Question

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

Answer 1

Answer:

Step-by-step 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

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

$AP = \frac{5k-1}{2}$

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)

$AC^2 = 9k^2 = AP(AM) = (\frac{5k-1}{2} )AM\\=5k(\frac{5k-1}{2} )\\25k^2-5 = 18k^2\\7k^2=5\\k^2 =5/7\\k =\sqrt \frac{5}{7} \\AM=\sqrt \frac{5}{7}$

Answer 2

Answer:

AM = 25/7

Step-by-step 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