RD Sharma Class 10 Solutions Chapter 14 Co-ordinate Geometry Exercise 14.4 (Updated for 2021-22)

RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4

RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4: The main topic of this exercise is section formula applications. The RD Sharma Class 10 Solutions can be used to look out solutions to problems in any chapter of the book. Students can also use the links below to access the RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry Exercise 14.4 PDF.

Download RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4 Free PDF

 


RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4

Access answers to RD Sharma Solutions Class 10 Maths Chapter 14 Exercise 14.4- Important Question with Answers

Question 1.
Find the centroid of the triangle whose vertices are :
(i) (1, 4), (-1, -1), (3, -2)
(ii) (-2, 3), (2, -1), (4, 0)
Solution:
RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry

Question 2.
Two vertices of a triangle are (1, 2), (3, 5) and its centroid is at the origin. Find the Co-ordinates of the third vertex.
Solution:
Centroid of a triangle is O(0, 0) ….(i)
Co-ordinates of two vertices of a ∆ABC are A (1, 2) and B (3, 5)
Let the third vertex be (x, y)
RD Sharma Class 10 Chapter 14 Co-Ordinate Geometry

Question 3.
Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
Solution:
Let two vertices of a ∆ABC be A (-3, 1) and B (0, -2) and third vertex C be (x, y)
Centroid of the ∆ABC is O (0, 0)
Co-Ordinate Geometry Class 10 RD Sharma

Question 4.
A (3, 2) and B (-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (53 , −13) . Find the coordinates of the third vertex C of the triangle. [CBSE 2004]
Solution:
A (3, 2) and B (-2, 1) are the two vertices of ∆ABC whose centroid is G (53 , −13)
Let third vertex C be (x, y)
Co-Ordinate Geometry Class 10 RD Sharma

Question 5.
If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the co-ordinates of its centroid.
Solution:
In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively.
The co-ordinates of D are (-2, 3), of E are (4,-3) and of F are (4, 5)
Let the co-ordinates of A, B and C be (x1, y1), (x2, y2), (x3, y3) respectively
RD Sharma Class 10 Solutions Co-Ordinate Geometry
RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.4
RD Sharma Class 10 Solutions Co-Ordinate Geometry
RD Sharma Class 10 Solutions Co-Ordinate Geometry Exercise 14.4

Question 6.
Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.
Solution:
In ∆ABC,
D and E are the mid points of the sides AB and AC respectively
RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry

Question 7.
Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the middle points of its diagonals meet in a point and bisect one another.
Solution:
Let A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4) be the vertices of quadrilateral ABCD
E and F are the mid points of side BC and AD respectively and EF is joined G and H are the mid points of diagonal AC and BD.
GH are joined
RD Sharma Class 10 Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Chapter 14 Co-Ordinate Geometry
RD Sharma Solutions Class 10 Chapter 14 Co-Ordinate Geometry

Question 8.
If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA² + PB² + PC² = GA² + GB² + GC² + 3GP².
Solution:
In AABC, G is the centroid of it Let P (h, x) is any point in the plane
Let co-ordinates of A are (x1, y1) of B are (x2, y2) and of C are (x3, y3)
Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry
Learncbse.In Class 10 Chapter 14 Co-Ordinate Geometry
Hence proved.

Question 9.
If G be the centroid of a triangle ABC, prove that AB² + BC² + CA² = 3 (GA² + GB² + GC²)
Solution:
Let the co-ordinates of the vertices of ∆ABC be A (x1, y1), B (x2, y2), C (x3, y3) and let G be the centroid of the triangle
Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry
RD Sharma Class 10 Pdf Free Download Full Book Chapter 14 Co-Ordinate Geometry
Class 10 RD Sharma Solutions Chapter 14 Co-Ordinate Geometry
Hence proved.

Question 10.
In the figure, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A and B.
Solution:
RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry
In right ∆OAB, co-ordinates of O are (0, 0) of A are (2a, 0) and of B are (0, 2b)
C is the mid-point of AB
Co-ordinates of C will be
RD Sharma Class 10 Solution Chapter 14 Co-Ordinate Geometry
We see that CO = CA = CB
Hence C is equidistant from the vertices O, A and B.
Hence proved.

We have provided complete details of RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4. If you have any queries related to CBSE Class 10, feel free to ask us in the comment section below.

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Yes, students can trust the RD Sharma Class 10 Solutions Chapter 14 Exercise 14.4 on the Kopykitab website since subject matter experts have created the solutions in accordance with the most recent CBSE guidelines and exam patterns. Furthermore, students can swiftly study the main concepts of the chapters in both online and offline formats at any time and from anywhere.

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