Question

The hypotenuse of a right angled triangle is 2 cm more than the base and 1 cm more than twice the altitude, then find the sides of the triangle.​

Answer 1

Step-by-step explanation:

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Let b = the length of the base. Let the other sides be h and p

h = b+2 ---> b = h-2

h = 2p+1 ---> p = (h-1)/2

we also know h^2 = b^2 + p^2

h%5E2+=+%28h-2%29%5E2+%2B+%28%28h-1%29%2F2%29%5E2+

h%5E2+=+h%5E2+-4h+%2B+4+%2B+%28h%5E2+-2h+%2B+1%29%2F4+

4h%5E2+=+4h%5E2+-+16h+%2B+16+%2B+h%5E2+-2h+%2B1

0+=+h%5E2+-18h+%2B17

0+=+%28h-17%29%28h-1%29

h = 17 or 1.

Since base is h-2, h cannot be one since then b would be negative

so h = 17

b = 15

p = 8