Given: - OA = 13 cm, OB = OC = 5 cm (radius), - Tangent length AB = AC = √(OA² - OB²) = √(13² - 5²) = 12 cm. Area of ΔABO = (1/2) × AB × OB = (1/2) × 12 × 5 = 30 cm². Similarly, Area of ΔACO = 30 cm². Total area of ABOC = 30 + 30 = 60 cm². This question pertains to Chapter 10 of the Class 10th NCERTRead more
Given:
– OA = 13 cm, OB = OC = 5 cm (radius),
– Tangent length AB = AC = √(OA² – OB²) = √(13² – 5²) = 12 cm.
Area of ΔABO = (1/2) × AB × OB = (1/2) × 12 × 5 = 30 cm².
Similarly, Area of ΔACO = 30 cm².
Total area of ABOC = 30 + 30 = 60 cm².
This question pertains to Chapter 10 of the Class 10th NCERT Mathematics textbook, specifically the chapter on “Circles.” Provide your answer in accordance with the knowledge and understanding you have gained from studying this chapter.
Given: - Radius of circle = 4 cm, - Tangents PA and PB are perpendicular. The tangents form a square with the radii. The diagonal of the square is the hypotenuse: Diagonal = √(4² + 4²) = √32 = 4√2. Side (tangent length) = Diagonal / √2 = 4√2 / √2 = 4 cm. This question is associated with Chapter 10 oRead more
Given:
– Radius of circle = 4 cm,
– Tangents PA and PB are perpendicular.
The tangents form a square with the radii. The diagonal of the square is the hypotenuse:
Diagonal = √(4² + 4²) = √32 = 4√2.
Side (tangent length) = Diagonal / √2 = 4√2 / √2 = 4 cm.
This question is associated with Chapter 10 of the Class 10th NCERT Mathematics textbook, which deals with the topic of “Circles.” Answer the question based on your understanding of the concepts covered in this chapter.
We are given: - PA and PB are tangents to the circle with center O, - Angle ∠APB = 50°. Key property of tangents The tangents PA and PB are equally inclined to the line joining the external point P to the center O. Therefore: ∠APO = ∠BPO = ∠APB / 2. Substitute ∠APB = 50°: ∠APO = ∠BPO = 50° / 2 = 25°Read more
We are given:
– PA and PB are tangents to the circle with center O,
– Angle ∠APB = 50°.
Key property of tangents
The tangents PA and PB are equally inclined to the line joining the external point P to the center O. Therefore:
∠APO = ∠BPO = ∠APB / 2.
Relationship between angles
In triangle OAB:
– OA and OB are radii of the circle, so triangle OAB is isosceles.
– The angle ∠OAB is equal to ∠APO because the tangent PA is perpendicular to the radius OA.
Thus:
∠OAB = ∠APO = 25°.
This question is linked to Chapter 10 of the Class 10th NCERT Mathematics textbook, which is on the topic of “Circles.” Respond with an answer that reflects your understanding of the chapter.
Given: - AB = 5 cm, BC = 12 cm, AC = 13 cm (using Pythagoras theorem). Radius of incircle: r = (a + b - c) / 2, where a = 5, b = 12, c = 13. Substitute: r = (5 + 12 - 13) / 2 = 4 / 2 = 2 cm. This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer accRead more
Given:
– AB = 5 cm, BC = 12 cm, AC = 13 cm (using Pythagoras theorem).
Radius of incircle:
r = (a + b – c) / 2,
where a = 5, b = 12, c = 13.
Substitute:
r = (5 + 12 – 13) / 2 = 4 / 2 = 2 cm.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
Given: - Length of tangent = 24 cm, - Distance from Q to center = 25 cm. Using Pythagoras theorem: 25² = 24² + r². Simplify: 625 = 576 + r². Solve for r: r² = 49 ⇒ r = 7 cm. This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your unRead more
Given:
– Length of tangent = 24 cm,
– Distance from Q to center = 25 cm.
Using Pythagoras theorem:
25² = 24² + r².
Simplify:
625 = 576 + r².
