Homework help-due soon?

Question

I just don't understand this problem:
One of the legs of a right triangle has the length of 4 cm. Express the length of the altitude (h) prependicular to the hypotenuse (c) as a function of the length of the hypotenuse. (Label the additional leg as b) Make a sketch to support the given information before working the problem.

I got the sketch-the easy part. I think the answer might be h(c)=1/c
but I honestly have no idea...that just seemed too easy.

Answer 1

One of the legs of a right triangle has a length of 4 cm.

Express the length of the altitude (h) perpendicular to the hypotenuse (c)

as a function of the length of the hypotenuse.

(Label the additional leg as b)



The right triangle has (c), the hypotenuse,

'b' one leg and 'a' the other leg which measures 4 cm.

'h' is the perpendicular to the hypotenuse (c)



Theorem:

In a right-angled triangle, if a perpendicular

is drawn from the right angle to the hypotenuse, the

triangles on each side of the perpendicular are

similar to the whole triangle (and are also similar

to each other.)



This theorem has a corollary:

The perpendicular CD is a mean proportional between

the segments of the base.

i.e length (AD) * length( DB) = length (CD)^2



c^2 = b^2 + 16

b^2 = c^2 - 16



h^2 = b^2 - x^2

h^2 = 16 - (c - x)^2

b^2 - x^2 = 16 - (c - x)^2

b^2 - x^2 = 16 - (c^2 - 2cx + x^2)

b^2 - x^2 - 16 + c^2 - 2cx + x^2 = 0

c^2 - 16 - 16 + c^2 - 2cx = 0

2c^2 - 2cx - 32 = 0

c^2 - cx - 16 = 0

cx = c^2 - 16

x = c - 16/c



h^2 = 16 - [c - (c - 16/c)]^2

h^2 = 16 - (16/c)^2

h^2 = 16(c - 4)(c + 4) / c^2



h = 4[(c - 4)(c + 4)]^(1/2) / c

Answer 2

Ok a right triangle. The altituide perpendicular basically splits the large triangle into 2 smaller right triangle where 1/2 the hypotenuse is one leg of the right triangle with the altituide being the other leg.



The altituide is going to be the same length as half the hypotenuse.



So h(c)=1/2c



And also hypotenuse is a2+b2=c2 where a and b are 4



C^2=32 so altituide is sqrt(32)/2 or (4sqrt2)/2 or 2sqrt2



Putting thatback in h(a)=sqrt(2a^2)/2

Answer 3

The pythagorean theorem is a^2 + b^2 = c^2

a = 4 cm

b= h cm

c= c cm



16 + h^2 = c^2

h^2 = c^2 -16

*take the square root on both sides*



h = sqrt( c^2 - 16)



since the altitude h is a function of the hypotenuse c, we could right this function as

h(c) =sqrt(c^2 - 16)



Hope that helps.

Answer 4

Divide side c into (x) and (c - x), either side of the altitude, h.

Then you have the 2 internal right triangles with sides (h, x, 4) and (h, c - x, b).

Then you have the 3 equations to the 3 triangles :



(1) b^2 + 16 = c^2

(2) h^2 + x^2 = 16

(3) h^2 + (c - x)^2 = b^2



From (2), x^2 = 16 - h^2, so, x = √(16 - h^2)



From (1), b^2 = c^2 - 16



Substitute these into equation (3) to get :



h^2 + [c - √(16 - h^2)]^2 = c^2 - 16



Now, expand that out, simplify and solve for h in terms of c.

I'll let you do that and hopefully you'll get h = 4√(c^2 - 16) / c.