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

Question

Given: △ACM, m∠C=90°, 

CP

  ⊥ 

AM

 .

Chapter Reference

b

AC:CM=3:4, MP-AP=1. 

Find AM.

Answer 1

Hello Jack,

In this problem you've to use geometric mean (leg) theorem, which states that in a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments.

The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg.

AC² = AP * AM and CM² = MP * AM

AC² : CM² = (AP * AM) : (MP * AM) = AP : MP --------- (1)

Given AC : CM = 3 : 4

Therefore AC² : CM² = 3² : 4² = 9 : 16 = AP : MP ----------- BY (1) --------------(2)

AP = (9/16)*MP ------------ (3)

Given : MP - AP = 1-------------------(4)

MP - (9/16)*MP = 1 --------------- substitute value of AP from (3)

Solve for MP

MP = 16 / 7

Plug-in the value of MP in equation 4 and solve for AP.

AP = 9 /7

AM = AP + MP

= 9/7 + 16/7 = 25/7

Answer is AM = 25/7

I hope this helps. If you need further help myself or any other tutor can help.

Thank You,

Shefali J