# CBSE NCERT Solutions for Class 10 Maths (Updated for 2019-20) हिंदी & English Medium

CBSE NCERT Solutions for Class 10 Maths in PDF & Video form, हिंदी मीडियम as well as English Medium session 2019-20 for CBSE, UP, Gujrat Board, MP Board, Bihar, Uttarakhand board and all other boards following new CBSE Curriculum. Download Offline Apps based on NCERT Sols updated as per the latest NCERT Books for 2019-20. Download CBSE Date Sheet 2019-2020 Exams Starting from February 15, 2020. Join the Discussion forum to share your knowledge.

 Class 10: Maths – गणित Medium: English – हिंदी

## CBSE NCERT Solutions for Class 10 Maths

NCERT – CBSE Solutions for class 10 all chapters are given below. You may download to use it offline or use online as it is. Download 10 Maths App in Hindi for offline use or 10 Maths in English for offline free.

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#### 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.

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.

#### A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Height of conical part (h) = radius of conical part (r) = 1 cm
Radius of conical part (r) = radius of hemispherical part (r) = 1 cm
Volume of Solid = Volume of conical part + Volume of hemispherical part
= 1/3 πr^2 h + 2/3 πr^3
=1/3 π〖.1〗^2.1 + 2/3 π〖.1〗^3 = π〖cm〗^3

#### What is the empirical relationship between the three measures of central tendency?

The empirical relationship between the three measures of central tendency:

3 Median = Mode + 2 Mean

#### एक थैले में 5 लाल गेंद और कुछ नीली गेंदे हैं यदि इस थैले में से नीली गेंद निकालने की प्रायिकता लाल गेंद निकालने की प्रायिकता की दुगुनी है, तो थैले में नीली गेंदों की संख्या ज्ञात कीजिए।

माना नीली गेंदों की कुल संख्या = x
लाल गेंदों की कुल संख्या = 5
कुल गेंदे = x + 5
P (लाल गेंद) = 5/(5 + x)
P (नीली गेंद) = x/(5 + x)
दिया है,
2(5/(5 + x)) = x/(5 + x)
⇒ 10(x + 5) = x^2 + 5x
⇒ x^2-5x-50=0
⇒ x^2 – 10x + 5x – 50 = 0
⇒ x(x – 10) + 5 (x – 10) = 0
⇒ (x – 10)(x + 5) = 0
⇒ x – 10 = 0 या x + 5 = 0
⇒ x = 10 या x = – 5
क्योंकि गेंदों की संख्या ऋणात्मक नहीं हो सकती, अतः नीली गेंदों की कुल संख्या 10 है ।

#### How to draw a tangent at a point of a circle?

To draw a tangent at a point of a circle, simply draw the radius through this point and draw a line perpendicular to this radius through this point and this will be the required tangent at the point.

#### एक वृत्त की कितनी स्पर्श रेखाएँ हो सकती हैं?

एक वृत्त की अनन्त स्पर्श रेखाएँ हो सकती हैं क्योंकि एक वृत्त में उस पर असीम बिंदुओं की संख्या होती है और हर बिंदु पर एक स्पर्शरेखा खींची जा सकती है।

#### What is meant by angle of depression?

The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

#### If tan 2A = cot(A-18°), where 2A is an acute angle, find the value of A.

Given that:
tan 2A = cot(A – 18°)
⇒ cot(90° – 2A) = cot(A – 18°) ∵ cot⁡〖(90° – θ) = tan⁡θ 〗 ] ⇒ 90° – 2A = A – 18°
⇒ 90° + 18° = 3A
⇒ 3A = 108°
⇒ A = 36°
Hence, A = 36°

#### y का वह मान ज्ञात कीजिए, जिसके लिए बिंदु P(2, -3) और Q(10, y) के बीच की दूरी 10 मात्रक है।

बिंदु P(2, -3) और Q(10, y) के बीच की दूरी 10 मात्रक है।
⇒ √((10 – 2)^2 +[y – (-3)]^2 ) = 10
⇒ √(64 + y^2 + 9 + 6y) = 10
दोनों ओर वर्ग करने पर
64 + y^2 + 9 + 6y = 100
⇒ y^2 + 6y – 27 = 0
⇒ y^2 + 9y – 3y – 27 =0
⇒ y(y + 9) – 3(y + 9) = 0
⇒ (y + 9)(y – 3) = 0
⇒ (y + 9) = 0 या (y – 3) = 0
⇒ y = – 9 या y = 3

#### यदि दो समरूप त्रिभुजों के क्षेत्रफल बराबर हों तो सिद्ध कीजिए कि वे त्रिभुज सर्वांगसम होते हैं।

माना, ∆ABC ~ ∆DEF, इसलिए
ar(∆ABC)/ar(∆DEF) =〖AB〗^2/〖DE〗^2 =〖BC〗^2/〖EF〗^2 =〖AC〗^2/〖DF〗^2
…(1)
दिया है, ar(∆ABC) = ar(∆DEF)
इसलिए, ar(∆ABC)/ar(∆DEF) = 1
समीकरण (1) से,
〖AB〗^2/〖DE〗^2 =〖BC〗^2/〖EF〗^2 =〖AC〗^2/〖DF〗^2 =1
⇒ AB = DE, BC = EF और AC = DF
∴ ∆ABC ≅ ∆DEF [SSS सर्वांगसम प्रमेय से]

#### A.P.: 121,117,113,…, का कौन-सा सबसे पहला ऋणात्मक पद होगा?

यहाँ, a = 121 तथा d = 117-121 = -4 है,
माना, A.P. का nवाँ पद पहला ऋणात्मक पद है।
〖⇒a〗_n < 0 ⇒ a + (n - 1)d < 0 ⇒ 121 + (n - 1)(-4) < 0 ⇒ 121 - 4n + 4 < 0 ⇒ 125 < 4n ⇒ n > 125/4
⇒ n > 31.25
⇒ n = 32
अतः, इस A.P. का 32वाँ पद सबसे पहला ऋणात्मक पद होगा।

#### एक समकोण त्रिभुज की ऊँचाई इसके आधार से 7 cm कम है। यदि कर्ण 13 cm का हो, तो अन्य दो भुजाएँ ज्ञात कीजिए।

माना आधार = x cm
इसलिए, ऊँचाई = x – 7 cm
दिया है, कर्ण = 13 cm
पाइथागोरस प्रमेय से, x^2 + (x – 7)^2 =〖13〗^2
⇒ x^2 + x^2 – 14x + 49 = 169
⇒〖2x〗^2 – 14x – 120 = 0
⇒ x^2 – 7x – 60 = 0
⇒ x^2 – 12x + 5x – 60 = 0
⇒ x(x – 12) +5(x – 12) = 0
⇒(x – 12)(x + 5) = 0
⇒(x – 12) = 0 या (x + 5) = 0
अर्थात x = 12 या x = -5
लेकिन x ≠ -5 , क्योंकि x त्रिभुज की भुजा है।
इसलिए, x = 12 और दूसरी भुजा x-7 = 12 – 7 = 5
अतः, अन्य दो भुजाएँ 12 cm और 5 cm हैं।

#### The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Let the larger angle = x
Let the smaller angle = y
According to question,
x = y + 18 … (1)
Both angles are supplementary, therefore
x + y = 180 … (2)
Putting the value of x in equation (2), we get
y + 18 + y = 180
⇒ 2y = 162
⇒ y = 81
Putting the value of y in equation (1), we get
x = 81 + 18 = 99
Hence, one angle is 81° and the other one is 99°.