Question

solve this 2nd question
waste answers for me will be reported...
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Answer 1

Answer:

$\huge\bold{answer :-}$

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

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