**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.

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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:

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)

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)

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)

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 (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) respectively

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

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 (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3}) and D (x_{4}, y_{4}) 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

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 (x_{1}, y_{1}) of B are (x_{2}, y_{2}) and of C are (x_{3}, y_{3})

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 (x_{1}, y_{1}), B (x_{2}, y_{2}), C (x_{3}, y_{3}) and let G be the centroid of the triangle

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:

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

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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