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In the kite shown alongside, ABCD is a square. BCD is a quadrant of a circle of radius 14 cm and CEF is an isosceles right-angled triangle with CE = CF = 3 cm. Find the area of the shaded portion. - Mathematics

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Question

In the kite shown alongside, ABCD is a square. BCD is a quadrant of a circle of radius 14 cm and CEF is an isosceles right-angled triangle with CE = CF = 3 cm. Find the area of the shaded portion.

Sum
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Solution

Given:

  • ABCD is a square
  • BCD is a quadrant of a circle with radius 14 cm
  • CEF is an isosceles right triangle with CE = CF = 3 cm
  • π = 3.14

Stepwise Calculation:

1. Since ABCD is a square and BCD is a quadrant based on the same circle radius 14 cm, then side BC = 14 cm.

2. Area of quadrant BCD a quarter circle of radius 14 cm:

Area of quadrant = `(πr^2)/4`

= `(3.14 xx 14 xx 14)/4`

= `(3.14 xx 196)/4`

= `(615.44)/4`

= 153.86 cm2

3. Area of triangle CEF isosceles right triangle with legs 3 cm each:

Area = `1/2 xx 3 xx 3`

Area = 4.5 cm2

4. The shaded portion is the sum of the quadrant BCD and the triangle CEF because the shaded area includes both shapes the kite is formed by combining these shapes, not just subtracting the triangle from the quadrant.

Thus, total shaded area = Area of quadrant + Area of triangle

= 153.86 + 4.5

= 158.36 cm

Rounded off to 158.5 cm2.

The area of the shaded portion is approximately 158.5 cm2, as the shaded region comprises both the quadrant and the small triangle combined, not as the difference.

So the difference in area in your first calculation was due to subtracting the triangle’s area instead of adding it.

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Chapter 17: Mensuration - MISCELLANEOUS EXERCISE [Page 218]

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B Nirmala Shastry Mathematics [English] Class 9 ICSE
Chapter 17 Mensuration
MISCELLANEOUS EXERCISE | Q 17. | Page 218
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