Exercise 7.3

Question 1:

Find the area of the triangle whose vertices are:

(i) (2, 3), (− 1, 0), (2, − 4) (ii) (− 5, − 1), (3, − 5), (5, 2)

(i) Area of a triangle is given by

(ii)

Question 2:

In each of the following find the value of ‘k’, for which the points are collinear.

(i) (7, − 2), (5, 1), (3, − k) (ii) (8, 1), (k, − 4), (2, − 5)

(i) For collinear points, area of triangle formed by them is zero.
Therefore, for points (7, −2) (5, 1), and (3, k), area = 0

(ii) For collinear points, area of triangle formed by them is zero.
Therefore, for points (8, 1), (k, −4), and (2, −5), area = 0

Question 3:

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, − 1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

Let the vertices of the triangle be A (0, −1), B (2, 1), C (0, 3).
Let D, E, F be the mid-points of the sides of this triangle. Coordinates of D, E, and F are given by

Question 4:

Find the area of the quadrilateral whose vertices, taken in order, are (− 4, − 2), (− 3, − 5), (3, − 2) and (2, 3)

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Question 5:

You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, − 6), B (3, − 2) and C (5, 2)

Let the vertices of the triangle be A (4, −6), B (3, −2), and C (5, 2).
Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

However, area cannot be negative. Therefore, area of ΔADC is 3 square units.
Clearly, median AD has divided ΔABC in two triangles of equal areas.