the hypotenuse of a triangle is 26 feet long.?

Question

one leg of the triangle is 14 feet longer than the other leg. find the lengths of the legs of the triangle

Answer 1

Ah you just need to know how to set this question up.

Remember A^2 + B^2 =C^2

So use X as the length of one of the unknown sides.

X^2 + (x +14)^2 = 26^2

so you end up with 2x^2 +196 = 676

then solve it out

Hope that helps!

Answer 2

Given: c = 26 feet, hypotenuse of a acceptable triangle a = x, length of one leg b = x + 14, length of the different leg discover: lengths of the legs answer: utilizing the Pythagorean Theorem: c^2 = a^2 + b^2 (26)^2 = x^2 + (x + 14)^2 676 = x^2 + x^2 + 28x + 196 x^2 + x^2 + 28x + 196 = 676 2x^2 + 28x + 196 - 676 = 0 2x^2 + 28x - 480 = 0 Divide with the help of two both left and acceptable of the equations.... x^2 + 14x - 240 = 0 Factoring: (x + 24)(x - 10) = 0 Equate each binomial aspect to 0.... x + 24 = 0 x = -24 overlook this answer because it truly is detrimental. x - 10 = 0 x = 10 ANS the different leg is, x + 14 = 10 + 14 = 24 ANS teddyboy

Answer 3

Let x= length of the smaller side of a triagle and
x+14= lengh ofthe other side of a trianle
It is given that the hypotenuse of a right triangle is= 26
Using pythogorian theorm
c^2=a^2 + b^2
(26)^2= (x)^2 + (x+14)^2
676 = x^2 +x^2 +28x+ 196
2x^2 +28x+ 196 -676 =0
2 (x^2 +14x- 240) =0
on factoring (x^2 +14x- 240) =0 we get
(x+24) (x-10) =0
Therefore x= -24 and x=10
Neglect the negative sign, because lenth cannot be negative
Therefore x=10 is the answer
Hence one side is 10 unit and the other side will be 10+14=24units

Answer 4

Let x be the length of the short leg. Then the long leg is x+14 and (from Pythagoreas Theorem)

x² + (x+14)² = 26² so that

2x² + 28x -480 = 0

(2x - 20)*(x + 24) = 0 So

x = 10 and x = -24 are solutions

Negative length is an extraenous root, so the lengths of the sides are 10 ft. and 24 ft.


Doug

Answer 5

We have a right triangle which must form a pythagorean triplet as follows

(x,x+14,26)

So that 26^2 = x^2 + (x+14)^2

676 = x^2 + x^2 + 28x + 196

2x^2 + 28x -480 = 0

x^2 + 14x - 240 = 0

Factoring we get

(x + 24)(x-10) = 0

Which means our sides must be 10 and 24

x = -24 is not a valid solution since a length can not be negative

The the triplet is (10,24,26)

Answer 6

a = b + 14
c = 26

If this is a right triangle, then

a^2 + b^2 = c^2
(b + 14)^2 + b^2 = 26^2
((b + 14)(b + 14)) + b^2 = 676
(b^2 + 14b + 14b + 196) + b^2 = 676
b^2 + 28b + 196 + b^2 = 676
2b^2 + 28b + 196 = 676
2b^2 + 28b - 480 = 0
2(b^2 + 14b - 240) = 0
b^2 + 14b - 240 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-14 ± sqrt(14^2 - 4(1)(-240)))/(2(1))
x = (-14 ± sqrt(196 + 960))/2
x = (-14 ± sqrt(1156))/2
x = (-14 ± 34)/2
x = (-56/2) or (20/2)
x = -28 or 10

b = 10

a = 10 + 14 = 24

ANS : 24ft and 10ft

Answer 7

(leg1)^2 + (leg2)^2 = (hypotenuse)^2
or
a^2 + b^2 = c^2

leg1 = leg2 + 14 feet
or
a = b + 14

(b + 14)^2 + b^2 = 26^2
b^2 + 28b + 196 + b^2 = 676
2b^2 + 28b - 480 = 0
2(b^2 + 14b - 240) = 0
b^2 + 14b - 240 = 0

b = [-14 +/- sqrt( (14^2) - (4)(1)(-240)) ] / 2
b = [-14 +/- sqrt( (196 + 960) ] / 2
b = [-14 +/- 34 ] / 2
b = -24 or 10 feet

Since length cannot be negative, b = 10 feet

leg 2 = 10 feet
leg 1 = 24 feet

Answer 8

x^2 + (x + 14)^2 = 26^2
x^2 + x^2 + 28x + 196 = 676
2x^2 + 28x = 480
x^2 + 14x = 240
x^2 + 14x + 49 = 240 + 49 = 289
(x + 7)^2 = sqrt(289) = +/-17
x = 17 - 7 = 10

Check:
10^2 + 24^2 = 100 + 576 = 676 = 26^2

Answer 9

the legs are 10 ft and 24 ft as 24^2+10^6=26^2

Answer 10

x^2 + (x+14)^2 = 26^2

2x^2 + 28x -480 = 0

2( x^2 + 14x -240) = 0

10 and 24 are the answers