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It is given in the question that, The vertices of a parallelogram PQRS taken in order are P(3,4), Q(-2,3) and R(-3,-2) Let, the fourth vertices is S (x , y). The parallelogram's PQRS's vertices, listed in order. Because of this, a parallelogram's diagonals are split in half. We know that the diagonals of a parallelogram bisects with each other. ∴ Coordinates of the PR's midpoint = coordinates for QS's midpoint \begin{gathered}(\frac{3-3}{2} ,\frac{4-2}{2} ) = (\frac{-2+x}{2}, \frac{3+y}{2} )\\ (0,1) = (\frac{-2+x}{2}, \frac{3+y}{2} )\end{gathered}(23−3,24−2)=(2−2+x,23+y)(0,1)=(2−2+x,23+y) Now, 0 = \frac{-2+x}{2}0=2−2+x .............(1) Simplify the equation, \begin{gathered}0 = -2+x\\x=2\end{gathered}0=−2+xx=2 Again, 1=\frac{3+y}{2}1=23+y ...............(2) Simplify the equation, \begin{gathered}1=\frac{3+y}{2}\\2=3+y\\y=-1\end{gathered}1=23+y2=3+yy=−1 the coordinates of its fourth vertex S are (2,-1).

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Case study of sample t

Posted by Sana Khan 1 year, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Alok Kumar 1 year, 11 months ago

Hello

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Given below is the conversation between Gayatri and a mason. Gayatri : Here are some 90 cm x 90 cm tiles. I want to use them in flooring of our courtyard.' Mason: 'If I use all of these tiles, there will be four more coloumns than rows.' Gayatri: I want to make a square floor.' Mason: 'Either some tiles will be left unsued or 35 additional tiles are required.' Gayatri: 'I do not think 35 additional tiles will be required to make a square courtyard.' Mason: 'Okay, that is possible, but you still have to buy some more.' (i) How many tiles did Gayatri have? (ii) What is the minimum number of tiles she has to purchase to make a square courtyard? (iii) For the minimum number of tiles purchased, what can be the side length (in m) of the square courtyard?

Posted by Anshu Gupta 1 year, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Preeti Dabral 1 year, 11 months ago

Area of one tile= (10x10)-100cm².

Area of kitchen floor = l×b

=(220x180) cm²

=(200 +20) (200-20)

=(200)² - (20)²

=40000-400

= 39,600 cm²

Total no of tiles needed=area of tiles needed

area of one tile

=39.600

100

therefore, 396 tiles will be needed.

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Value of cos15

Posted by Nikhil Pal 1 year, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Jessika Chauhan 1 year, 11 months ago

√3/2

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Shubh Lakshna Pandey 1 year, 11 months ago

0.96592582

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Ankit Class 1 year, 11 months ago

√3+1/2√2

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Unknown 1947 1 year, 11 months ago

√3/2

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Please share the syllabus of class 10 maths for 2022-23

Posted by Basant Pandey 1 year, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Sãrthâk Dïxít 1 year, 11 months ago

UNIT I: NUMBER SYSTEMS REAL NUMBER (15) Periods Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality of UNIT II: ALGEBRA POLYNOMIALS (8) Periods Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES (15) Periods Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems. QUADRATIC EQUATIONS (15) Periods Standard form of a quadratic equation ax2 + bx + c = 0, (a 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities to be incorporated. ARITHMETIC PROGRESSIONS (10) Periods Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems. UNIT III: COORDINATE GEOMETRY Coordinate Geometry (15) Periods Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division). UNIT IV: GEOMETRY TRIANGLES (15) Periods Definitions, examples, counter examples of similar triangles. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. (Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. (Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. (Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar. CIRCLES (10) Periods Tangent to a circle at, point of contact (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. (Prove) The lengths of tangents drawn from an external point to a circle are equal. UNIT V: TRIGONOMETRY INTRODUCTION TO TRIGONOMETRY (10) Periods Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45° and 60°. Relationships between the ratios. TRIGONOMETRIC IDENTITIES (15) Periods Proof and applications of the identity sin2A + cos2A = 1. Only simple identities to be given. HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. (10) Periods Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45° and 60°. UNIT VI: MENSURATION AREAS RELATED TO CIRCLES (12) Periods Area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only. SURFACE AREAS AND VOLUMES (12) Periods Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones. UNIT VII: STATISTICS AND PROBABILITY STATISTICS (18) Periods Mean, median and mode of grouped data (bimodal situation to be avoided). PROBABILITY (10) Periods Classical definition of probability. Simple problems on finding the probability of an event.

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8*8*8*8*8=?

