Answer:
a rectangle is twice as long as it is wide . if both its dimensions are increased 4 m , its area is increaed by 88 m squared make a sketch and find its original dimensions of the original rectangle
Step-by-step explanation:
Let l = the original length of the original rectangle
Let w = the original width of the original rectangle
From the description of the problem, we can construct the following two equations
l=2*w (Equation #1)
(l+4)*(w+4)=l*w+88 (Equation #2)
Substitute equation #1 into equation #2
(2w+4)*(w+4)=(2w*w)+88
2w^2+4w+8w+16=2w^2+88
collect like terms on the same side of the equation
2w^2+2w^2 +12w+16-88=0
4w^2+12w-72=0
Since 4 is afactor of each term, divide both sides of the equation by 4
w^2+3w-18=0
The quadratic equation can be factored into (w+6)*(w-3)=0
Therefore w=-6 or w=3
w=-6 can be rejected because the length of a rectangle can't be negative so
w=3 and from equation #1 l=2*w=2*3=6
I hope that this helps. The difficult part of the problem probably was to construct equation #1 and to factor the equation after performing all of the arithmetic operations.
Answer:
2 different perimeters, 32 cm and 40 cm
Step-by-step explanation:
Given the area of a rectangle is 64 cm squared.
We know, Area of the rectangle is = length x breath
∴ 64 = 8 x 8
Hence the sides can be 8 cm by 8 cm.
So the perimeter of the rectangle is = 8 cm + 8 cm + 8 cm +8 cm
= 32 cm
Also, Area of the rectangle is = length x breath
∴ 64 = 16 x 4
So the perimeter of the rectangle is = 16 cm + 4 cm + 16 cm +4 cm
= 40 cm
<h3>Hello There!!</h3>
Given figure is a <u>parallelogram</u>
<u>To </u><u>Find</u>
Value of"x"
<h3><u>Solution</u></h3>
<h3>Hope this helps</h3>
