A toy manufacturer needs a piece of plastic in the shape of a right triangle with the longer leg 2 cm more than twice as long as the shorter leg, and the hypotenuse 1 cm more than the longer leg. How long should the three sides of the triangular piece be?
2006-11-13 01:43:56 · 10 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Let x = short leg
2x +2 = long leg
2x + 3 = hypotenuse
Since this has to be a right triangle:
x^2 + (2x+2)^2 = (2x+3)^2
x^2 + 4x^2 + 8x + 4 = 4x^2 + 12x +9
x^2 + 4x^2 - 4x^2 + 8x - 12x + 4 - 9 = 0
x^2 - 4x - 5 = 0
(x+1)(x-5) = 0
x = -1 or x = 5
Since x cannot be a negative number, the short leg must be 5 cm, the long leg 12 cm and the hypotenuse 13 cm
2006-11-13 01:57:37 · answer #1 · answered by T 5 · 0⤊ 0⤋
Let x = shorter leg
Then 2x+2 = longer leg
And 2x+3 = hypotenuse
Since this is a right triangle we must have:
(2x+3)^2 = (2x+2)^2 +x^2
4x^2 +12x +9 = 4x^2 + 8x + 4 + x^2
x^2-4x-5=0
(x-5)(x+1) = 0
x= 5 or -1
X = -1 is rejected because can't have a negative length
Therefore x=5 = shorter leg
2x+2=2(5)=12 = longer leg
2x+3 = 2(5)+3 = 13 = hypotenuse
This is a unique solution; no other solution is possible
2006-11-13 10:07:29 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
According to Pythagoras a^2 + b^2 = c^2 =>
a^2 + (2*a + 2)^2 = ((2*a + 2) +1)^2 =>
a^2 + 4*a^2 + 8*a + 4 = 4*a^2 + 12*a + 9 =>
a^2 - 4*a - 5 = 0 =>
(a + 1) * (a - 5) = 0 =>
a = -1 or a = 5,
since the length of a leg can't be negative => a = 5
Short leg: 5
Long leg: 2*5+2 = 12
Hypothenusa: (2*5+2)+1 = 13
2006-11-13 09:59:33 · answer #3 · answered by sunnyboy 3 · 0⤊ 0⤋
You know. I went through the trouble of solving the problem, only to find out that it doesn't matter which length the shorter leg has, it will always full fill the condition. The longer leg Will always have 2 cm more and the hypotenuse will always have 3 cm more. The procedure performed by CD actually serves as a demonstration in case you need one.
2006-11-13 11:44:28 · answer #4 · answered by Paul G 5 · 0⤊ 0⤋
let shorter leg=x cm
longer leg=2x+2
hypotenuse=2x+3
Since it is a right
angled triangle we have
[2x+3]^2=[2x+2]^2+x^2
4x^2+12x+9-4x^2-8x-4-x^2=0
-x^2+4x+5=0
x^2-4x-5=0
x^2-5x+x-5=0
x[x-5]+1[x-5]=0
[x-5][x+1]=0
x=5,-1
Discard the -ve answer.
small side=5
longer side=12
hypotenuse=13
verify
13^2
=169
5^2+12^2
=25+144
=169
ok
2006-11-13 10:00:06 · answer #5 · answered by openpsychy 6 · 0⤊ 0⤋
Triangle ABC where angle B is a right angle:
AB is short leg; BC is longer leg; AC is hypotenuse.
BC = 2 + 2(AB)
AC = 1 + BC
AC = 1 + 2 + 2AB
AB^2 + BC^2 = AC^2
BC^2 = (2AB + 2))^2
AC^2 = (2AB + 3)^2
AB^2 + 4AB^2 + 8AB +4 = 4AB^2 + 12AB + 9
5AB^2 + 8AB + 4 = 4AB^2 + 12AB + 9
AB^2 - 4AB - 5 = 0
(AB + 1) (AB -5) = 0
AB = 5 or -1
therefore AB = 5 cm
BC = 12 cm
AC = 13 cm
2006-11-13 09:55:44 · answer #6 · answered by C D 3 · 0⤊ 0⤋
Let S=short leg, L=long leg, H=hypotenuse. S^2 + L^2 = H^2
L = 2S + 2
H = L + 1, so H^2 = S^2 + L^2, S^2 + L^2 = (L + 1)^2
S^2 + L^2 = (L + 1)^2
S^2 + (2S+2)^2 = (2s+2 + 1)^2
S^2 + (4S^2 +4S +4) = 4S^2 + 12S +9
S^2 - 8S - 5 =0
solve with quadratic formula
2006-11-13 09:53:33 · answer #7 · answered by fcas80 7 · 0⤊ 1⤋
let x=shorter leg.
longer leg=2x+2
hypotenuse=2x+2+1=2x+3
(2x+3)^2=x^2+(2x+2)^2
4x^2+12x+9=x^2+4x^2+8x+4
x^2-4x-5=0
(x-5)(x+1)=0
only the positive answer has physical meaning so
x-5=0
x=5
2x+2=12
2x+3=13
check 13^2=12^2+5^2
169=144+25
169=169
2006-11-13 11:13:35 · answer #8 · answered by yupchagee 7 · 0⤊ 0⤋
You can make the shorter leg be anything you want.
I made it 2 and so then the longer leg is 6 and the hypotnuse is 7
2006-11-13 09:47:48 · answer #9 · answered by Olha K 2 · 0⤊ 1⤋
sides are x, (2x+2), hypothenuse (2x+3)
(2x+3)sqd = ((2x+2)sqd + xsqd
simplyfy out and factorise gives
(x+1)(x-5)=0
x=-1 (not possible)
or x= 5.
Sides are 5, 12 and 13
The other answers are crap, by the way
2006-11-13 09:56:34 · answer #10 · answered by andyoptic 4 · 1⤊ 0⤋