Question

the perimeter of a rectangle plot is 62m and its area is 228sq metres. find the dimension of the plot​

Answer 1

Answer:

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Step-by-step explanation:

perimeter=2(l+b)

we know that: perimeter=62

2(l+b) = 62

⇒(l+b) = 31

⇒l = (31 -b)

also, area=l×b

we know that: area=228

l×b = 228

⇒(31-b)b = 228

⇒b²- 31b +228 =0

⇒b² - 12b - 19b +228 =0

⇒b(b-12) - 19(b-12) =0

⇒(b-12)(b-19)=0

∴hence b = 12 m and l = 19 m

and length is greater than breadth

Answer 2

Answer:

Step-by-step explanation:

ANSWER:-

Given that:-

• Perimeter = 62 m

• Area = 228 $\sf{m^2}$

We need to find the dimensions of the plot.

We know that:-

$\bf{Perimeter \ of \ Rectangle = 2 \times(Length+Breadth)}$

$\sf{Area \ of \ Rectangle = Length \times Breadth}$

• Let length be "L".
• Let breadth be "B".

With perimeter,

$\sf{2(L+B)=62}$

$\sf{L+B = 31 }$

$\sf{L = 31 - B}$ ____________________(1)

With area,

$\sf{LB = 228}$ _________________________________(2)

Substituting (1) in (2)

$\sf{(31-B)(B) = 228}$

$\sf{31B - B^2 = 228}$

$\sf{B^2 - 31B + 228 = 0}$

Dividing 31 in 12 and 19 (Suitable factors),

$\sf{B^2 - 12B - 19B + 228 = 0}$

$\sf{B(B-12)-19(B-12)=0}$

$\bf{(B-12)(B-19)=0}$

As we have got two values , so there will be 2 values of length also.

(1) L = 31-12 = 19.

(2) L = 31-19 = 12.

The two answers will be:-

• Length is 12 metre and Breadth 19 metre
• Length will be 19 m and Breadth is 12 m.