Question

Given: ΔACM, m∠C = 90º

Answer 1

Answer:

CM=20 and CP=12

Step-by-step explanation:

The given triangle ΔACM has the measurements as follows:

m∠C=90°, CP⊥AM, AC=15, AP=9, PM=16.

To Find: CP and CM

We can use Pythagoras theorem to calculate the sides CP and CM.

Pythagoras theorem gives a relation between hypotenuse, base and height/perpendicular of a right angled triangle which is as follows:

$h^{2}=p^{2}+b^{2}$

where h is hypotenuse of triangle, b is base and p is perpendicular of triangle.

The figure shows that in ΔACM is a right angled triangle at C where,

AM --> hypotenuse

CM --> base

AC --> height

So substituting values into formula:

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

$25^{2}=15^{2}+CM^{2}$

$625-225=CM^{2}$

$400=CM^{2}$

$\sqrt{400} =CM$

$CM=20$, which is required answer.

Similarly, we can see that triangle ΔCPM is also a right angled triangle at P and thus Pythagoras theorem can again be applied to calculate CP. Since CM is the side opposite to right angle P, it is the hypotenuse.

So we have,

$CM^{2}=PM^{2}+CP^{2}$

$20^{2}=16^{2}+CP^{2}$

$400-256=CP^{2}$

$144=CP^{2}$

$\sqrt{144} =CP$

$CP=12$, which is required answer.