Your answer is awaiting moderation. Radius of first circle (r₁) = 19 cm, Radius of second circle (r₂) = 9 cm Let, the radius of the third circle = r Circumference of the first circle = 2πr₁ = 2π (19) = 38π Circumference of the second circle = 2πr₂ = 2π (9) = 18π Circumference of the third circle = 2Read more
Your answer is awaiting moderation.
Radius of first circle (r₁) = 19 cm,
Radius of second circle (r₂) = 9 cm
Let, the radius of the third circle = r
Circumference of the first circle = 2πr₁ = 2π (19) = 38π
Circumference of the second circle = 2πr₂ = 2π (9) = 18π
Circumference of the third circle = 2πr
According to question,
Circumference of the third circle = Circumference of the first circle + Circumference of the second circle
⇒ 2πr = 38π + 18π ⇒ 2πr = 56π ⇒ r = 56π/2π = 28
Hence, the radius of the circle, which has circumference equal to the sum of the circumferences of the two circles, is 28 cm.
Here is the Video Solution of the above question ✔
Radius of first circle (r₁) = 8 cm, Radius of second circle (r₂) = 6 cm Let the radius of the third circle = r Area of the first circle = πr₁² = π(64) = 64π Area of the second circle = πr₂² = π(36) = 36π Area of the third circle = πr² According to question, Area of the third circle = Area of the firRead more
Radius of first circle (r₁) = 8 cm,
Radius of second circle (r₂) = 6 cm
Let the radius of the third circle = r
Area of the first circle = πr₁² = π(64) = 64π
Area of the second circle = πr₂² = π(36) = 36π
Area of the third circle = πr²
According to question,
Area of the third circle = Area of the first circle + Area of the second circle
⇒ πr² = 64π + 36π ⇒ πr² = 100π ⇒ r = √100 = 10 cm
Hence, the radius of the circle, having area equal to the sum of the areas of the two circles, is 10 cm.
Environmental Studies, or EVS, is a subject that helps students understand the world around them. In grade 2, students learn about various topics related to their daily life like sense organs, different clothes to wear in different seasons, different types of festivals and so on. Some of the benefitRead more
Environmental Studies, or EVS, is a subject that helps students understand the world around them. In grade 2, students learn about various topics related to their daily life like sense organs, different clothes to wear in different seasons, different types of festivals and so on.
Some of the benefits of learning about EVs in Class 2 include:
EVS helps the students to understand the importance of moral values and its role in our lives. It helps students to understand their surroundings.
Yes, it is possible to learn Environmental Studies (EVS) in an easy way. There are many resources available that can make learning about EVS engaging and enjoyable.
Some tips to learn about EVS in an easy way:
Relate EVS concepts to your daily life and experiences to make them more relevant and easier to understand. Encourage curiosity and encourage your child to ask questions about the environment and how things work.
I know the answer 😎, Let a be any positive integer and b = 3. Using Euclid’s Division Lemma, a = 3q + r for some integer q ≥ 0 where r = 0, 1, 2 because 0 ≤ r < 3. Therefore, a = 3q or 3q + 1 or 3q + 2 a² = (3q)² or (3q + 1)² or (3q + 2)² =(3q)² or 9q² + 6q + 1 or 9q² + 12q + 4 = 3×(3q²) or 3×(3qRead more
I know the answer 😎,
Let a be any positive integer and b = 3. Using Euclid’s Division Lemma, a = 3q + r for some integer q ≥ 0 where r = 0, 1, 2 because 0 ≤ r < 3.
Therefore, a = 3q or 3q + 1 or 3q + 2
a² = (3q)² or (3q + 1)² or (3q + 2)²
=(3q)² or 9q² + 6q + 1 or 9q² + 12q + 4
= 3×(3q²) or 3×(3q² + 2q) + 1 or 3×(3q² + 4q + 1) + 1
= 3k₁ or 3k₂ + 1 or 3k₃ + 1
Where k₁, k₂ and k₃ are some positive integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
A round table cover has six equal designs as shown in If the radius of the cover is 28 cm, find the cost of making the designs at the rate of ` 0.35 per cm2. (Use √3 = 1.7)
6 design of table cover = 6 sectors of circle Therefore, angle of each sector = 360°/6 = 60° In △OAB, ∠OAB = ∠OBA [Because OA = OB] ∠AOB = 60° ∠0AB + ∠OBA + ∠AOB = 180° 2∠0AB = 180° - 60° = 120° ∠0AB = 60° Similarly, △OAB is an equilateral triangle. Area of sector OAPB 60°/360° × πr² = 1/6 × π(28)²Read more
6 design of table cover = 6 sectors of circle
See lessTherefore, angle of each sector = 360°/6 = 60°
In △OAB,
∠OAB = ∠OBA [Because OA = OB]
∠AOB = 60°
∠0AB + ∠OBA + ∠AOB = 180°
2∠0AB = 180° – 60° = 120°
∠0AB = 60°
Similarly, △OAB is an equilateral triangle.
