Solution: i The point which lies on X and Y-axes both is origin whose coordinates are 0, 0. Question Taking 0. Solution: Here, in point 4 1, 3 both x and y-coordinates are positive, so it lies in I quadrant. In point 8 -3, -1 ,both x and y-coordinates are negative, so it lies in III quadrant. In point C 1, -4 , x-coordinate is positive and y-coordinate is negative, so it lies in IV quadrant.
In point D -2, 3 , x-coordinate is negative and y-coordinate is positive, so it lies in II quadrant. In point E 0,-8 x-coordinate is zero, so it lies on Y-axis and in point F 1,0 y-coordinate is zero, so it lies on X-axis. On plotting the given points, we get the following graph. Plot these points on a graph paper and hence, find the coordinates of the vertex C. Thinking Process i Firstly, plot the given points on a graph and join in order.
So, abscissa of C should be equal to abscissa of B i. Hence, the coordinates of C are -2, — 4. Question 2: Write the coordinates of the vertices of a rectangle whose length and breadth are 5 and 3 units respectively, one vertex at the origin, the longer side lies on the X-axis and one of the vertices lies in the third quadrant.
So, the length of the rectangle is 5 units in the negative direction of X-axis and then vertex is A -5, 0. Also, the breadth of the rectangle is 3 units in the negative,direction of y-axis and then vertex is C 0, The fourth vertex B is -5, — 3. Question 3: Plot the points P 1, 0 , Q 4, 0 and 5 1, 3. Solution: In point P 1, 0 , y-coordinate is zero, so it lies on X-axis.
In point Q 4, 0 , y-coordinate is zero so it lies on X-axis. In point S 1, 3 , both coordinates are positive, so it lies in I quadrant. On plotting these points, we get the following graph.
Then, all sides will be equal i. So, abscissa of R should be equal to abscissa of Q i. Hence, the coordinates of R are 4, 3. Question 4: From the given figure, answer the following questions i Write the points whose abscissa is 0. So, the required points whose abscissa is 0 are A, L and O.
So, the required points, whose ordinate is 0 are G,l and O. So, the required points whose abscissa is -5, are D and H. Note: We know that, origin O is the intersection point of both axes. So, we can consider it on X-axis as well as on Y-axis. Question 5: Plot the points A 1, — 1 and B 4, 5. Is this statement true? Give reason for your answer. Solution: No, diagonals of a parallelogram are not perpendicular to each other, because they only bisect each other.
Why or why not? What special name can be given to this quadrilateral? Question 5: All the angles of a quadrilateral are equal. What special name is given to this quadrilateral? Hence, given quadrilateral is a rectangle. Question 6: Diagonals of a rectangle are equal and perpendicular. Solution: No, diagonals of a rectangle are equal but need not be perpendicular.
Question 7: Can all the four angles of a quadrilateral be obtuse angles? Solution: No, all the four angles of a quadrilateral cannot be obtuse. By mid-point theorem, DE AC and. Since, ABCD is a parallelogram, then opposite angles of a parallelogram are equal.
Question Can all the angles of a quadrilateral be acute angles? Solution: No, all the angles of a quadrilateral cannot be acute angles. So, maximum of three acute angles will be possible. This chapter tells how to find the linear equations which present the linear equation into two variables in the graphical method by solving equations that are parallel to the x-axis and y-axis. This chapter has different theorems as well postulated which are presented by Euclid.
The student will understand how to use the different theorems and postulates via solving the questions. The student who wants to find the solution for this chapter can easily go to Vedantu and access the solutions. Chapter 6 - Lines and Angles. This chapter included lines, angles and its types. Chapter 7 - Triangles. This is a very important chapter to look forward to. This chapter tells about triangles and their properties. Here students can learn some conditions based on triangles as well as inequalities of triangles.
Chapter 8 - Quadrilaterals. This chapter includes solving sums that are based on angle sum property. Also, they can learn about different types of present in a quadrilateral. NCERT exemplar problems not only helps students in school level exams but they have a great role in competitive exams as well. The students who prepare from ncert exemplar books are most likely to solve all questions in exams like NTSE, Olympiads, etc. This chapter contains problems based on topics like representation of real numbers on number line, real numbers and their decimal expansion, irrational number problems, exponent laws for real numbers, etc.
The problems in this chapter of ncert exemplar book are based on topics like remainder theorem, factorization of polynomials, polynomial in one variable, finding zeroes of polynomial etc. Students will learn to solve problems based on Cartesian system and questions based on plotting a point in the plane if its coordinates are given only x-y coordinates and on other related topics.
In this chapter, students will learn to find the solution for linear equations, also to represent the linear equation with two variables graphically and solving equations of lines parallel to x-axis and y-axis. This chapter have problems based on lines intersecting and non-intersecting , pairs of angles, parallel lines, transversals, two lines parallel to the same line and also questions based on angle sum property of a triangle.
Triangles is one of the important chapter for class 9 which consist of topics like congruency of triangles and its conditions, also you will learn some of the basic properties of triangle and inequalities in triangles, etc. In this chapter, we are going to solve exemplar problems based on the angle sum property of a quadrilateral.
Also, learn different types of quadrilaterals here. Know about properties of a parallelogram and condition for a quadrilateral to be a parallelogram along with mid-point theorem.
From the chapter name itself, it is clear that we are going to solve exemplar problems, finding the areas of parallelogram and triangles, where the parallelograms and triangles are on the same base and between the same parallels.
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