Question

Given: △ABC, CM ⊥ AB , BC=5, AB=7, CA=4√ 2 Find: CM

Answer 1

Answer:

Given:

In ΔABC , $CM \perp AB$ , BC = 5 unit , AB = 7 unit and CA = $4\sqrt{2}$

You can see the figure as shown in the attachment below;

Let the length of AM = x.

and the length of MB = AB-AM=7-x.

Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Using Pythagoras theorem in ΔBMC,

$BM^2+CM^2 = BC^2$

Substitute the value of BC and BM in above expression;

$(7-x)^2+CM^2=(5)^2$ or

$CM^2 =25 -(7-x)^2$     ,......[1]

Similarly, in ΔAMC

Using Pythagoras theorem;

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

Substitute the value of AC and AM above we get;

$x^2+CM^2=(4\sqrt{2})^2$

$x^2+CM^2 = 32$ or

$CM^2 = 32-x^2$     .......[2]

Equate [1] and [2] to solve for x;

$25-(7-x)^2=32-x^2$

$25-(49+x^2-14x)=32-x^2$ or

$25-49-x^2+14x=32-x^2$

On simplify:

$14x =32+49-25 =56$

Divide 14 to both sides of an equation we get;

x= 4.

∵ $CM^2 = 32-x^2$ = $32-(4)^2 = 32-16 =16$

Therefore, the length of CM = 16 units.

Answer 2

Thank you for asking this question. Here is your answer:

b² = a² + c² - 2ac·cos(B)

(4√2)² = 5² + 7² - 2·5·7·cos(B)

32 = 74 -70·cos(B)

cos(B) = (74 -32)/70 = 3/5

BM = BC·cos(B)

= 5·(3/5)

= 3

BC² = BM² + CM²

5² = 3² + CM²

CM = √(25 -9) = 4

So the final answer for this question is 4

If there is any confusion please leave a comment below.