Area of sector OAPB
60°/360° × πr²
= 1/6 × π(28)²
= 1/6 × 28 × 28
= 1232/3 cm²
Area of equilateral triangle OAB
= √3/4(28)² = 196√3
= 196 x 1.7 = 333.2 cm²
Area of segment
= Area of sector OAPB – area of equilateral triangle OAB
= (1232/3 – 333.2) cm²
Area of 6 designs of table cover
= 6 x (1232/2 – 333.2) cm²
= (2464 – 1999.2) cm² = 464.8 cm²
Cost of making 1 cm² design = ₹ 0.35
Therefore, the cost of making 464.8 cm² design = ₹ 0.35 × 464.8 = ₹ 162.68
The radii of two circles are 19 cm and 9 cm respectively. Find the radius of the circle which has circumference equal to the sum of the circumferences of the two circles.
Your answer is awaiting moderation. Radius of first circle (r₁) = 19 cm, Radius of second circle (r₂) = 9 cm Let, the radius of the third circle = r Circumference of the first circle = 2πr₁ = 2π (19) = 38π Circumference of the second circle = 2πr₂ = 2π (9) = 18π Circumference of the third circle = 2Read more
Your answer is awaiting moderation.
Radius of first circle (r₁) = 19 cm,
Radius of second circle (r₂) = 9 cm
Let, the radius of the third circle = r
Circumference of the first circle = 2πr₁ = 2π (19) = 38π
Circumference of the second circle = 2πr₂ = 2π (9) = 18π
Circumference of the third circle = 2πr
According to question,
Circumference of the third circle = Circumference of the first circle + Circumference of the second circle
⇒ 2πr = 38π + 18π ⇒ 2πr = 56π ⇒ r = 56π/2π = 28
Hence, the radius of the circle, which has circumference equal to the sum of the circumferences of the two circles, is 28 cm.
Here is the Video Solution of the above question ✔
See lessThe radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the areas of the two circles.
Radius of first circle (r₁) = 8 cm, Radius of second circle (r₂) = 6 cm Let the radius of the third circle = r Area of the first circle = πr₁² = π(64) = 64π Area of the second circle = πr₂² = π(36) = 36π Area of the third circle = πr² According to question, Area of the third circle = Area of the firRead more
Radius of first circle (r₁) = 8 cm,
Radius of second circle (r₂) = 6 cm
Let the radius of the third circle = r
Area of the first circle = πr₁² = π(64) = 64π
Area of the second circle = πr₂² = π(36) = 36π
Area of the third circle = πr²
According to question,
Area of the third circle = Area of the first circle + Area of the second circle
⇒ πr² = 64π + 36π ⇒ πr² = 100π ⇒ r = √100 = 10 cm
Hence, the radius of the circle, having area equal to the sum of the areas of the two circles, is 10 cm.
Video Explanation for this question, see here👀🙌
See lessHow to download Class 2 EVS study material?
Environmental Studies, or EVS, is a subject that helps students understand the world around them. In grade 2, students learn about various topics related to their daily life like sense organs, different clothes to wear in different seasons, different types of festivals and so on. Some of the benefitRead more
Environmental Studies, or EVS, is a subject that helps students understand the world around them. In grade 2, students learn about various topics related to their daily life like sense organs, different clothes to wear in different seasons, different types of festivals and so on.
Some of the benefits of learning about EVs in Class 2 include:
EVS helps the students to understand the importance of moral values and its role in our lives. It helps students to understand their surroundings.
Yes, it is possible to learn Environmental Studies (EVS) in an easy way. There are many resources available that can make learning about EVS engaging and enjoyable.
Some tips to learn about EVS in an easy way:
Relate EVS concepts to your daily life and experiences to make them more relevant and easier to understand. Encourage curiosity and encourage your child to ask questions about the environment and how things work.
For More details and Study Material Click here 👀👇
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-2/evs/
Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.
I know the answer 😎, Let a be any positive integer and b = 3. Using Euclid’s Division Lemma, a = 3q + r for some integer q ≥ 0 where r = 0, 1, 2 because 0 ≤ r < 3. Therefore, a = 3q or 3q + 1 or 3q + 2 a² = (3q)² or (3q + 1)² or (3q + 2)² =(3q)² or 9q² + 6q + 1 or 9q² + 12q + 4 = 3×(3q²) or 3×(3qRead more
I know the answer 😎,
Let a be any positive integer and b = 3. Using Euclid’s Division Lemma, a = 3q + r for some integer q ≥ 0 where r = 0, 1, 2 because 0 ≤ r < 3.
Therefore, a = 3q or 3q + 1 or 3q + 2
a² = (3q)² or (3q + 1)² or (3q + 2)²
=(3q)² or 9q² + 6q + 1 or 9q² + 12q + 4
= 3×(3q²) or 3×(3q² + 2q) + 1 or 3×(3q² + 4q + 1) + 1
= 3k₁ or 3k₂ + 1 or 3k₃ + 1
Where k₁, k₂ and k₃ are some positive integers.
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.
See this for Video Explanation 🤓
See less