Let the each side of equilateral triangle = a Area of equilateral triangle = 17320.5 cm² ⇒ √3/4 a² = 17320.5 ⇒ 1.73205/4 a² ⇒ 17320.5 ⇒ a² = 40000 ⇒ a = 200 Area of sector ADEF = θ/360° × πr² = 60°/360° × 22/7(100)² = 1/6 × 3.14 × 100 × 100 = 15700/3 cm² Area of shaded region = Area of equilateral tRead more
Let the each side of equilateral triangle = a
Area of equilateral triangle = 17320.5 cm²
⇒ √3/4 a² = 17320.5 ⇒ 1.73205/4 a² ⇒ 17320.5 ⇒ a² = 40000 ⇒ a = 200
Area of sector ADEF
= θ/360° × πr² = 60°/360° × 22/7(100)² = 1/6 × 3.14 × 100 × 100 = 15700/3 cm²
Area of shaded region = Area of equilateral triangle – Area of three sectors
= 17320.5 cm² – 3 x 15700/3 cm²
= 17320.5 cm² – 15700 cm²
= 1620.5 cm²
Radius of circle = 7 cm Area of one circular design πr² = π(7)² = 22/7 × 7 × 7 = 154 cm² Side of square = 42 cm Area of squre = (Side)² = (42)² = 1764cm² The area of the remaining portion = Area of square - Area of 9 circular designs = 1764 - 9 × 154 = 196 1386 = 378 cm²
Radius of circle = 7 cm
Area of one circular design
πr² = π(7)² = 22/7 × 7 × 7 = 154 cm²
Side of square = 42 cm
Area of squre = (Side)² = (42)² = 1764cm²
The area of the remaining portion = Area of square – Area of 9 circular designs
= 1764 – 9 × 154 = 196 1386 = 378 cm²
In A0AB, OB² = OA² + AB² ⇒ OB² = (20)² + (20)² ⇒ OB² = 400+ 400 ⇒ OB² = 800 ⇒ OB = √800 Radius of quadrant = 20√2 cm Area of quadrant = 90°/360° × πr² = 1/4 × π(20√2)² = 1/4 × 3.14 × 20√2 × 20√2 = 628 cm² Area of square = (Side)² = (20)² = 400 cm² Area of shaded region = Area of quadrant - Area of sRead more
In A0AB,
OB² = OA² + AB² ⇒ OB² = (20)² + (20)² ⇒ OB² = 400+ 400
⇒ OB² = 800 ⇒ OB = √800
Radius of quadrant = 20√2 cm
Area of quadrant
= 90°/360° × πr² = 1/4 × π(20√2)²
= 1/4 × 3.14 × 20√2 × 20√2 = 628 cm²
Area of square
= (Side)² = (20)² = 400 cm²
Area of shaded region = Area of quadrant – Area of square
= 628 – 400 = 228 cm²
Radius of smaller circle = 7/2 cm Area of smaller circle = πr² = π(7/2)² = 22/7 × 7/2 × 7/2 = 77/2 cm² Radius of larger circle = 7 cm Area of semicircle AECFB = 1/2 × πr² = π(7)² = 1/2 × 22/7 7 × 7 = 77 cm² Area of triangle ACB = 1/2 × AB × OC = 1/2 × 14 × 7 = 49 cm² Area of Shaded Region = Area ofRead more
Radius of smaller circle = 7/2 cm
Area of smaller circle = πr² = π(7/2)² = 22/7 × 7/2 × 7/2 = 77/2 cm²
Radius of larger circle = 7 cm
Area of semicircle AECFB
= 1/2 × πr² = π(7)² = 1/2 × 22/7 7 × 7 = 77 cm²
Area of triangle ACB
= 1/2 × AB × OC = 1/2 × 14 × 7 = 49 cm²
Area of Shaded Region
= Area of smaller circle + Area of semicircle AECFB – Area of triangle ACB
= (77/2 + 77 – 49) cm² = (38.5 + 28) cm² = 66.5 cm²
Here you can see the video explanation of this question for better understanding😄👇
जब कूलम्ब के नियम को कूलम्ब (संक्षिप्त रूप में "C" के रूप में जाना जाता है) नामक आवेश की इकाई का उपयोग करके व्यक्त किया जाता है, तो आमतौर पर यह माना जाता है कि दोनों कणों की प्रकृति समान है, या उन दोनों में एक ही प्रकार का आवेश है (या तो दोनों धनात्मक या दोनों नकारात्मक)। ऐसा इसलिए है क्योंकि कूलम्बRead more
जब कूलम्ब के नियम को कूलम्ब (संक्षिप्त रूप में “C” के रूप में जाना जाता है) नामक आवेश की इकाई का उपयोग करके व्यक्त किया जाता है, तो आमतौर पर यह माना जाता है कि दोनों कणों की प्रकृति समान है, या उन दोनों में एक ही प्रकार का आवेश है (या तो दोनों धनात्मक या दोनों नकारात्मक)। ऐसा इसलिए है क्योंकि कूलम्ब को एक कण के आवेश के रूप में परिभाषित किया जाता है जो 1 C के आवेश वाले दूसरे कण पर 9 x 10⁹ N का बल उत्पन्न करने में सक्षम होता है, जब दो कणों को 1 मीटर की दूरी से अलग किया जाता है।
