The figure shown is made up of rectangle ABCD, a circle with FG as the diameter, 2 identical small semi-circles with diameters EF and GH and 2 larger semi-circles with diameters EG and FH. The radius of the circle FG is 18 cm. EH is a straight line. The length of BC is 60 cm.

  1. (a)

    Find the length of EF.

  2. (b)

    Find the total area of all the shaded parts. (Take π=3.14\pi = 3.14)

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Worked solution

15(a)

1. Identify the total length of the line segment EH

Since ABCDABCD is a rectangle and EHEH is a horizontal line across it, the length of EHEH is equal to the length of BCBC (or ADAD).
EH=BC=60EH = BC = 60 cm.

2. Express EH in terms of its segments

The line segment EHEH is composed of EFEF, FGFG, and GHGH.
EH=EF+FG+GHEH = EF + FG + GH

3. Use the given information to find EF

We are given that the radius of the circle FGFG is 1818 cm, so its diameter FG=2×18=36FG = 2 \times 18 = 36 cm.
We are also given that the two small semi-circles are identical, so their diameters are equal: EF=GHEF = GH.
Substituting these into the equation for EHEH:
60=EF+36+EF60 = EF + 36 + EF
2EF=60−36=242EF = 60 - 36 = 24
EF=12EF = 12 cm.

Answer

1212 cm

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