the area of a right triangle is 92 sq.ft. If the hypotenuse os 20 ft., find the lengths of the two legs.?

Question

with solutions

Answer 1

area of triangle=1/2*b*h
92=1/2*b*h...(i)
184/h=b...(ii)
1/2*b*h=area of triangle................substitude (ii) into (i)
1/2*184/h*h=92..............multiply with 2h
h*368*2h^2=184h
736h^3=184h............... divide by 184h
4h^2=1
h^2=1/4
h=1/2

184/h=b...(ii)
184/1/2=b
368=b

thus the lengths of the two legs is 1/2 and 368 respectively

Answer 2

The area of a triangle is 1/2 times base times height, but for a right triangle, that is the same as 1/2 times leg1 times leg2
If you let leg 1 be x and leg 2 be y, then the area would be 1/2xy which would be equal to 92
1/2xy=92

We can also assume x^2+y^2=400 by the Pythagorean theorem (the sum of the squares of the legs is equal to the square of the hypotenuse)

So that leaves us with a system of equations.

1/2xy=92
x^2+y^2=400

If we multiply everything in the top one by 2, we will have something a little more manageable to work with:
xy=184

Now the only thing we have to do is substitution to find the values of x and y. I will let you figure it from here.

Answer 3

let the base be x ft and the height h ft
for a rt angled triangle
hypotenuese ^2 = base^2 + height^2
20^2=x^2+h^2
h=sqrt(400-b^2)
area of any triangle = 1/2* base* height
=1/2 *b * sqrt(400-b^2)
in this case the area is 92 sqft
92= 1/2 * b * sqrt(400-b^2)
squaring both sides we get
92^2 = 1/4 * b^2 * (400-b^2)
8464 = 1/4 * b^2(400-b^2)
8464=100b^2 -1/4 b^4
1/4b^4 - 100b^2 + 8464 = 0
let b^2= x
1/4 x^2 - 100x +8464 = 0
multiply the entire equation by 4
x^2 -400x + 33856 = 0
x= 400+- sqrt(400^2 - 4*1*33856) / 2*1
=400 +- 156.77/ 2
= 278.39 or 121.62
since x=b^2
b^2=278.39 or 121.62
b = 16.69 or 11.03
h= sqrt(400-b^2)
= sqrt(400-278.39) or sqrt(400-121.62)
= 11.03 or 16.69
so its eveident that the two legs are 11.03 and 16.69 respectively

Answer 4

A = 1/2bh, the h doesn't mean hypotenuse. The h is the height, which in the case of a right triangle means one of the legs.

Answer 5

let the lengths of the two legs be a&b
Area of the right triangle=a*b/2=92, ab=184
hipotenuse^2 =a^2+b^2=20^2=400
(a+b)^2=400+368=768
a+b=27.71
a-b)^2=400-368=32
a-b=5.66
2a=33.37
a=16.69ft
2b=22.05
b=11.025ft.
so the lengths of the legs are: 16.69ft &11.0025ft

Answer 6

legs are 12 and 16 from a 3,4,5 triangle that is magnified by 4.

Or solve
b= base
h= height
400 = b^2 +h^2
92 = 1/2 bh

Answer 7

A = (1/2)bh
92 = (1/2)bh
b = 184/h
b^2 + h^2 = 20^2
(184/h)^2 + h^2 = 400
33856/h^2 + h^2 = 400
Let x = h^2
33856/x + x^2 = 400
33856 + x^2 = 400x
x^2 - 400x + 33856 = 0
x = (-B =/- √[B^2 - 4AC])/2A
x = (400 +/- √[400^2 -4(33856)])/2
x = (400 +/- 156.767)/2
x = 2078.38, 121.617
h = √x
h = √2078.38 = 45.589 too big
h = √121.617 = 11.028
b = 184/h = 184/11.028 = 16.685

Answer 8

A = 1/2bh

Answer 9

11.03 and 16.68.It cant be 12 and 16 as area will 96 not 92.

Answer 10

A=1/2bh
92=1/2b(20)
92=20b/2
184/20=20b/20
b=9.2ft

to get the other leg:
c(squared)=a(squared)+b(squared)
20^2=a^2+9.2^2
400=a^2+84.64
400-84.64=a^2
315.36=a^2
a=17.75ft