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Zorluk: Çok zorPerimeter and Area of Plane Shapes

A composite plane figure is formed by joining a major sector of a circle of radius 14 cm14\text{ cm} (having a central angle of 270270^\circ and center OO) to a square OACBOACB. The vertices AA and BB of the square lie on the circle such that OAOA and OBOB serve as the bounding radii of the major sector. What is the total area of the combined figure in cm2\text{cm}^2? (Take π=227\pi = \frac{22}{7})

  1. A
    350 cm2350\text{ cm}^2
  2. B
    504 cm2504\text{ cm}^2
  3. C
    560 cm2560\text{ cm}^2
  4. 658 cm2658\text{ cm}^2Cevap

Cevap

658 cm2658\text{ cm}^2
The total area of the composite figure is the sum of the non-overlapping regions: the major sector of central angle 270270^\circ (462 cm2462\text{ cm}^2) and the square of side length 14 cm14\text{ cm} (196 cm2196\text{ cm}^2), giving 462+196=658 cm2462 + 196 = 658\text{ cm}^2.

Adım Adım Çözüm

1
Calculate the area of the major sector of central angle 270270^\circ and radius 14 cm14\text{ cm}.
\text{Area}_{\text{sector}} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times \frac{22}{7} \times 14^2 = \frac{3}{4} \times 616 = 462\text{ cm}^2
The major sector covers 270360=34\frac{270}{360} = \frac{3}{4} of the full circle.
2
Calculate the area of the square OACBOACB with side length s=OA=14 cms = OA = 14\text{ cm}.
\text{Area}_{\text{square}} = s^2 = 14^2 = 196\text{ cm}^2
The radii OAOA and OBOB form two adjacent sides of the square OACBOACB of length 14 cm14\text{ cm}.
3
Sum the areas of the major sector and the square to find the total composite area.
\text{Total Area} = 462\text{ cm}^2 + 196\text{ cm}^2 = 658\text{ cm}^2
The major sector and square share boundaries OAOA and OBOB without interior overlap.

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Perimeter and Area of Plane Shapes
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