Ncert solutions class 12 maths chapter 6 exercise 6.5
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Therefore, is only the turning point. Register for our free webinar class with best mathematics tutor in India. Let meters be the side of square and meters be the radius of the circle. Therefore, option A is correct. Of all the closed cylindrical cans right circular , of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area.

About 12 Maths Exercise 5. Since the diagonals of a rhombus bisect each other at right angles. At what points on the interval does the function attain its maximum value? In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes. Ans : In a triangle, the sum of the lengths of either two sides is always greater than the third side. . Let denotes the volume of the open box. The topics and sub-topics included in the Applications of Derivatives chapter are the following: Section Name Topic Name 6 Applications of Derivatives 6.

Find each of its altitudes. We know that the mid-point of the hypotenuse of a right triangle is equidistant from the vertices. To prove : i ii iii Proof : i Since Therefore and each triangle is similar to Because Therefore ii Since, Therefore iii Since, Q. . By using this property, these problems can be solved as follows. Show that the semi-vertical angle of the right circular cone of given surface area and maximum volume is Ans.

Right angle will be in front Of the side Of 2. Download Vedantu to get a better understanding of all the exercises questions. This solution contains questions, answers, images, explanations of the complete chapter 6 titled The Triangle and its Properties of Maths taught in class 7. Find two positive integers whose sum is 16 and sum of whose cubes is minimum. Let the two positive numbers are and ………. Let denotes the volume of the open box.

The maximum value of is: A B C 1 D Ans. He has 5 years of experience in the education field. For a better understanding of this chapter, you should also see summary of Chapter 6 The Triangle and its Properties , Maths, Class 7. Then dimensions of the open box formed by folding the flaps after cutting off squares are and cm. Hence, these are the sides Of a right-angled triangle. Q6 Does Miscellaneous Exercise question comes in board exam? The point on the curve which is nearest to the point 0, 5 is: A B C 0, 0 D 2, 2 Ans. Let cm be the side of each of the four squares cut off from each corner.

Two poles of heights 6 m and 11m stand on a plane ground. Therefore, those two measures are 27cm and 3 cm. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. What should be the side of the square to be cut off so that the volume of the box is the maximum possible? Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is of the volume of the sphere. Find two positive integers and such that their sum is 35 and the product is a maximum.

All the best and feel free to leave a comment below. At At At Therefore, maximum value of is 89. In this exercise the symbols t and θ are used as parameters. Find the distance of the foot of the ladder from base of the wall. How far apart will be the two planes after 1 hours? Ncert solution class 12 Maths includes text book solutions from both part 1 and part 2.

Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals. Variables x and y are connected with the help of the third variable called parameter. In case of a right triangle, write the length of its hypotenuse. How far apart will be the two planes after 1½ hours? Also find the local maximum and local minimum values. Go through them and get a clear idea about how to approach the problems so that you can solve them in the most efficient way. To keep the wire taut, let it be fixed to a stake at A. Hence, these sides are not Of a right-angled triangle.

Show that of all the rectangles inscribed in a given fixed circle, the square has maximum area. Since, , therefore, Volume is maximum at From eq. A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. Also find the minimum value of the sum of their volumes. Let cm be the side of each of the four squares cut off from each corner. Let be the radius of the circular base and be the height of closed right circular cylinder.

Here, the third side will be lesser than the sum of these two i. Therefore, option C is correct. What is the maximum value of the function Ans. . Also, the solution for miscellaneous exercise are provided.