In triangle △abc, ∠abc=90°, bh is an altitude. find the missing lengths. ah=4 and hc=1, find bh.

Side BC= √5, AB = 2√5, altitude BH = 2

Step-by-step explanation:

In the figure attached let the side BH = h , AC = x , AB = y

We have to find the missing lengths of AB, AC, and BH.

Given sides are AH =4 and CH = 1

In ΔABC ⇒ x²+y²=5²

                 x²+y²=25------(1)

In ΔHBC ⇒ h² = x²-1² = (x²-1) ( Pythagoras theorem)

In ΔABH ⇒ h² = y²-16 ( Pythagoras theorem)

So h² = x²-1 = y²-16

x²-y² = 1-16 = -15------(2)

By addition of equation 1 and 2.

(x²+y²)+(x²-y²) = 25-15

2x² = 10

x² = 5 ⇒ x = √5

By putting the value of x in equation 1.

5 + y² = 25

y² =25-5 = 20

y = √20 = 2√5

Therefore side BH = h = √(x²-1) = √(5-1) = √4 = 2

-2

7 - 9 = -2

reversed:

-2 + 9 = 7