A right triangle has a hypotenuse 12cm long.?

Question

A right triangle has a hypotenuse 12cm long. Suppose the perimeter is 26cm. Determine the lengths of the legs of the triangle. Round to hundredths. Explanation as well.

Answer 1

Let one leg be x
other leg = 26-x-12 = 14-x (becuase sum of sides 12,x and rest = 16)
because it is right angled triangle

x^2+(14-x)^2 = 12^2
x^2+196+x^2 - 28 x = 144
or 2x^2-28x + 196-144 =0

or 2x^2-28x + 52 =0
or x^2-14x+26=0
(x-7)^2 -23 =0
x= 7 +/- sqrt(23)
so the 2 sides are 7+ sqrt(23) and 7- sqrt(23)
= 7+4.795 and 7- 4.795
= 11.80 and 2.20 (2 decimal places)

Answer 2

Let x and y be the legs.
By Pythagoras, x^2 + y^2 = 12^2 = 144

Perimeter is 26 cm, so x + y + 12 = 26
Therefore, x + y = 26 -12 = 14 cm
From this, we have : y = 14 - x

Substituting into the Pythagorean equation gives :

x^2 + (14 - x)^2 = 144

Expand the LHS :
x^2 + 196 - 28x + x^2 = 144

Add like terms :
2x^2 + 196 - 28x = 144

Subtract 144 from both sides and rearrange terms :
2x^2 - 28x + 52 = 0

Divide both sides by 2 :
x^2 - 14x + 26 = 0

Apply the quadratic formula to obtain x :
x = {-(-14) ± sqrt[(-14)^2 - 4(1)(26)]} / (2 * 1)
= [14 ± sqrt(196 - 104)] / 2
= [14 ± sqrt(92)] / 2
= [14 ± 2 * sqrt(23)] / 2
= 7 ± sqrt(23)
= 2.20 or 11.80 rounded to hundredths.

Answer 3

hypotenuse = a longer cathetus = a√3/2 other cathetus = a/2 (in 30-60-90 triangle) The proof is simple. Only amendment given triangle to an equilateral triangle.

Answer 4

OK. It has sides a and b and hypotenuse c we know
c = 12 cm and a+b+c = 26cm. => a+b=14 We also know c²=a²+b². Since b=14-a, substitute and get
c²=(14-b)²+b² or
144=196-28b+2b² => 2b²-28b+52=0 or
b²-14b+26=0
Now solve the quadratic (using the quadratic formula) for b and back substitute into a=14-b to get the value of a.


Doug

Answer 5

let the sides x y and z. since perimeter is 26cm therefore x+y=14.
now x^2+y^2=z^2. (x+y)^2=x^2+2xy+y^2=196. 2xy=196-144 (i.e. z^2). or xy = 26
x=1 y=26 , x=2 y=13, or the reverse of these. but the only problem is that one side is longer than the hypotenuese. can you solve this mystery.

Answer 6

got it! n its a perfect answer :D
ok.so let hypotenuse be H, perpendicular be P and base be B
now H=12
perimeter=26
hence, H+B+P=26
so, 12+B+P=26
therefore, B+P=26-12=14;
now you know that H square(raised to the power two) = B square +P square;
hence, 12 square = B square + P square
144= B square + P square... let this be equation one.
now we don't know the value of B and P seperately
we just know that B+P=14
now u know that (a+b) square= a square + b square +2ab
now, here's the trick..watch it closely
a square + b square can be written as equal to (a+b) square - 2ab;
(open the bracket n check if u're unsure)
so B square + P square = ( B+P) square -2BP
putting its value in equation one
we get
144= (B+P) square-2BP
as B+P=14
144= 14 square -2BP
144=196 -2BP
-52=-2BP
hence BP=26
now, B+P=14
also B=26/P
therefore, 26/P +P =14
P square -14P +26 =0
solving this we get ( solve urself its getting way long, u know how to solve a quadratic equation, right?)
P=12,2
B=approx. 2,13
*phew* got it??
if u still have a prob. tell me!

um... yeah one side is longer than hypotenuse, dunno why! :(

Answer 7

let the length of legs be x,y
x^2+y^2 =12^2 =144-------1
perimeter= 26=x+y+12
x+y=14----------------2

solve 1,2 4 values

Answer 8

about 11.8 and 2.2 but I don't know how to work it out exactly!

Answer 9

math kp.