अधिक जानकारी के लिए आप इस विडियो को देख सकते है.
The angle of a prism is the measure of the inclination between two adjacent faces of the prism. In other words, it is the angle between two planes that intersect on a line that is perpendicular to both planes. For better explanation see here.
The angle of a prism is the measure of the inclination between two adjacent faces of the prism. In other words, it is the angle between two planes that intersect on a line that is perpendicular to both planes.
For better explanation see here.
Find the area of the shaded region.
Area of shaded region = Area of sector OAEB - Area of sector OCFD = 30°/360° × π × (21)² - 30°/360° × π × 7² = 1/12 × π[441 - 49] = 1/12 × 22/7 × 392 = 308/3 cm²
Area of shaded region = Area of sector OAEB – Area of sector OCFD
See less= 30°/360° × π × (21)² – 30°/360° × π × 7²
= 1/12 × π[441 – 49]
= 1/12 × 22/7 × 392 = 308/3 cm²
The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see Figure).
Let the each side of equilateral triangle = a Area of equilateral triangle = 17320.5 cm² ⇒ √3/4 a² = 17320.5 ⇒ 1.73205/4 a² ⇒ 17320.5 ⇒ a² = 40000 ⇒ a = 200 Area of sector ADEF = θ/360° × πr² = 60°/360° × 22/7(100)² = 1/6 × 3.14 × 100 × 100 = 15700/3 cm² Area of shaded region = Area of equilateral tRead more
Let the each side of equilateral triangle = a
See lessArea of equilateral triangle = 17320.5 cm²
⇒ √3/4 a² = 17320.5 ⇒ 1.73205/4 a² ⇒ 17320.5 ⇒ a² = 40000 ⇒ a = 200
Area of sector ADEF
= θ/360° × πr² = 60°/360° × 22/7(100)² = 1/6 × 3.14 × 100 × 100 = 15700/3 cm²
Area of shaded region = Area of equilateral triangle – Area of three sectors
= 17320.5 cm² – 3 x 15700/3 cm²
= 17320.5 cm² – 15700 cm²
= 1620.5 cm²
In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, fiIn Fig. 12.30, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm,
Radius of circle = 7 cm Area of one circular design πr² = π(7)² = 22/7 × 7 × 7 = 154 cm² Side of square = 42 cm Area of squre = (Side)² = (42)² = 1764cm² The area of the remaining portion = Area of square - Area of 9 circular designs = 1764 - 9 × 154 = 196 1386 = 378 cm²
Radius of circle = 7 cm
See lessArea of one circular design
πr² = π(7)² = 22/7 × 7 × 7 = 154 cm²
Side of square = 42 cm
Area of squre = (Side)² = (42)² = 1764cm²
The area of the remaining portion = Area of square – Area of 9 circular designs
= 1764 – 9 × 154 = 196 1386 = 378 cm²
In Figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm,
In A0AB, OB² = OA² + AB² ⇒ OB² = (20)² + (20)² ⇒ OB² = 400+ 400 ⇒ OB² = 800 ⇒ OB = √800 Radius of quadrant = 20√2 cm Area of quadrant = 90°/360° × πr² = 1/4 × π(20√2)² = 1/4 × 3.14 × 20√2 × 20√2 = 628 cm² Area of square = (Side)² = (20)² = 400 cm² Area of shaded region = Area of quadrant - Area of sRead more
In A0AB,
See lessOB² = OA² + AB² ⇒ OB² = (20)² + (20)² ⇒ OB² = 400+ 400
⇒ OB² = 800 ⇒ OB = √800
Radius of quadrant = 20√2 cm
Area of quadrant
= 90°/360° × πr² = 1/4 × π(20√2)²
= 1/4 × 3.14 × 20√2 × 20√2 = 628 cm²
Area of square
= (Side)² = (20)² = 400 cm²
Area of shaded region = Area of quadrant – Area of square
= 628 – 400 = 228 cm²
In Figure, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.