Posted by Yash Dawas 1 year, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Riti .. 1 year, 11 months ago

(8)⁵ = 32768

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Sãrthâk Dïxít 1 year, 11 months ago

(8)⁵

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Basant Pandey 1 year, 11 months ago

32768

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Manash Mishra 1 year, 11 months ago

32768

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A vendor sells glasses of juice, which is in the form of a cylinder mounted by a hollow hemisphere. The diameter of the hemisphere and the cylinder is 8.4 cm. If the total height of the glass is 15 cm, which of these is closest to the volume, in cubic centimeters of 8 such glasses

Posted by Anil Yadav 1 year, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

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The distance between the two

Posted by Kuldeep Agrawal 1 year, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Manash Mishra 1 year, 11 months ago

the length of the line segment

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Nikhil Pal 1 year, 11 months ago

Point is call row

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ABC is an equilateral triangle inscribed in a circle of radius 4cm.Find the area of the shaded portion

Posted by Priyanshu Jha 1 year, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Manash Mishra 1 year, 11 months ago

we have,R=4cm AB=BC=CA=R√3=4√3cm [R=2/3h and h =√3/2a , R=a√3] ∠AOC=∠BC=2×60°=120° ∴Required area=1/3(Area of the circle×area of ∆ABC ) Required area = 1/3 [ πR²-√3/2×(sides) Required area =1/2[16π-√3/2×(4√3)² Required area=1/3(16π-12√3)cm² 4/3(4π-3√3cm²

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Find the sum and product of zeroes of p(x)=3(x2-5)+x and then form a quadratic polynomial whose zeros are 3alpha and 3 beta

Posted by Jashan Kaur 1 year, 11 months ago

CBSE > Class 10 > Mathematics

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In the given figure, a circle inscribed in a triangle abc touches the sides ab,bc,ac at point d,e,f respectively if ab=12 bc=8 and ac10 find the length of ad ,be and cf

Posted by Janvi Tyagi 1 year, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Vaibhav Tiwari 1 year, 11 months ago

x + y = 12, y + z = 8, z + x = 3 . Solving, we get x =7 , y = 5, z= 3 So AD =x=7, BE =y=5, CF = z=3

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Vaibhav Tiwari 1 year, 11 months ago

Hii

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56823379556+789456742

Posted by Shaini S Gowda 1 year, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Shubh Lakshna Pandey 1 year, 11 months ago

57,61,28,36,298

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Manash Mishra 1 year, 11 months ago

57612836298

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Arpita And Ishika Dhandhi 1 year, 11 months ago

57,612,836,298

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Curve surface area of cylinder

Posted by Gaurav Kumar 1 year, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Shubh Lakshna Pandey 1 year, 11 months ago

2πrh

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Rohan K Gupta 1 year, 11 months ago

2πrh

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Warisha Rahman 1 year, 11 months ago

2πrh

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Ankit Class 1 year, 11 months ago

2πrh

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2*2-1+2*1*5%9/2+5-5

Posted by Yash Dawas 1 year, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Shubh Lakshna Pandey 1 year, 11 months ago

3.45

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Manash Mishra 1 year, 11 months ago

3.5

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If hcf of (336,54)=6 find lCM of 336,54

Posted by Krish Yadav 1 year, 11 months ago

CBSE > Class 10 > Mathematics

3 answers

Kafi Malik 1 year, 11 months ago

3024

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Shubh Lakshna Pandey 1 year, 11 months ago

---:HCF * LCM = product of two numbers ---:6*LCM = 336*54 ---: LCM = (336*54)/6 ---: LCM = 3024

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Naina Chaudhary 1 year, 11 months ago

3024

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4x-4x-3

Posted by Basanti Raut 1 year, 11 months ago

CBSE > Class 10 > Mathematics

2 answers

Yash Dawas 1 year, 11 months ago

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Yash Dawas 1 year, 11 months ago

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If the sum of n terms of an ap is 3n^2+5n and it's mth term is 164 find the value of m

Posted by Aditya Mukesh B. K. 1 year, 11 months ago

CBSE > Class 10 > Mathematics

1 answers

Preeti Dabral 1 year, 11 months ago

S_n = 3n² + 5n
Put n = 1, 2, 3,....
S₁ = 8
{tex}{S_2} = 3 \times 4 + 10 = 22{/tex}
a₁ = S₁ = 8
a₂ = S₂ - S₁
= 22 - 8 = 14
d = a₂ - a₁ = 14 - 8 = 6
a_m = 164
{tex} \Rightarrow {/tex} a + (m - 1)d = 164
{tex} \Rightarrow {/tex} 8 (m - 1) (6) = 164
{tex} \Rightarrow {/tex} 8 + 6m - 6 = 164
{tex} \Rightarrow {/tex} 6m = 164 - 2
{tex} \Rightarrow {/tex} 6m = 162
{tex} \Rightarrow m = \frac{{162}}{6} = 27{/tex}