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4 Use the rule of Pythagoras to find the diameter of the circle. 12.5 Calculate the area of the triangle. 12.6 Find the shaded area, to 1 decimal place.

Mathematics

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3 parts
12.4

13 cm13 \text{ cm}

12.5

30 cm230 \text{ cm}^2

12.6

36.4 cm236.4 \text{ cm}^2

Step-by-step solution

9 steps

Part 12.4· 3 steps

Use the rule of Pythagoras to find the diameter of the circle.

  1. Identify the sides of the right-angled triangle.

    The triangle inscribed in the semicircle is a right-angled triangle. The two given sides are the legs, and the hypotenuse is the diameter of the semicircle.
    Leg 1 (aa) = 5 cm5 \text{ cm}
    Leg 2 (bb) = 12 cm12 \text{ cm}
    Hypotenuse (cc) = Diameter (dd)

  2. Apply the Pythagorean theorem (a2+b2=c2a^2 + b^2 = c^2).

    52+122=d25^2 + 12^2 = d^2 25+144=d225 + 144 = d^2 169=d2169 = d^2

  3. Solve for the diameter dd.

    d=169d = \sqrt{169} d=13 cmd = 13 \text{ cm}
    • 13 cm\boxed{13 \text{ cm}}

Part 12.5

30 cm230 \text{ cm}^2

Calculate the area of the triangle.

  1. Identify the base and height of the right-angled triangle.

    For a right-angled triangle, the two legs can be considered the base and height.
    Base = 12 cm12 \text{ cm}
    Height = 5 cm5 \text{ cm}

  2. Use the formula for the area of a triangle (A=12×base×heightA = \frac{1}{2} \times \text{base} \times \text{height}).

    A=12×12 cm×5 cmA = \frac{1}{2} \times 12 \text{ cm} \times 5 \text{ cm} A=6 cm×5 cmA = 6 \text{ cm} \times 5 \text{ cm} A=30 cm2A = 30 \text{ cm}^2
    • 30 cm2\boxed{30 \text{ cm}^2}

Part 12.6

36.4 cm236.4 \text{ cm}^2

Find the shaded area, to 1 decimal place.

  1. Calculate the radius of the semicircle.

    The diameter d=13 cmd = 13 \text{ cm} (from 12.4).
    The radius r=d2=132=6.5 cmr = \frac{d}{2} = \frac{13}{2} = 6.5 \text{ cm}.

  2. Calculate the area of the semicircle.

    The area of a full circle is Acircle=πr2A_{circle} = \pi r^2.
    The area of a semicircle is Asemicircle=12πr2A_{semicircle} = \frac{1}{2} \pi r^2.

    Asemicircle=12×π×(6.5 cm)2A_{semicircle} = \frac{1}{2} \times \pi \times (6.5 \text{ cm})^2 Asemicircle=12×π×42.25 cm2A_{semicircle} = \frac{1}{2} \times \pi \times 42.25 \text{ cm}^2 Asemicircle=21.125π cm2A_{semicircle} = 21.125 \pi \text{ cm}^2

    Using π≈3.14159\pi \approx 3.14159:

    Asemicircle≈21.125×3.14159≈66.366 cm2A_{semicircle} \approx 21.125 \times 3.14159 \approx 66.366 \text{ cm}^2

  3. Calculate the shaded area.

    The shaded area is the area of the semicircle minus the area of the triangle.

    Ashaded=Asemicircle−AtriangleA_{shaded} = A_{semicircle} - A_{triangle} Ashaded=66.366 cm2−30 cm2A_{shaded} = 66.366 \text{ cm}^2 - 30 \text{ cm}^2 Ashaded=36.366 cm2A_{shaded} = 36.366 \text{ cm}^2

  4. Round the shaded area to 1 decimal place.

    Ashaded≈36.4 cm2A_{shaded} \approx 36.4 \text{ cm}^2
    • 36.4 cm2\boxed{36.4 \text{ cm}^2}
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