Question

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Answer 1

Answer:

6 cm

Step-by-step explanation:

From figure:

Let AB = 4 cm, AC = 5 cm, BC = 3 cm, AM = a, BM = b, CM = x.

Then, (a + b) = 4   ----- (i)

(ii) In ΔACM:

(AM)² + (CM)² = (AC)²

a² + x² = (5)²

a² = 25 - x²

(iii) In ΔBCM:

(BM)² + (CM)² = (BC)²

b² + x² = 3²

b² = 9 - x²

From (ii) & (iii), we get

⇒ a² - b² = (25 - x²) - (9 - x²)

⇒ (a + b)(a - b) = 25 - x² - 9 + x²

⇒ 4 * (a - b) = 16

⇒ a - b = 4

On solving (i) & (iii), we get

a + b = 4

a - b = 4

-------------------

2a = 8

a = 4.

Then, substitute a = 4 in (i), we get

a + b = 4

4 + b = 4

b = 0.

Then, from (ii), we get

⇒ x² = 25 - a²

⇒ x² = 25 - 4²

⇒ x² = 9

⇒ x = 3

∴ Length of common chord = 2CM

                                              = 2(3)

                                              = 6 cm.

Therefore, Length of common chord = 6 cm.

Hope it helps!

Answer 2

Step-by-step explanation:

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