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ABDC is a quadrant of a circle of radius 28 cm and a semi-circle BEC is drawn with BC as diameter. Find the area of the shaded region.Class - 10th chapter - Area related to circles ___________________________ ❌ SPAM WON'T BE TOLERATED ❌ ❌❌ NO COPY PASTE ❌❌ ❌❌JUNIORS STAY AWAY ❌❌___________________________STEP BY STEP EXPLANATION ✅ Seeking on expert's help~ ! ⭐ |
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Answer» ong>Answer: Hope it helps!! MARK this answer as brainliest if u found it useful and follow me for quick and ACCURATE answers... Step-by-step explanation: Radius of the quadrant ABC of circle = 28 cm AB = AC = 28 cm BC is DIAMETER of semicircle. ABC is right ANGLED triangle. By Pythagoras theorem in ΔABC, BC² = AB² +AC² ⇒ BC² = 28² +28² ⇒ BC = 28√2 cm Radius of semicircle = 28√2/2 cm = 14√2 cm Area of ΔABC = ½×AB×AC = ½×28×28 = 392 cm² Area of quadrant = ¼πr² = (¼)×(22/7)×(28×288) = 616 cm² Area of the semicircle =½πr² = (½)×(22/7)×14√2×14√2 = 616 cm² Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant = 616 +392- 616 cm² = 392cm² |
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