Solve for r:
r² = 49 ⇒ r = 7 cm.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
If from a point A which is at distance of 13 cm from the centre O of a circle of radius 5cm, the pair of tangent AB and AC to the circle are drawn, then the area of quadrilateral ABOC is
Given: - OA = 13 cm, OB = OC = 5 cm (radius), - Tangent length AB = AC = √(OA² - OB²) = √(13² - 5²) = 12 cm. Area of ΔABO = (1/2) × AB × OB = (1/2) × 12 × 5 = 30 cm². Similarly, Area of ΔACO = 30 cm². Total area of ABOC = 30 + 30 = 60 cm². This question pertains to Chapter 10 of the Class 10th NCERTRead more
Given:
– OA = 13 cm, OB = OC = 5 cm (radius),
– Tangent length AB = AC = √(OA² – OB²) = √(13² – 5²) = 12 cm.
Area of ΔABO = (1/2) × AB × OB = (1/2) × 12 × 5 = 30 cm².
Similarly, Area of ΔACO = 30 cm².
Total area of ABOC = 30 + 30 = 60 cm².
This question pertains to Chapter 10 of the Class 10th NCERT Mathematics textbook, specifically the chapter on “Circles.” Provide your answer in accordance with the knowledge and understanding you have gained from studying this chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If two perpendicular tangents PA and PB are drawn from an external point to a circle of radius 4cm, then the length of each tangent is
Given: - Radius of circle = 4 cm, - Tangents PA and PB are perpendicular. The tangents form a square with the radii. The diagonal of the square is the hypotenuse: Diagonal = √(4² + 4²) = √32 = 4√2. Side (tangent length) = Diagonal / √2 = 4√2 / √2 = 4 cm. This question is associated with Chapter 10 oRead more
Given:
– Radius of circle = 4 cm,
– Tangents PA and PB are perpendicular.
The tangents form a square with the radii. The diagonal of the square is the hypotenuse:
Diagonal = √(4² + 4²) = √32 = 4√2.
Side (tangent length) = Diagonal / √2 = 4√2 / √2 = 4 cm.
This question is associated with Chapter 10 of the Class 10th NCERT Mathematics textbook, which deals with the topic of “Circles.” Answer the question based on your understanding of the concepts covered in this chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If PA and PB are tangents to the circle with the centre O such that angle APB = 50°, then angle OAB is equal to
We are given: - PA and PB are tangents to the circle with center O, - Angle ∠APB = 50°. Key property of tangents The tangents PA and PB are equally inclined to the line joining the external point P to the center O. Therefore: ∠APO = ∠BPO = ∠APB / 2. Substitute ∠APB = 50°: ∠APO = ∠BPO = 50° / 2 = 25°Read more
We are given:
– PA and PB are tangents to the circle with center O,
– Angle ∠APB = 50°.
Key property of tangents
The tangents PA and PB are equally inclined to the line joining the external point P to the center O. Therefore:
∠APO = ∠BPO = ∠APB / 2.
Substitute ∠APB = 50°:
∠APO = ∠BPO = 50° / 2 = 25°.
Relationship between angles
In triangle OAB:
– OA and OB are radii of the circle, so triangle OAB is isosceles.
– The angle ∠OAB is equal to ∠APO because the tangent PA is perpendicular to the radius OA.
Thus:
∠OAB = ∠APO = 25°.
This question is linked to Chapter 10 of the Class 10th NCERT Mathematics textbook, which is on the topic of “Circles.” Respond with an answer that reflects your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
In a right angle triangle ABC, right angled at B, BC = 12 cm and AB = 5 cm. The radius of the circle inscribed in the triangle (in cm) is
Given: - AB = 5 cm, BC = 12 cm, AC = 13 cm (using Pythagoras theorem). Radius of incircle: r = (a + b - c) / 2, where a = 5, b = 12, c = 13. Substitute: r = (5 + 12 - 13) / 2 = 4 / 2 = 2 cm. This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer accRead more
Given:
– AB = 5 cm, BC = 12 cm, AC = 13 cm (using Pythagoras theorem).
Radius of incircle:
r = (a + b – c) / 2,
where a = 5, b = 12, c = 13.
Substitute:
r = (5 + 12 – 13) / 2 = 4 / 2 = 2 cm.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
From a point Q, the length of the tangent to to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
Given: - Length of tangent = 24 cm, - Distance from Q to center = 25 cm. Using Pythagoras theorem: 25² = 24² + r². Simplify: 625 = 576 + r². Solve for r: r² = 49 ⇒ r = 7 cm. This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your unRead more
Given:
– Length of tangent = 24 cm,
– Distance from Q to center = 25 cm.
Using Pythagoras theorem:
25² = 24² + r².
Simplify:
625 = 576 + r².
Solve for r:
r² = 49 ⇒ r = 7 cm.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/