The perimeter and area of a rectangle are 22cm and 30cm2.find the length and breadth of the rectangle?

Question

ah was easy i almost did it just had to think abit more

well thnx everyone

Answer 1

1*10 = 10
2* 9 = 18
3* 8 = 24
4* 7 = 28
5* 6 = 30


.... not that hard

Answer 2

Perimeter of rectangle formula:
P = 2L + 2W

Area of rectangle formula:

A = LW

We know that P = 22 and A = 30, so

22 = 2L + 2W
30 = LW

Divide the first equation by 2.

11 = L + W
30 = LW

To solve for L and W, use substitution; solve for L in the first equation, and then substitute into the second.

L = 11 - W

Plugging this into 30 = LW, we get

30 = (11 - W)W
30 = 11W - W^2
W^2 - 11W + 30 = 0
(W - 6)(W - 5) = 0

Therefore,

W = {5, 6}

Since L = 11 - W, we can derive values for L

If W = 5:
L = 11 - 5 = 6

If W = 6:
L = 11 - 6 = 5

We get the set of equations
L = 6, W = 5
L = 5, W = 6

But they are equivalent as far as rectangle dimensions are concerned.

The dimensions of the rectangle with perimeter 22cm and area 30cm^2 is 5 x 6.

Answer 3

22 = 2(l + w)
11 = l + w
l = 11 - w

30 = l*w
30 = (11 - w) w
0 = - w² + 11w - 30
0 = - (w² - 11w + 30)
0 = - (w - 6)(w - 5)
w = 6 or 5

if w = 6 then l = 5
if w = 5 then l = 6

so the dimentions are 6cm by 5cm

Answer 4

w + h = 11; wh = 30; by inspection, w = 5 and h = 6 (or vice versa). If the answer were not obvious, some algebra and a whack with the quadratic formula would bring it out.

Answer 5

let sides be x and y.
Then Area, a = xy = 30
Perimeter, 2x + 2y = 22
Then x + y = 11
x = 11-y
Substitute into area,
y(11-y)=30
y^2 -11y + 30 = 0
(y-5)(y-6)=0
y = 5 or 6
Then x = 6 or 5

So sides of rectangle are 5 and 6 cm

Answer 6

2L + 2B = 22
L.B = 30
L = 30 / B
60 / B + 2B = 22
60 + 2 B² = 22 B
B² - 11 B + 30 = 0
(B - 6).(B - 5) = 0
B = 6 , B = 5
L = 5 , L = 6
Dimensions are 6 cm by 5 cm