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In a given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the
In a given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
📘 K C Sinha Class 10 Maths | Area of Shaded Region in Circle | Mensuration
In this video, we solve an important mensuration problem involving a circle, chords, and sectors from K C Sinha Class 10 Mathematics.
Question:
In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
This question uses concepts of:
✔ Area of a circle
✔ Area of a sector
✔ Area of a triangle inscribed in a circle
✔ Diameter theorem
✔ Shaded region problems
Since CB passes through the centre O, CB is the diameter.
Using Pythagoras' theorem:
24² + 7² = 625 = 25²
Therefore,
CB = 25 cm = Diameter
Radius = 12.5 cm
A=πr^2
The shaded region consists of:
Area of the semicircle above CB minus area of △ABC
Area of sector BOD (90°)
Total shaded area
= (1/2 × π × 12.5² − 1/2 × 24 × 7)
(90/360 × π × 12.5²)
= (245.54 − 84) + 122.77
= 284.31 cm² (approximately)
🔹 K C Sinha Class 10 Maths Solutions
🔹 Circle and Sector Problems
🔹 Shaded Region Questions
🔹 Board Exam Preparation
🔹 Step-by-Step Explanation
#KCSinha #Class10Maths #Mensuration #AreaOfShadedRegion #Circle #Sector #Geometry #MathsSolution #BoardExam #Mathematics
Видео In a given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the канала padhai.Online
📘 K C Sinha Class 10 Maths | Area of Shaded Region in Circle | Mensuration
In this video, we solve an important mensuration problem involving a circle, chords, and sectors from K C Sinha Class 10 Mathematics.
Question:
In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the area of the shaded region.
This question uses concepts of:
✔ Area of a circle
✔ Area of a sector
✔ Area of a triangle inscribed in a circle
✔ Diameter theorem
✔ Shaded region problems
Since CB passes through the centre O, CB is the diameter.
Using Pythagoras' theorem:
24² + 7² = 625 = 25²
Therefore,
CB = 25 cm = Diameter
Radius = 12.5 cm
A=πr^2
The shaded region consists of:
Area of the semicircle above CB minus area of △ABC
Area of sector BOD (90°)
Total shaded area
= (1/2 × π × 12.5² − 1/2 × 24 × 7)
(90/360 × π × 12.5²)
= (245.54 − 84) + 122.77
= 284.31 cm² (approximately)
🔹 K C Sinha Class 10 Maths Solutions
🔹 Circle and Sector Problems
🔹 Shaded Region Questions
🔹 Board Exam Preparation
🔹 Step-by-Step Explanation
#KCSinha #Class10Maths #Mensuration #AreaOfShadedRegion #Circle #Sector #Geometry #MathsSolution #BoardExam #Mathematics
Видео In a given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and ∠BOD = 90°. Find the канала padhai.Online
K C Sinha Class 10 Maths K C Sinha Solution area of shaded region circle mensuration sector of circle shaded region problem geometry class 10 mensuration class 10 circle theorem diameter theorem area of sector area of triangle class 10 maths solution board exam maths mathematics tutorial K C Sinha English Medium circle questions geometry problems exam preparation shaded area questions
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18 июня 2026 г. 16:00:05
00:04:53
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