If the top of a rectangular table has an area of 35in^2, and its dimensions have a sum of 12, what are the dimensions?

Length is 8 m and width is 4 m.

Step-by-step explanation:

Given,

The area of the rectangular table = 35 sq m

Sum of its dimensions = 12 m

To find the length and width of the table.

Let, the length and width of the table is l and b respectively.

As per the problem,

lb = 35 and l+b = 12

Formula

(a-b)² = (a+b)²-4ab

Now,

(l-b)² = (l+b)²-4lb

= (12)²-4×35 = 4

so, (l-b) = 2

Now we will solve l+b = 12 and l-b = 4

By adding we get,

l+b+l-b = 12+4

or, 2l = 16

or, l = 8

And b= 12-l = 12-8 = 4

Hence,

Length is 8 m and width is 4 m.

(-1,-0.5)

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2