# Rd sharma class 10 exercise 2 1 solutions. RD Sharma Class 10 Maths Solutions Chapter 2 2018-07-08

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## RD Sharma Class 10 Solutions Chapter 14 Surface Areas and Volumes Revision Exercise

The sphere is melted and is drawn into a long wire of uniform circular cross section. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2. Find the area of the triangle and the length of the hypotenuse. Malik's Newton Classes are the answer to all your nightmares related to Maths. A metallic sphere 1 dm in diameter is beaten into a circular sheet of uniform thickness equal to 1 mm. Solution: Let two vertices of an equilateral triangle are A 3,4 , and B -2,3 and let the third vertex be C x, y Question 52.

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## Polynomials R.D. Sharma Solution Chapter 2

Therefore, Substituting we get, Hence, the condition for the given polynomial is. The breadth of rectangle is units. If a point A 0, 2 is equidistant from the points B 3, p and C p, 5 , then find the value of p. Find the point on x-axis which is equidistant from the points -2, 5 and 2, -3. Assuming that her string from the tip of her rod to the fly is taut, how much string does she have out? Solution 12 : Answer : We know that if is a zero of a polynomial, and then is a factor of. Therefore is a factor of Now, we divide by to find the value of k Now, Remainder Hence, the value of k is Solution 36 : Answer : If a quadratic polynomial is factorized into linear polynomials then the total number of real and distinct zeros of will be.

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## Polynomials R.D. Sharma Solution Chapter 2

Find the centre of the circle passing through 5, -8 , 2, -9 and 2, 1. Parabola open upwards above the x-axis or open downwards below x-axis where or Hence, if quadratic polynomial is not factorizable into linear factors then it has no real zeros. Find a relation between x and y such that the point x, y is equidistant from the points 3, 6 and -3, 4. Let us assume that x is greater than or equal to y. Find the number of cones so formed. Find the radius of the base of new cylinder. Now, we divide by to find the other zeros of.

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## RD Sharma Class 10 Maths Solutions Chapter 2

Given, The product of the zeros of the quadratic polynomial. Thus, we have So, we have two equations Here x and y are unknowns. The inner and outer radii of a hollow cylinder are 15 cm and 20 cm, respectively. Solution: Let the co-ordinates of the vertices A, B, C and D of a rhombus are A -3, 2 , B -5, -5 , C 2, -3 and D 4, 4 Question 27. Therefore Solution 18 : Answer : The graph of a polynomial touches x - axis at two points We know that if a curve touches the x-axis at two points then it has two common zeros of.

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## RD SHARMA CLASS 10 CH

Prove that the points 7, 10 , -2, 5 and 3, -4 are the vertices of an isosceles right triangle. The curve cuts X axis at two coincident points that is exactly at one point. The sum of the two numbers is 8. Graph the equation Putting in equation we get: Putting in equation we get: Use the following table to draw the graph. Then, The required polynomial of is given by , where k is any non-zero real number. Now, we divide by to find the other zeros of. So the coordinates of P are.

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## RD SHARMA CLASS

This will certainly improve your ability to grasp the content as well as enhance your speed in solving questions. Find the volume of water. In each of the following figures, you find two triangles. The areas of two similar triangles are 36 cm² and 100 cm². Also, difference between two odd and two even integers is even Hence it is proved that if a and b are two odd positive integers is even.

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## RD Sharma Class 10 Solutions Maths Chapter 2 Polynomials Exercise 2.2

Our free reference resources include Interactive Simulations Enhancing Visual Understanding , Concept Maps , Previous Years Solved Papers Upto 12 Years. The two lines intersects at point B Hence is the solution Exercise 3. Hence, the correct choice is Solution 7 : Answer : Since and are the zeros of quadratic polynomial We have The value of c is. Sharma Solution: We first compute L. Determine the values of x, y, z in each case. If the total height be 22 cm, the diameter of the cylindrical portion 8 cm and the diameter of the top of the funnel 18 cm, find the area of the tin required.

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## RD Sharma Class 10 Solutions Maths Chapter 2 Polynomials Exercise 2.1

Two equal conical holes each of radius 3 cm and height 4 cm arc cut off. Find the height of the bucket. Thus, we have The sum of the two numbers is four times their difference. We have to solve the above equations for x and y. Now, we divide by to find the other zeros of. If the length of the whole solid is 108 cm and the diameter of the cylinder is 36 cm, find the cost of polishing the surface at the rate of 7 paise per cm2.

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