In the diagram above, XZ is the diameter of the circle XZW, with centre O and radius 15/2 cm. If XY = 12 cm, find the area of the triangle XYZ

  • A 54 cm2
  • B 45 cm2
  • C 27 cm2
  • D 75 cm2

The correct answer is A. 54 cm2

\( o \, (\text{< semi circle}) \)

Since the radius \( r \) is given by \( \frac{15}{2} \) cm, the diameter is \( 2 \times \frac{15}{2} = 15 \) cm.

\( \text{XYZ is a right-angled triangle} \)

Therefore, using the Pythagorean theorem \( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse:

\(15^2 = 12^2 + (YZ)^2 \)

\( 225 = 144 + (YZ)^2 \)

\(225 - 144 = (YZ)^2 \)

\(81 = (YZ)^2 \)

Taking the square root of both sides:

\(\sqrt{81} = YZ \)

\(9 = YZ \)

The area (\( A \)) of the triangle \( XYZ \) is given by the formula \( A = \frac{1}{2} \times \text{base} \times \text{height} \):

\(A = \frac{1}{2} \times 9 \times 12 \)

\(A = 9 \times 6 = 54 \, \text{cm}^2 \)

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