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Mrac [35]
3 years ago
8

I need help with these questions (see image). Please show workings.​

Mathematics
2 answers:
Eduardwww [97]3 years ago
6 0

Answer:

r = \frac{1}{2} \sqrt{l^2+4h^2}

Step-by-step explanation:

h bisects l at right angles.

There is a right triangle formed with legs h and \frac{1}{2} l and hypotenuse r

Using Pythagoras' identity, then

r² = (\frac{1}{2} l )² + h²

   = \frac{1}{4} l² + h² ← factor out \frac{1}{4}

    = \frac{1}{4} (l² + 4h²)

Take the square root of both sides

r = \sqrt{\frac{1}{4}(l^2+4h^2) } = \frac{1}{2} \sqrt{l^2+4h^2}

LuckyWell [14K]3 years ago
3 0

Answer:

  • r = \sqrt{(l/2)^2 + h^2}

Step-by-step explanation:

Connect the endpoints of the with the center. The formed triangle is isosceles with sides l, r and r.

The height of the triangle h, is the distance between the center of the circle and the midpoint of the chord.

<u>As per Pythagorean theorem we have:</u>

  • (l/2)² + h² = r²
  • \sqrt{(l/2)^2 + h^2} = r
  • r = \sqrt{(l/2)^2 + h^2}

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