Question

In ΔABC, AB=10,AC=7,BC=9 then find the length of the median drawn from point C to side AB

Answer 1

Let E is the point on AB in such a way that AE = EB then, CE is median from point C to side AB . Here we are also drawn CF ⊥ AB
Let FE = x
Now, from right angled ∆AFC
so, AC² = CF² + AF²
⇒7² = CF² + (AE - FE)² [ see figure ]
⇒49 = CF² + (5 - x)²
⇒49 = CF² + 25 + x² - 10x
⇒CF² = -x² + 10x + 24 -----(1)

Again, from right angled ∆BFC
BC² = CF² + BF²
CF² = BC² - BF²
= 9² - (5 + x)² [ see figure ]
= 81 - 25 - x² - 10x
= 56 - x² - 10x ------(2)

From equation (1), and (2),
-x² + 10x +24 = 56 - 10x - x²
20x = 56 - 24 = 32
x = 8/5 = 1.6 m so, FE = 1.6 m
Now, CF² = 56 - (1.6)² - 10(1.6) = 56 - 2.56 - 16 = 40 - 2.56 = 37.44 m

Now, From right angled ∆CFE ,
CF² + FE² = CE²
37.44 + (1.6)² = CE²
37.44 + 2.56 = CE²
CE² = 40 ⇒CE = √40 = 2√10

Hence, Median , CE = 2√10