Question

Volume question in the attachment. Help!

Answer 1

In this diagram there are two figures, Δ EDF and Semi circle AENDC

Height of Δ EDF = Height of Δ ABC-2×0.5 =10.5 - 1 cm= 9.5 cm

Base= AC - 2×0.5 =7-1=6cm

Area of Δ EDF = 1÷2 × Base × Height

=0.5 × 6 × 9.5

=28.5 cm²

Radius of semi circle=3.5 cm

Area of semi circle=πr²/2              (Since semi circle is half of circle)

22/7×3.5×3.5/2=19.25 cm²

Area of unshaded part = Area of Δ EDF + Area of Semi circle

=28.5 + 19.25= 47.75 cm²

Hope it helps!

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