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If the circumference of a circle is diminished by 10 %, then its area is diminished by
Given: - Circumference decreased by 10% ⇒ New radius = 0.9r. New area: A' = π(0.9r)² = 0.81πr² = 0.81A. Decrease in area: A - A' = A - 0.81A = 0.19A. Percentage decrease: (0.19A / A) × 100 = 19%. This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to CiRead more
Given:
– Circumference decreased by 10% ⇒ New radius = 0.9r.
New area:
A’ = π(0.9r)² = 0.81πr² = 0.81A.
Decrease in area:
A – A’ = A – 0.81A = 0.19A.
Percentage decrease:
(0.19A / A) × 100 = 19%.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
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The area of incircle of an equilateral triangle is 154 cm². The perimeter of the triangle is
Given: - Area of incircle = 154 cm². Step 1: Find radius of incircle. πr² = 154 ⇒ (22/7)r² = 154 ⇒ r² = 49 ⇒ r = 7 cm. Step 2: Relate radius to side of equilateral triangle. r = (a√3)/6 ⇒ 7 = (a√3)/6 ⇒ a = 42/√3 = 14√3 cm. Step 3: Find perimeter. Perimeter = 3a = 3 × 14√3 ≈ 3 × 14 × 1.732 ≈ 72.7 cm.Read more
Given:
– Area of incircle = 154 cm².
Step 1: Find radius of incircle.
πr² = 154 ⇒ (22/7)r² = 154 ⇒ r² = 49 ⇒ r = 7 cm.
Step 2: Relate radius to side of equilateral triangle.
r = (a√3)/6 ⇒ 7 = (a√3)/6 ⇒ a = 42/√3 = 14√3 cm.
Step 3: Find perimeter.
Perimeter = 3a = 3 × 14√3 ≈ 3 × 14 × 1.732 ≈ 72.7 cm.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The perimeter of a triangle is 30πcm and the circumference of its incircle is 88 cm. The area of the triangle is
Given: - Perimeter of triangle = 30π cm ⇒ Semiperimeter, s = 15π cm. - Circumference of incircle = 88 cm ⇒ Radius, r = 88 / (2π) = 14 cm. Area of triangle = s × r = 15π × 14 = 210π cm². Approximate π ≈ 22/7: Area ≈ 210 × (22/7) = 660 cm². This question related to Chapter 11 Mathematics Class 10th NCRead more
Given:
– Perimeter of triangle = 30π cm ⇒ Semiperimeter, s = 15π cm.
– Circumference of incircle = 88 cm ⇒ Radius, r = 88 / (2π) = 14 cm.
Area of triangle = s × r = 15π × 14 = 210π cm².
Approximate π ≈ 22/7:
Area ≈ 210 × (22/7) = 660 cm².
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The area of a circle is 220 cm². The area of a square is inscribed in it is
Given: - Area of circle = 220 cm² ⇒ πr² = 220 ⇒ r² = 70. Diameter of circle = 2r = 2√70. Diagonal of square = Diameter = 2√70. Side of square, a = Diagonal / √2 = (2√70) / √2 = √140. Area of square = a² = (√140)² = 140 cm². This question related to Chapter 11 Mathematics Class 10th NCERT. From the CRead more
Given:
– Area of circle = 220 cm² ⇒ πr² = 220 ⇒ r² = 70.
Diameter of circle = 2r = 2√70.
Diagonal of square = Diameter = 2√70.
Side of square, a = Diagonal / √2 = (2√70) / √2 = √140.
Area of square = a² = (√140)² = 140 cm².
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
What are the six limbs of art according to the Vishnudharmottara Purana?
According to the Vishnudharmottara Purana, the six limbs of art are essential for mastering visual art. They include knowledge of appearance (understanding form), proportion (balance and scale), emotions (expressing feelings), aesthetics (visual beauty), similarity (realistic representation) and skiRead more
According to the Vishnudharmottara Purana, the six limbs of art are essential for mastering visual art. They include knowledge of appearance (understanding form), proportion (balance and scale), emotions (expressing feelings), aesthetics (visual beauty), similarity (realistic representation) and skillful use of tools and colors. These principles help artists create harmonious, expressive and technically sound artworks that convey meaning while maintaining artistic integrity.
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