CBSE Class 10 Answered

Asked by chaudahryanshika | 03 Jan, 2023, 10:04: AM

Expert Answer

Question (5)

Quadratic equation :- a x² + b x + c = 0

for real roots , b² = ( 4 a c )

Given quadratic equation :- x² + 4x + k= 0

a =1 , b = 4 , c = 1

b² = 16

4 ac = 4 k = 16

Hence k = 4

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Question (6)

As shown in figure , we gwt the sides of triangle as 4 unit, 3 unit and 5 unit.

Hence perimeter = 12 unit

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Question (7)

In a given similar triangles ABC and DEF

begin mathsize 14px style fraction numerator A B over denominator D E end fraction equals fraction numerator B C over denominator F D end fraction space space space space t h e n space angle B space equals space angle D end style

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Question (8) Partially hidden, unable to answer

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Question (9)

begin mathsize 14px style angle A O B space equals space 130 degree end style

because

begin mathsize 14px style angle P A O space equals space angle P B O space equals space 90 degree end style

ΔOAB is isoceless because OA = OB

Hence

begin mathsize 14px style angle O A B space equals space angle O B A space equals space 25 degree end style

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Question (10)

begin mathsize 14px style s e c space A space equals space fraction numerator 1 over denominator cos space A end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus space sin squared A end root end fraction space equals space fraction numerator 1 over denominator square root of 1 space minus space open parentheses begin display style 1 half end style close parentheses squared end root end fraction space equals space fraction numerator 2 over denominator square root of 3 end fraction end style

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Question (11)

begin mathsize 14px style square root of 3 space cos squared A space plus space square root of 3 space sin squared A space equals space square root of 3 space open parentheses cos squared A plus sin squared A close parentheses equals space square root of 3 end style

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Question (12)

cos1^ocos2^o cos3^o ..........................cos88^ocos89^o cos90^o = 0

because cos90 = 0 and ita ppears in product

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Question (13)

let r be the radius of circle. perimeter is ( 2

begin mathsize 14px style pi end style

r )

If perimeter of square is ( 2

begin mathsize 14px style pi end style

r ) then side of square ( 2

begin mathsize 14px style pi end style

r )/4 = (

begin mathsize 14px style pi end style

r )/2

Area of circle : area of square =

begin mathsize 14px style pi end style

r² : (1/4)

begin mathsize 14px style pi end style

² r²

Let us use

begin mathsize 14px style pi end style

= 22/7 , then if simplify above ratio , we get areas are in the ratio 14 : 11

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Question (14)

area of circle is proportioanl to square of radius.

If radii of circle are in the ratio 4:3 , then areas are in the ratio 16:9

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Question (15)

Total surface area of solid hemisphere = 3

begin mathsize 14px style pi end style

r² = 3

begin mathsize 14px style pi end style

×49 = 147

begin mathsize 14px style pi end style

cm²

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Question 16

Modal class is between 15 to 20 , because frequency is highest for this class.

Upper limit is 20

Answered by Thiyagarajan K | 03 Jan, 2023, 02:26: PM

CBSE Class 10 Answered

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CBSE Class 10 Answered

Application Videos

Introduction to Quadratic Equations

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Introduction to Quadratic Equations ...

Introduction to Quadratic Equations

Introduction to Quadratic Equations

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