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Ask QuestionPosted by Anurag Singh 6 years, 11 months ago
- 1 answers
Posted by Anurag Singh 6 years, 11 months ago
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Posted by Kartik Sharma 6 years, 11 months ago
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Posted by Wtsp_Status_Official Goomic 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
Three semicircles each of diameter 3 cm, a circle of diameter 4.5 cm and a semicircle of radius 4.5 cm are drawn in the given figure. Find the area of the shaded region.
Answer
Area of semi-circle PQR = ½π(9/2)2
= 81π/8 cm2
Area of region circle, A = π(9/4)2
= 81π/16 cm2
Area of region (B + C) = π(3/2)2
= 9π/4 cm2
Area of region D = ½π(3/2)2
= 9π/8 cm2
Area of shaded region = Area of semicircle - Area of circle - Area of region (B + C) + Area of region D
= 81π/8 - 81π/16 - 9π/4 + 9π/8
= 63π/16 cm2
= 99/8 cm2
Posted by Saumya Guru 6 years, 11 months ago
- 2 answers
Drashti Vaish 6 years, 11 months ago
Posted by Priyanshi Miglani 6 years, 11 months ago
- 3 answers
Yogita Ingle 6 years, 11 months ago
- Learn sin and cos formula properly.
- Then Value of tan is value of sin divide by value of cos.
- Cot is inverse of tan.
- Sec is inverse of cos.
- cosec is inverse of sin.
Music_ Lover?? 6 years, 11 months ago
Music_ Lover?? 6 years, 11 months ago
Posted by Piyush Bothra 6 years, 11 months ago
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Nikita Sharma 6 years, 11 months ago
Posted by Aishwardha Choudhary 6 years, 11 months ago
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Posted by Ayush Bandhu 6 years, 11 months ago
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Lavina Rajput ??? 6 years, 11 months ago
Posted by Saloni Gupta 6 years, 11 months ago
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Avinash Saigal 6 years, 11 months ago
Posted by Ayush Bandhu 6 years, 11 months ago
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Posted by Himanshu Bangari 6 years, 11 months ago
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Posted by Akshat Shah 6 years, 11 months ago
- 3 answers
Ram Kushwah 6 years, 11 months ago
You can download from
https://www.topperlearning.com/learn/previous-year-question-papers/cbse/class-10/b101c2e8
Posted by Aditya Veer 6 years, 11 months ago
- 4 answers
Gaurav Seth 6 years, 11 months ago
We know that, the lengths of tangents drawn from an external point to a circle are equal.
∴ TP = TQ
In ΔTPQ,
TP = TQ
⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)
∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)
∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))
⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)
We know that, a tangent to a circle is perpendicular to the radius through the point of contact.
OP ⊥ PT,
∴ ∠OPT = 90º
⇒ ∠OPQ + ∠TPQ = 90º
⇒ ∠OPQ = 90º – ∠TPQ
⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)
From (1) and (2), we get
∠PTQ = 2∠OPQ
Posted by Anshika & Aliens???? 6 years, 11 months ago
- 4 answers
Ram Kushwah 6 years, 11 months ago
{tex}\begin{array}{l}\frac{\cos45}{sec30}+\cos ec30\\=\frac{\displaystyle\frac1{\sqrt2}}{\displaystyle\frac2{\sqrt3}}+2=\frac1{\sqrt2}\times\frac{\sqrt3}2+2\\=\frac{\sqrt3}{2\sqrt2}+2\\=\frac{\sqrt3{\displaystyle+}{\displaystyle4}{\displaystyle\sqrt2}}{2\sqrt2}=\frac{1.73+4\times1.41}{2\times1.41}=\frac{1.73+5.64}{2.82}\\=\frac{7.37}{2.82}=2.613\\\end{array}{/tex}
Anshika & Aliens???? 6 years, 11 months ago
Puja Sahoo?♀️ 6 years, 11 months ago
Posted by Anshika & Aliens???? 6 years, 11 months ago
- 1 answers
Yogita Ingle 6 years, 11 months ago
cos 45 = 1/ √2
sec 30= 2/ √3
Cosec 30 = 2
Cos 45/sec 30 + cosec 30
=1/ √2 / 2/ √3 + 2
= 1/ √2 * √3/2 + 2
= √3/2 √2 + 2
Posted by Tushar Tyagi 6 years, 11 months ago
- 1 answers
Ram Kushwah 6 years, 11 months ago
No, two numbers can't have 18 as their HCF and 380 as LCM because HCF of the numbers must be a factor of the LCM.
Here 380 is not divisible with 18.
Posted by Wentures Dance Crew S 6 years, 11 months ago
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Posted by Ranvijay Singh 6 years, 11 months ago
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? ? 6 years, 11 months ago
Posted by Rabikanta Barman 6 years, 11 months ago
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Gaurav Seth 6 years, 11 months ago
Q.ABC is a right triangle, right angled at C. Let BC=a, CA=b, AB=c and let p be length of perpendicular from C on AB. Prove that
(i) cp=ab
(ii) 1/p2=1/a2+1/b2
<hr />Solution:
Posted by Sajeesh George 6 years, 11 months ago
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Posted by Rohit Singh 6 years, 11 months ago
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Lavina Rajput ??? 6 years, 11 months ago
Posted by Rajendra Kumar Yadav 5 years, 8 months ago
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Avinash Saigal 6 years, 11 months ago
Posted by Ayush Badoni 6 years, 11 months ago
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Aarju Khan 6 years, 11 months ago
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Vidhi Lalwani? 6 years, 11 months ago
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Kanchan Obra 6 years, 11 months ago
1Thank You