The breadth of a rectangle is 3cm less than its length...?

Question

The breadth of a rectangle is 3cm less than its length. The diagonal of the rectangle is 15cm. Find its length and breadth. What is the length of its other diagonal?

Answer 1

Say the length in centimeters is l
Then the bredth(b) is 3 less than that,
b = l - 3

Now the bredth, length and diagonal(d) form a right triangle,
so we can say d^2 = b^2 + l^2 by pythagorus.
d^2 = (l-3)^2 + l^2
15^2 = (l-3)(l-3) + l^2
225 = l^2 - 6l + 9 + l^2
216 = 2l^2 - 6l div by 2
108 = l^2 - 3l move over the 108
0 = l^2 - 3l - 108
You could use the quadratic formula, or factor.
0 = (l-12)(l+9)
l = 12

b = 12 - 3 = 9
The bredth is 9cm and the length is 12cm.

Since the other diagonal is just the diagonal flipped along the center, it is the same length: 15 cm.

Answer 2

Let's call x the length of the rectangle, so the breadth is equal to x-3
You have to use the Pythagorean theorem:
x^2 + (x-3)^2 = 15^2
If you develop and divide by two you get:
x^2 -3x -108 = 0 which is equivalent to (x-12)(x+9)=0
There are two solutions to this equation: x=12 and x= -9
Since a length cannot be negative, the only possible solution is x=12.
So your rectangle has a length of 12 and a breadth of 9
One of the proprieties of a rectangle is that its diagonals have the same length. So its other diagonal is 15cm long.

Answer 3

breadth 9 cms x length 12 cms
Pythagorean triple is 9^2 + 12^2 = 15^2
(which is simply three times the well-known primitive triple 3^2 + 4^2 = 5^2)
15 cms is the length of both diagonals.

recognising the (9, 12, 15) triple makes it quick and easy (which is how come I worked it out first!) but the tedious long-winded algebra loved by maths teachers goes:

L = B + 3
L^2 + B^2 = 15^2
substitute first equation in second:
B^2 + 6B + 9 + B^2 = 225
2B^2 + 6B = 216
divide by 2 throughout:
B^2 + 3B = 108
B^2 + 3B - 108 = 0
solve the quadratic equation:
(B + 12) (B - 9) = 0

so breadth = -12 or 9
and length = -9 or 12

as real-world rectangles cannot have negative-length sides, discard the -12 and -9 solution,

Answer 4

b=l-3
l^2+(l-3)^2=225
l^2+l^2-6l+9-225=0
2l^2-6l-216=0
dividing by 2
l^2-3l-108=0
(l-12)(l+9)=0
l=12
so b=9