Question

The length of the hypotenuse of a right angled triangle exceeds the length of the base by 2cm and exceeds twice the length of the altitude by 1cm.Find the length of each side of the triangle​

Answer 1

Answer: a = 8;  b = 15;  c = 17

Step-by-step explanation:

c = b + 2  and c = 2a + 1

b = c − 2 and a =  $\frac{c-1}{2}$

​

c² = b² + a²  

$c^{2} = (c-2)^{2} + \frac{c-1}{2}$

$c^{2}= \frac{4 (c-2)^{2}+(c - 1)^{2} }{4}$

$4c^{2}={4 (c^{2} + 4 -4c)+c ^{2} + 1-2c$

$4c^{2}={4 c^{2} + 16 -16c)+c ^{2} + 1-2c$

$c^{2}-18c+17=0$

$c(c -17)-1(c-17)=0$

​$(c -1)(c-17)=0$

c = 1 or c = 17

If c = 1 then b = 1 - 2 = -1, that's not possible

So x = 17

b = 17 - 2 = 15

a = $\frac{17-1}{2} = 8$

Length sides of the triangle are 17 cm, 15 cm and 8 cm.

I saw some of this on a site but I want to make sure its correct so we're gonna apply the pythagorean theory and see if its correct.

$a^{2} + b^{2} = c^{2} \\8^{2} + 15^{2} = 17^{2} \\64 + 225 = 289\\289 = 289$

This shows that this is correct.

Hope this helped! Again, some of this was from a site, not from me.