Maths help. mathematicians Pls i Help me Out with this problems.?

Question

1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.
Find the sides of the triangle. (Pythagorean Theorem)

2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides
is 2 cm. Find the sides of the triangle. (Pythagorean Theorem

Answer 1

1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.
Find the sides of the triangle. (Pythagorean Theorem)

What we can take from this problem is that one LEG is 12 cm long, while the other LEG is unknown (we will call it X) and the HYPOTENUSE -- the longest side of a right triangle -- is also unknown (we will designate it X+2, since it differs 2 cm from the unknown leg).

Setting up Pythagorean Theorem as always:

"Leg squared" + "Leg squared" = "Hypotenuse squared"

12^2 + x^2 = (x+2)^2

Expanding:

144 + x^2 = x^2 + 4x + 4

Collecting like terms:

4x = 140

Divide by 4:

x = 35

The unknown leg is 35 cm, while the hypotenuse is 37. Thus, we can also call the triplet (12,35,37) a "Pythagorean Triple" (of which there are infinitely many!).


2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides
is 2 cm. Find the sides of the triangle. (Pythagorean Theorem

In this case, we know the hypotenuse, but do not know the lengths of the legs. For convenience, we will again use the expressions X and (X+2) for each side's length.

By P.T.:

x^2 + (x+2)^2 = 58^2

Expanding:

x^2 + x^2 + 4x + 4 = 3364

Collect like terms:

2 x^2 + 4x - 3360 = 0

Divide by the GCF (2):

x^2 + 2x - 1680 = 0

Can we factor this trinomial easily? We could, if we knew integer factors of -1680 (one positive, one negative) that combine to make 2.

Luckily, -1680 can be factored into -2*2*2*2*3*5*7, and these can be grouped into two sets:
2*3*7 = 42
-2*2*2*5 = -40

Factor:

(x+42) (x-40) = 0

Although two solutions come from these factors (-42 and 40), we may only consider the positive answer, since lengths of triangle's legs are always positive!

Therefore, x=40 (Leg 1), and Leg 2 is 42 cm.

Answer 2

1)One leg of a right triangle is 12 cm long. The diference between the other two sides is 2 cm.
Find the sides of the triangle. (Pythagorean Theorem)

let x=unknown lef
12^2+x^2=(x+2)^2
144+x^2=x^2+4x+4
140=4x
x=35cm
hypotenuse=37cm
check
12^2+35^2=144+1225=1369
√1369=37cm

2) The hypotenuse of a right triangle is 58 cm long. The diference between the other two sides
is 2 cm. Find the sides of the triangle. (Pythagorean Theorem)

let x=shorter side
58^2=x^2+(x+2)^2
3364=2x^2+4x+4
x^2+2x-1680=0
(x-40)(x+42)=0

only
x-40=0
x=40 is physically possible
legs are 40cm & 42cm.

Answer 3

1)
Let the other two sides be x and x+2
So by the Pythagorean Theorem,
12^2 + x^2 = (x+2)^2
144 + x^2 = x^2 + 4x + 4
144 = 4x + 4
140 = 4x
x = 35
and x+2 = 37
So the sides are 12, 35, 37

You can recheck at this point that 12^2 + 35^2 = 37^2.
144 + 1225 = 1369
1369 = 1369. So the answer checks out.

2)
Let the sides be x and x+2
So by the Pythagorean Theorem,
x^2 + (x+2)^2 = 58^2
x^2 + x^2 + 4x + 4 = 3364
2x^2 + 4x + 4 = 3364
x^2 + 2x + 2 = 1682
x^2 + 2x - 1680 = 0
(x+42)(x-40) = 0
So x = 40, or -42, and we can reject -42, so x = 40
x+2 = 42
So the sides are 40, 42, 58

Recheck by showing 40^2 + 42^2 = 58^2.

Answer 4

1. Let the leg be x and the hypotenuse be x+2. Use the Pythagorean Theorem

x^2+144=(x+2)^2
x^2+144=x^2+4x+4
144=4x+4
140=4x
35=x
37=x+2

2. Same thing

Let the legs be x and (x+2) You could also use x and (x-2) but I like to work with positives wherever posssible.

x^2+(x+2)^2=58^2
x^2+x^2+4x+4=3364
2x^2+4x+4=3364
2x^2+4x-3360=0
x^2+2x-1680=0
(x-40)(x+42)=0
x=-42 is not possible so
x=40 and x+2 =42 are teh answers

Answer 5

a^2 + b^2 = c^2

We know that 'a' (one leg) is 12.

12^2 + b^2 = c^2

We also know that the difference between b & c is 2. Since the hypotenuse must be the longest side of the triangle, c must be greater than b, so c = b + 2. By substituting this into our equation, we get:

12^2 + b^2 = (b+2)^2

simplify:

144 + b^2 = b^2 + 4b + 4

subtract b^2 from both sides of the equation:

144 + b^2 - b^2 = b^2 + 4b + 4 - b^2
144 = 4b + 4

subtract 4 from both sides

144 - 4 = 4b + 4 - 4
140 = 4b

divide both sides by 4

140/4 = 4b/4
35 = b

So, side b must be 35. Side c would be b + 2 or 37.


You can solve the second one by the same basic principle, except that you know the length of c, and that b = a + 2.

Good luck.

Answer 6

1) Let x be the other leg
Then the hypotenus must be x=2
So x^2+12^2 =(X+2)^2
x^2 +144 X^2 = 4X +4
4x=140
x= 35 (one side)
x+2= 37 (hypotenuse)
Third side =12 (other side)

2) Let x = side 1
Then side 2 = x+2
So x^2 + (x+2)^2 = 58^2
x^2 +x^2 +4x + 4 = 3364
2x^2 +4x -3360 = 0
x^2 + 2x - 1680
(x+42)(x-40)=0
x=40 [value of x =-42 is rejected because length cannot be - .]
therefore the two sides are 40 and 42

Answer 7

1)
let one of the other sides be x and the next to be x+2
12^2 + (x)^2 = (x+2)^2
** It cannot be 12sq + (x+2)sq = x sq cos the hypotenus must be longer than any sides of the triangle.

so,
144 + x^2 = x^2 + 4x + 4
x = 35
x+2 = 37

2)
let the sides be y and y+2
similarly, you get 2y^2 + 4y + 4 = 3364
and y^2 +2y +2 = 1682
y^2 + 2y - 1680 = 0
(y+42)(y-40)=0

so y = 40. (-42 cant be accepted)
and y +2 = 42

HOPE THIS HELPS=)

Answer 8

1) let the length of the other side = x
144=(x+2)^2 - x^2 then x = 35 cm and the hyp = 37