the area of a right trangle is 44m squared. find the lengths of its legs if one of the legs is 3m longer?

Question

than the other

Answer 1

area = 1/2 base times height
Base times height = 88 (44*2)
Lenght of the legs are 11 and 8

Answer 2

let b = base length and h = height length and 3 m longer than the base length = b+3. The area of a right triangle is 1/2 base x height = 44 m so 44 = 1/2 * (b)(b+3)
So you get a quadratic equation b^2 + 3b - 88 = 0. So solving by factoring into (b+11)(b-8)=0 we get b = 8 or -11 but a length can not be negative so we discard -11. We can also use the quadratic formula to solve this.

Thus the base is 8 m and height is 11m

Answer 3

Let x = one leg

Let x+3 = longer leg

So x(x+3)/2=44 [area of rt triangle = product of legs/2]

x^2 + 3x = 44(2)

x^2 + 3x -88 = 0

By factorization (x+11)(x-8) =0

x-8=0, so x=8 = shorter leg

substitute x=8
x+3= 8+3 =11 = longer leg

Answer 4

Let x = one leg
Let x+3 = longer leg
So x(x+3)/2=44 [area of rt triangle = product of legs/2]
x^2 + 3x = 88
x^2 + 3x -88 = 0
(x+11)(x-8) =0
x-8=0, so x=8 = shorter leg
x+3= 8+3 =11 = longer leg

Answer 5

assume leg 1 = x and leg 2 = y
y = x + 3

Area = 0.5 * x * y
44 = 0.5 * x * (x + 3) = 0.5 * (x2 + 3 x)
88 = x2 + 3 x
x2 + 3x - 88 = 0
(x + 11) (x - 8) = 0
x = -11 or x = 8

cause x must be a positif x = 8 m and y = 8 + 3 = 11 m

Answer 6

(1/2)xy = 44
xy = 88
y = x + 3

(x + 3)x = 88
x^2 + 3x = 88
x^2 + 3x - 88 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-3 ± sqrt(3^2 - 4(1)(-88)))/(2(1))
x = (-3 ± sqrt(9 + 352))/2
x = (-3 ± sqrt(361))/2
x = (-3 ± 19)/2
x = (-22/2) or (16/2)
x = -11 or 8
x = 8

y = 8 + 3 = 11

ANS : 8m and 11m

Answer 7

1/2*s*(s+3)=44

s=8

l1=8
i2=11