A problem using quadratic eqns again...please answer..?

Question

one side of a rectangle is 1 dm less than its diagonal. if the perimeter of the rectangle is 62 dm, find the area of the triangle. show ur solution plz. tnx

Answer 1

let a&b be the sides of the rectangle. so a+b=31.
and we have a^2+b^2 = (a+1)^2 which a+1 is the lenght of diagonal. if you make the equation simple the result will be
b^2 = 2.a+1.if you put 31-b instead of a you'll have
b^2 + 2.b - 63 = 0. this is a simple equation that you can find b from it an therefor you'll find a. from a & b you can find the area.

Answer 2

Define:
D = diagonal of rectangle
L = length of rectangle and let it be = D - 1
W = width of rectangle

Area = L*W so you need to find L and W

Perimeter = sum of all sides = 62
Perimeter = L + W + L + W = 2L + 2W = 2(D - 1) + 2W = 62
And: W = 32 - D

The diagonal along with L and W form a right triangle with D as the longest side. And so:
D^2 = L^2 + W^2 = (D - 1)^2 + W^2
D^2 = D^2 - 2D + 1 + W^2
W = SQRT(2D - 1)

From the perimeter calculation above:
W = 32 - D
Now use: W = SQRT(2D - 1)
SQRT(2D - 1) = 32 - D
2D - 1 = 1024 - 64D + D^2
D^2 - 66D + 1025 = 0

D = (66 +/- SQRT(4356 - 4100))/2
D = (66 +/- 16)/2 = 50/2 = 25 (the other value is not a good solution since it gives a negative width)

So D = 25
W = 32 - 25 = 7
L = D - 1 = 24

Area = L*W = 24*7 = 168

Answer 3

let x be the width, y be the length and z be the diagonal

x = d - 1

P = 2(x + y)
62 = 2(d - 1 + y)
31 = d + y - 1

pythagarian theorem.
a^2 + b^2 = z^2

the width and the length are the legs
x^2 + y^2 = z^2
(z - 1)^2 + y^2 = z^2
z^2 - 2z + 1 + y^2 = z^2
-2z + 1 + y^2 = 0

so you have three equations
x = z - 1
31 = z + y - 1
-2z + 1 + y^2 = 0


y = 32 - z

-2z + 1 + (32 - z)^2 = 0
-2z + 1 + 1024 - 64z + z^2 = 0
z^2 -66z + 1025 = 0
(z - 41) (z - 25) = 0
z = 41 or 25

y = 32 - z
y = 32 - 41
y = -9 (can not have a negative distance so 41 is not the answer)


the diagonal is 25, the width is 7 and the length is 24

Area = xy
Area = 7 x 24
Area = 168 dm^2

Answer 4

diagonal = d
side 1 = d -1
let the second side = x
perimeter = 62 = 2 ( d -1 + x) = 2d -2 + 2x
x = (64 - 2d) / 2 = 32 -d
by pythagorean theorem
d^2 = (d - 1)^2 + ( 32 - d )^2
d^2 = ( d^2 - 2d +1) + ( 1024 - 64 d + d^2 )
d^2 - 66d + 1025 = 0
(d - 25) (d - 41 ) = 0
d = 25 . . . . this is the diagonal
first side = 24 dm
second side = 32 - d = 7 dm
area = 24 (7) = 168 sq.dm

Answer 5

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I felt abused

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