The length of a rectangle is 3 cms more than the width, The diagonal is 87 cms. What are its length and width?

Question

The width and length are both integers,

Answer 1

It helps if you know something about Pythagorean triples (integer solutions to the Pythagorean equation for right-angled triangles).

If you do, you will know that there are no primitive triples in which the difference between the legs is 3, The smallest such differences are 1, 7 and 17 (always prime, but not all primes will serve).

Therefore this must be a non-primitive triple which is a multiple by three of a primitive triple with the legs differing by 1. There is a series of such triples, starting with (3, 4, 5) which can be generated from one another.

So we divide 87 by 3 and get 29 and recognise that this must be the triple (20, 21, 29), the second in the series.

So plainly the answer is length = 63 cms and width = 60 cms (i.e. 3 x 21 and 3 x 20)

THE ALGEBRA

L = B + 3
L^2 + B^2 = 87^2
2 B^2 + 6B + 9 = 7569
B^2 + 3B - 3780 = 0
(B - 60) (B + 63) = 0

As rectangles in the real world have sides of positive value, ignore B = -63 and take B = 60, yielding L = 63,

But none of that is necessary if you know your triples!

Answer 2

here,we are dealing in cm

let L=length
W=width=L-3

using pythagoras

L^2+(L-3)^2 = 87^2
L^2+L^2-6L+3^2=87^2
2L^2-6L+3^3-87^2=0
2L^2-6L +(3+87)(3-87)=0
L^2-3L-90*42=0
(L-63)(L+60)=0

>>>L=63 or -60

take+ve value for L

but W=L-3 =63-3=60

therefore, length=63cm
width=60cm

the other answer makes sense-you
are simply rotating the object by pi
about the origin (multiply by i^2)
from the 1st to the 3rd quadrant and
interchanging the length and width

{primitive form of this triangle is 20,21,29
triangles of the form
(m^2-n^2)^2+(2mn)^2
=(m^2+n^2)^2
are called pythagorean triangles
here m=5 and n=2}

i hope that this helps

Answer 3

you said,,length = 3 and width is 2, as it is a rectangle length of the diagonal = square root of 3^2 + 2^2= sqrt of 13

Answer 4

...LET...
length = x
width = y
diagonal = z

...WE KNOW THAT...
x = y + 3
z = 87

...SO, EQUATIONS...
from pythagoras, x^2 + y^2 = z^2
but z = 87 and x = y+3
therefore: (y+3)^2 + y^2 = 87^2
...open out brackets and simplify...
y^2 + 6y + 3^2 + y^2 = 87^2
2y^2 + 6y + 9 = 7569
2y^2 + 6y = 7569-9 = 7560
... divide both sides by 2...
y^2 + 3y = 3780
y^2 + 3y - 3780 = 0

now we can solve this quadratically using the quadratic equation y = (-b+- (b^2-4ac)^0.5)/2a
y = (-3 +- (3^2 - 4(1)(-3780)^0.5)/2(1)
y = (-3 +- (9--15120)^0.5)/2
y = (-3 +- square root of 15129)/2
y = 1/2(-3+123) .... ignore the negative result as a length cannot be less than zero!!...
y = 1/2 (120)
y = 60cm.

therefore...
x = 60 + 3 = 63 cm

Hope that makes sense!! Can't set out the formulas the easy way to visualise them on computer!
...SO...
length = 60cm
width = 63cm

Answer 5

The length is 63 cm and the width is 60 cm.


To solve it you use the Pythagorean theorem to solve for the width, x and the length, x+3. You use the quadratic formula and solve, then check.

Answer 6

Another good way to solve is:
A^2+B^2=C^2
If one side is 3cm greater
set your short side to X and the side 3cm greater to X+3
Then plug it into the formula
(x+3)^2+(X)^2=(87)^2 Then solve for X
X can = -63 or 60 Take the positive side.X= 60
Then add 3 to it.
So the length = 63cm, and the width = 60cm

Answer 7

L = W + 3 cm
L^2 + W^2 = 87^2
(W + 3)^2 + W^2 = 7,569
2W^2 + 6W + 9 = 7,569
W^2 + 3W - 3,780 = 0
(W + 63)(W - 60) = 0
Negative length not being permitted,
W = 60 cm
L = 63 cm

3,600 + 3,969 = 7,569

Answer 8

x^2 + (x+3)^2 = 87^2 by pythagoras

2x^2 + 6x - 7560 = 0

x^2 + 3x - 3780 = 0

The square root of 3780 is 61.48.. The numbers need to be 3 apart, so it should be -60 and 63

(x + 63)(x - 60) = 0

So x = 60 is the answer (the other answer is negative)

Answer 9

It's a 20 X 21 X 29 triangle, 3times the size

60x63x87

Answer 10

The square of the hypotenuse is equal to the sum of the squares of the opposite sides.
The hypotenuse is 87, the width is X and the length is X+3
Work it out!