Maths problem....?

Question

The perimeter of a right angled triangle is 70 and the sum of the squares of its three sides is 1682. Find the lengths of the three sides

Answer 1

Firstly, consider Pythagoras' theorem - the square of the hypotenuse equals the sum of the squares on the other two sides. Therefore the sum of the squares of the three sides (in this case 1682) will be twice the square of the hypotenuse, and so the square of the hypotenuse will be half this amount.

1682/2 = 841, square root of 841 = 29, therefore the hypotenuse is 29 units long. Total length of sides = 70, therefore the other two sides add up to 70-29=41 units; the squares adding to 841. These two sides are therefore 20 & 21 units in length (20+21=41, 20^2=400, 21^2=441, 400+441=841).

So there you have it; the lengths of the sides are 20, 21 and 29.

I hope this helps, but please feel free to drop me a line if you'd like to discuss this or any other figure-work further.

Answer 2

There are only two givens, so we use only two variables.

Let a = one side, not the hypotenuse
b = other side, not the hypotenuse

the hypotenuse is c = 70 - a - b.

from the Pyth. ID, a^2 + b^2 = c^2 or

1) a^2 + b^2 = (70-a-b)^2

We need another equation.

2) a^2 + b^2 + (70-a-b)^2 = 1682.

Write
1) a^2 + b^2 - (70-a-b)^2 = 0
2) a^2 + b^2 + (70-a-b)^2 = 1682, add them
----------------------------------------------
2a^2 + 2b^2 = 1682 or
a^2 + b^2 = 841.

Since a^2 + b^2 = c^2, we have c^2 = 841 or
c = 29, the hypotenuse.

So we have
A) a+b+29 = 70, and
B) a^2 + b^2 = 841.

if we rewrite A) as a = 41-b, and square it,
a^2 = 1681 - 82b + b^2

from B), we have a^2 = - b^2 + 841, put them together

a^2 = 1681 - 82b + b^2 = - b^2 + 841, or

1681 - 82b + b^2 = - b^2 + 841, gather
2b^2 - 82b + 840 = 0, div by 2
b^2 - 41b + 420 = 0, factor or quadratic formula
(b - 21)(b - 20) = 0.
b = 21 | b=20.

You will get two values for a, see which one works.

Answer 3

a+b+c = 70 <-- Eq 1
a^2+b^2+c^2 = 1682 <-- Eq 2
-a^2-b^2 +c^2 = 0 <-- Eq 3
Add Eq 2 and Eq 3 getting:
2c^2 = 1682 --> c^2 = 841 --> c = 29
Replace c with 29 in Eq 1 and Eq 2 getting:
a+b = 41 --> a = 41-b
a^2 + b^2 = 841
(41-b)^2 +b^2 = 841
1681 -82b +b^2 +b^2 = 841
2b^2 -82b +840 = 0
b^2-41b+420 = 0
(b-20)(b-21) = 0
b = 20 or 21
29+20 +a =70
a= 21
So a=21, b=20, c=29

Answer 4

The square of the hypoteneus is equal to the sum of the squares of the other two sides i.e. 1682/2 = 841. The hypoteneus is 29.
a+b=41
a^2+b^2=841
erm....

Answer 5

(u^2-v^2) + (2uv) + (u^2+v^2) = 70

(u^2-v^2)^2 + (2uv)^2 + (u^2+v^2)^2 = 1682

Simplifying,

u^2 + uv = 35 = u(u+v). Therefore u=5 and v=2. Sides are

(5^2-2^2) = 21, 2(5)(2) = 20, and 5^2+2^2) = 29

Check: 21^2+20^2+29^2=441+400+841 = 1682.

Sides are 20,21,and 29.

Answer 6

1682 - 70 = 1612.
1612 divided by 2 = 806.
so the answer is 806.. I think.