements

प्रश्न

In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m2 [see figure]. Find the length and breadth of the pond.

बेरीज
Advertisements

उत्तर

Given that a rectangular pond has to be constructed in the centre of a rectangular lawn of dimensions 50 m × 40 m

So, the distance between pond and lawn would be same around the pond.

Say x m.

Now, length of rectangular lawn (l1) = 50 m and breadth of rectangular lawn (b1) = 40 m

Length of rectangular pond (l2)= 50 – (x + x) = 50 – 2x

And breadth of rectangular pond (b2) = 40 – (x + x)= 40 – 2x

Also, area of the grass surrounding the pond = 1184 m2

Area of rectangular lawn – Area of rectangular pond = Area of grass surrounding the pond

l1 × b1 – l2 × b2 = 1184   ......[∵ Area of rectangle = length × breadth]

⇒ 50 × 40 – (50 – 2x)(40 – 2x) = 1184

⇒ 2000 – (2000 – 80x – 100x + 4x2) = 1184

⇒ 80x + 100x – 4x2 = 1184

⇒ 4x2 – 180x + 1184 = 0

⇒ x2 – 45x + 296 = 0   

⇒ x2 – 37x – 8x + 296 = 0  ....[By splitting the middle term]

⇒ x(x – 37) – 8(x – 37) = 0

⇒ (x – 37)(x – 8) = 0

∴ x = 8

At x = 37,

Length and Breadth of pond are – 24 and – 34, respectively but length and breadth cannot be negative.

So, x = 37 cannot be possible

∴ Length of pond = 50 – 2x

= 50 – 2(8)

= 50 – 16

= 34 m

And breadth of pond = 40 – 2x

= 40 – 2(8)

= 40 – 16

= 24 m 

Hence, required length and breadth of pond are 34 m and 24 m, respectively.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 4: Quadatric Euation - Exercise 4.4 [पृष्ठ ४३]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 10
पाठ 4 Quadatric Euation
Exercise 4.4 | Q 7 | पृष्ठ ४३

संबंधित प्रश्‍न

Solve the following quadratic equations by factorization:

(x − 4) (+ 2) = 0


Solve the following quadratic equations by factorization:

4x2 + 4bx - (a2 - b2) = 0


The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find two numbers.


A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for a journey. Find the original speed of the train.


If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?


Solve:

`1/(x + 1) - 2/(x + 2) = 3/(x + 3) - 4/(x + 4)`


Solve the following quadratic equation by factorisation.

 2y2 + 27y + 13 = 0


If ax2 + bx + c = 0 has equal roots, then c =


If a and b are roots of the equation x2 + ax + b = 0, then a + b =


A farmer wishes to grow a 100m2 rectangular vegetable garden. Since he was with him only 30m barbed wire, he fences 3 sides of the rectangular garden letting the compound of his house to act as the 4th side. Find the dimensions of his garden .


Five years ago, a woman’s age was the square of her son’s age. Ten years hence, her age will be twice that of her son’s age. Find:

  1. the age of the son five years ago.
  2. the present age of the woman.

Solve the following equation by factorization

x (2x + 1) = 6


Solve the following equation by factorization

`sqrt(3)x^2 + 10x + 7sqrt(3)` = 0


Solve the following equation by factorization

`(1)/(2a + b + 2x) = (1)/(2a) + (1)/b + (1)/(2x)`


Sum of two natural numbers is 8 and the difference of their reciprocal is `2/15`. Find the numbers.


A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.


The hotel bill for a number of people for an overnight stay is Rs. 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight. 


Forty years hence, Mr. Pratap’s age will be the square of what it was 32 years ago. Find his present age.


Solve the following quadratic equation by factorization method.

3p2 + 8p + 5 = 0


The product of two successive integral multiples of 5 is 300. Then the numbers are:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×