Question

Al and cm are two medians of triangle ABC in which B=90° . If AC=5cm and AL=3✓5 /2 cm. FIND CM.

Answer 1

Answer:

2√5 cm

Step-by-step explanation:

Given Al and cm are two medians of triangle ABC in which B=90° . If AC=5cm and AL=3✓5 /2 cm. FIND CM.

In ΔABL,

     AL² = AB² + BL²

     AL² = AB² + (1/2 BC)²

    AL² = AB² + 1/4 BC²-----------(1)

In Δ BCM,

    CM² = BC² + BM²

    CM² = BC² + (1/2 AB)²

    CM² = BC² + 1/4 AB²----------(2)

Adding (1) and (2)

     AL² + CM² = (AB² + BC²) + 1/4(AB²+ BC²)

                       = AC² + 1/4 AC²

                       = AC²(1 + 1/4)

                   =  5/4 AC²---------(3)

 Given AL = 3√5/2 and AC = 5 cm

Substituting in eqn (3) we get

 (3√5/2)² + CM² = 5/4 x 25

                   CM² = 125/4 - 45/4

                    CM² = 20

                   CM = √20

               CM = 2√5 cm

Answer 2

Answer: √20 cm

Step-by-step explanation:

Here, AL is the median of triangle ABC⇒ L is the midpoint of side BC⇒ BC = 2BL

And, CM is the median of triangle ABC ⇒ M is the mid point of side AB ⇒ MB = 2AB

Also, AC = 5 cm and AL = 3√5/2 cm

Now, triangle ABC and ABL are two right triangle,

Therefore, by the Pythagoras theorem,

$AC^2 = AB^2 + BC^2$

And, $AL^2 = AB^2+BL^2$

By solving the above two equation,

We get, $BC^2-BL^2 = AC^2 -AL^2$

$(2\times BL)^2-BL^2=5^2 -(\frac{3\sqrt{5}}{2})^2$

$3BL^2=25 -\frac{45}{4}$

$3BL^2=\frac{100-45}{4}$

$BL^2=\frac{55}{12}$

⇒ $BC^2 = \frac{55}{3}$

$AB^2+BL^2=AL^2\implies AB^2 +\frac{55}{12}=(\frac{3\sqrt{5}}{2})^2\implies AB^2=\frac{45}{4}-\frac{55}{12}=\frac{80}{12}=\frac{20}{3}\implies MB^2=\frac{20}{12}$

Now, Again BMC is a right triangle,

Therefore,

$CM^2 = MB^2 + BC^2 = \frac{20}{12} + \frac{55}{3} = \frac{240}{12} = 20$

⇒ CM = √20 cm