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primary 6 | Maths | Geometry
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Silin
Silin

primary 6 chevron_right Maths chevron_right Geometry chevron_right Singapore

Pls help me asap

Date Posted: 4 weeks ago
Views: 17

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For Q17(a):

To find area of one shaded part, we take area of big semicircle and minus area of small semicircle.

Diameter of big semicircle: 10 cm
Radius of big semicircle: 5 cm
Area: 3.14 x 5 x 5 ÷ 2 = 39.25 cm²

Diameter of small semicircle: 7 cm
Radius of big semicircle: 3.5 cm
Area: 3.14 x 3.5 x 3.5 ÷ 2 = 19.2325 cm²

Area of one shaded part:
39.25 - 19.2325 = 20.0175 cm²

Area of all shaded parts:
20.0175 x 2 = 40.035 cm² ≈ 40 cm²

For Q17(b):

Perimeter of one shaded part is the sum of:
(1) Arc length of big semicircle
(2) Arc length of small semicircle
(3) straight line at the bottom

Arc length of big semicircle is:
3.14 x 10 ÷ 2 = 15.7 cm

Arc length of small semicircle is:
3.14 x 7 ÷ 2 = 10.99 cm

Length of straight line:
10 - 7 = 3 cm

Perimeter of one shaded part:
15.7 + 10.99 + 3 = 29.69 cm

Perimeter of whole shaded part:
29.69 x 2 = 59.38 cm ≈ 59 cm
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