A rectangle is 8cm longer than it is wide. The perimeter of the rectangle is 72cm.
a) Find the dimensions of the rectangle.
b) Find the area of the rectangle.
THANKYOU!!!
2007-06-13 00:23:10 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x+8 = y
2x + 2y =72
2x + 2(x+8) = 72
4x + 16 = 72
4x =56
x =14
y= x + 8 = 22
Do some work to find the area
2007-06-13 00:28:33 · answer #1 · answered by El Gigante 4 · 2⤊ 0⤋
to solve this u set up an algebraic equation 4x+16=72. this is because there are 4 sides (4x) and two sides are 8 longer (+16). and the perimeter is 72. I solved this and found the dimensions are 14 cm by 22 cm. Thus the area is 308 cm^2.
your welcome
P.S. everybody said u need "2 equations" not true I think mine is the best answer by the way.
2007-06-13 01:37:03 · answer #2 · answered by Anonymous · 1⤊ 0⤋
a. Perimeter = distance around, so add up the sides. A rectangle has two widths + two lengths. The perimeter = 72 cm. (Given). If the length is 8 cm longer, then the length is x + 8 and the width is x, so the answer is
x + x (the two widths) + (x + 8) + (x+8) (the two lengths) = 72 cm.
b. Area = width X length, so x (the width) X (x + 8) (the length) = 72 cm
2007-06-13 00:41:40 · answer #3 · answered by alialoggi 7 · 1⤊ 0⤋
(a) length, L = width, W + 8 ---> Equation (1)
L + L + W + W = 72
---> 2L + 2W = 72 ---> Equation (2)
Substitute Equation (1) into Equation (2),
2 (W + 8) + 2W = 72
2W + 16 + 2W = 72
4W = 72 - 16 = 56
W = 56/4 = 14cm
Substitute the value of W into Equation (1),
L = 14 + 8 = 22cm
(b) Area of rectangle = L * W = 22 * 14 = 308 cm²
2007-06-13 01:48:43 · answer #4 · answered by jurassicko 4 · 1⤊ 0⤋
Simple, 2 unknowns, 2 equations. Let l=length and w=breadth
l=8+w, l-w=8
2l+2w=72
Then you can 'subtract' the 2 equations: (2l+2w)-2(l-w)=72-2*8
4w=56
Solve for w, sub back in to get l, multiply l with w.
:)
2007-06-13 00:32:06 · answer #5 · answered by tgypoi 5 · 1⤊ 0⤋