Radius of smaller circle = 7/2 cm Area of smaller circle = πr² = π(7/2)² = 22/7 × 7/2 × 7/2 = 77/2 cm² Radius of larger circle = 7 cm Area of semicircle AECFB = 1/2 × πr² = π(7)² = 1/2 × 22/7 7 × 7 = 77 cm² Area of triangle ACB = 1/2 × AB × OC = 1/2 × 14 × 7 = 49 cm² Area of Shaded Region = Area ofRead more
Radius of smaller circle = 7/2 cm
Area of smaller circle = πr² = π(7/2)² = 22/7 × 7/2 × 7/2 = 77/2 cm²
Radius of larger circle = 7 cm
Area of semicircle AECFB
= 1/2 × πr² = π(7)² = 1/2 × 22/7 7 × 7 = 77 cm²
Area of triangle ACB
= 1/2 × AB × OC = 1/2 × 14 × 7 = 49 cm²
Area of Shaded Region
= Area of smaller circle + Area of semicircle AECFB – Area of triangle ACB
= (77/2 + 77 – 49) cm² = (38.5 + 28) cm² = 66.5 cm²
Here you can see the video explanation of this question for better understanding😄👇
See lessWhat are the important chapters NCERT class 12th political science?
Where can I get 12th class political science solution in Hindi medium?
Where can I get 12th class political science solution in Hindi medium?
See lessCoulomb’s law
जब कूलम्ब के नियम को कूलम्ब (संक्षिप्त रूप में "C" के रूप में जाना जाता है) नामक आवेश की इकाई का उपयोग करके व्यक्त किया जाता है, तो आमतौर पर यह माना जाता है कि दोनों कणों की प्रकृति समान है, या उन दोनों में एक ही प्रकार का आवेश है (या तो दोनों धनात्मक या दोनों नकारात्मक)। ऐसा इसलिए है क्योंकि कूलम्बRead more
जब कूलम्ब के नियम को कूलम्ब (संक्षिप्त रूप में “C” के रूप में जाना जाता है) नामक आवेश की इकाई का उपयोग करके व्यक्त किया जाता है, तो आमतौर पर यह माना जाता है कि दोनों कणों की प्रकृति समान है, या उन दोनों में एक ही प्रकार का आवेश है (या तो दोनों धनात्मक या दोनों नकारात्मक)। ऐसा इसलिए है क्योंकि कूलम्ब को एक कण के आवेश के रूप में परिभाषित किया जाता है जो 1 C के आवेश वाले दूसरे कण पर 9 x 10⁹ N का बल उत्पन्न करने में सक्षम होता है, जब दो कणों को 1 मीटर की दूरी से अलग किया जाता है।
See lessअधिक जानकारी के लिए आप इस विडियो को देख सकते है.
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See lessWhat do you mean by angle of prism?
The angle of a prism is the measure of the inclination between two adjacent faces of the prism. In other words, it is the angle between two planes that intersect on a line that is perpendicular to both planes. For better explanation see here.
The angle of a prism is the measure of the inclination between two adjacent faces of the prism. In other words, it is the angle between two planes that intersect on a line that is perpendicular to both planes.
See lessFor better explanation see here.