Question

Can you help me with this assignment? ​

Answer 1

Given :

Hypotenuse of right angle triangle = n + 7

Base = n - 1

Altitude = 2n

Using Pythagoras theorem :

→ ( n + 7 )² = (n - 1)² + (2n)²

→ n² + 14n + 49 = n² - 2n + 1 + 4n²

→ n² + 14n + 49 = 5n² - 2n + 1 = 0

→ n² + 14n + 49 - 5n² + 2n - 1 = 0

→ -4n² + 16n + 48 = 0

→ -4(n² - 4n - 12) = 0

→ n² - 6n + 2n - 12 = 0

→ n(n - 6) + 2(n - 6) = 0

→ (n - 6)(n + 2) = 0

As n ≠ Negative

So, n ≠ -2

So, n = 6

Hypotenuse = n + 7 = 6 + 7 = 13

Base = n - 1 = 6 - 1 = 5

Altitude = 2n = 2 × 6 = 12

So,

AB = 13

AC = 5

BC = 12

Answer 2

AC =5

BC = 12

AB = 13

Explanation:

Since this is a right triangle, we can use the Pythagorean theorem

a^2 + b^2 = c^2

(n-1)^2 + (2n)^2 = (n+7)^2

FOILing

n^2-2n+1 + 4n^2 = n^2+14n+49

Combine like terms

5n^2-2n+1 = n^2+14n+49

Bring everything to the left

5n^2-2n+1 -n^2-14n-49 = n^2+14n+49-n^2-14n-49

4n^2 -16n -48 = 0

Divide by 4

n^2 -4n -12 = 0

Factor

( n-6)(n+2) =0

Using the zero product property

n-6 =0 n+2 =0

n=6 n=-2

Since the length cannot be negative

n=6

AC = n-1 = 6-1 =5

BC = 2n = 2*6 = 12

AB = n+7 = 6+7 = 13