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

the formula for the equation of a circle is (x – h)2+ (y – k)2 = r2, where (h, k) represents the coordinates of the center of the circle, and r represents the radius of the circle. if a circle is tangent to the x-axis at (3,0), this means it touches the x-axis at that point.

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