Question

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

Answer 1

Answer:

$AM=\frac{25}{7}$

Step-by-step explanation:

It is given that AC:CM = 3:4.

Therefore, let AC = 3k and CM = 4k

It is given that Δ ACM is right angled.

Therefore,

$AM^{2} =AC^{2} +CM^{2}$

$=(3k)^{2} +(4k)^{2}$

$=9k^{2} +16k^{2}$

$=25k^{2}$

AM = 5k

But, AM = MP + AP

Therefore, MP + AP = 5k --- (1)

It is given that MP - AP = 1 --- (2)

Multiply (1) and (2), we get.

(MP + AP)(MP - AP) = 5k × 1 = 5k

$MP^{2} -AP^{2} =5k$

Add and subtract $CP^{2}$ on the left side.

$MP^{2} + CP^{2} - CP^{2} -AP^{2} =5k$

$(MP^{2} + CP^{2}) - (CP^{2} +AP^{2}) =5k$ --- (3)

But, since CP ⊥ AM, Δ CMP and Δ CAP are right triangles. Therefore,

$MP^{2} +CP^{2} =CM^{2}$ and

$CP^{2} +AP^{2} =AC^{2}$

Now, (3) becomes,

$CM^{2} -AC^{2} =5k$

$(4k)^{2} -(3k)^{2} =5k$

$7k^{2} =5k$

7k = 5 or

$k=\frac{5}{7}$

AM = 5k

$=5(\frac{5}{7} )$

$=\frac{25}{7}$