Question

Given: △ACM, m∠C=90°, CP ⊥ AM , AC:CM=3:4, MP-AP=1. Find AM.

Answer 1

AM = 25/7

Explanation:

Lets say that PM = x

Then AP = x-1

They are in the scale of 3:4, so

CP = 3/4 *x or 3x/4

CP also equals 4/3 *(x-1) or 4x-4/3

That means that

4x-4 = 3x

3 4

Then just do some cross multiplication to get

16x-16 = 9x

Then subtract and get 7x = 16, or

x = 16/7

Then substitute x in and get AM = 32/7 - 7/7 or

AM = 25/7

Answer 2

Explanation:

Given that ACM is a right angled triangle.

Angle C =90 and also angle P =90

AC:CM=3:4

`MP-AP=1` ... i

Let AC=3k and CM=4k

Then by pythagorean theorem

`AM^2=(3k)^2+(4k)^2\\AM=5k`

From i, we get

`AM=AP+MP = AP+AP+1=5k`

`AP = (5k-1)/(2)`

Triangles APC and ACM are similar, since

angle A is common and one angle is right angle

AP/AC = AC/AM (Since triangles are similar)

`AC^2 = 9k^2 = AP(AM) = ((5k-1)/(2) )AM\\=5k((5k-1)/(2) )\\25k^2-5 = 18k^2\\7k^2=5\\k^2 =5/7\\k =\sqrt (5)/(7) \\AM=\sqrt (5)/(7)`