NCERT Solutions For Class 10 MATH CHAPTER – 7 EXERCISE- 7.3

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NCERT Solutions For Class 10 Math Chapter – 7 Exercise – 7.3

 

Q1. Find the area of the triangle whose vertices are:

  • (2,3) , (-1,0) , (-2,4)
  • (-5,-1) , (3,-5) , (5,2)

Solution:-

 

Area of triangle = 1/2 x [x1(y2-y3) +x2(y3 – y1) + x3(y1 – y2)]

  • Area of triangle =  1/2 [2(0-(-4) + (-1) (-4) -(3)) +2(3-0)]

= 21/2

  • Area of triangle = 1/2  [ -5(-5) – (-2) + 3(2-(-1) + 5(-1-(-5)]

= 32

 

Q2. In each of the following find the value of ‘k’ for which the points are collinear.

  • (7,-2) , (5,1) , (3,-k)
  • (8,1) , (k,-4) , (2,-5)

Solution:-

 

  • Area of triangle = 1/2 [ 7(1-k) +5(k-2) +3(-2-1)] = 0

-2k + 8 = 0

K = 4

  • Area of triangle = 1/2[8(-4-(-5)) + k (-5) – (1) + 2(1-(-4)] = 0

6k = 18

K = 3

 

Q3. 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 (3,0) Find the ratio of this area to the area of the given triangle.

Solution:-

 

NCERT Solutions for Class 10 Chapter 7-26

 

 

 

 

Area of ∆DEF = 1/2 [ 1(2-1) + 1(1-0) + 0(0-2)]

= 1/2[1+1]

= 1

 

Area of ∆ABC = 1/2[0(1-3) +2(3-(-1) +0(-1-1)]

= 1/2(8)

= 4

 

Therefore , the required ratio 1:4

 

Q4. Find the area of the quadrilateral whose vertices , taken in order , are

(-4,-2) , (-3,-5) , (3,-2) and (2,3).

Solution:-

 

NCERT Solutions for Class 10 Chapter 7-27

 

 

 

 

 

 

Area of ∆ABC = 1/2 [ -4  ((-5) – (-2) + (-3) (-2) – (-2) + 3 (-2) – (-5)]

= 21/2

 

Area of ∆ACD = 1/2 [ (-4) (-2) – (3) + 3 (3-(-2) +2 (-2) – (-2)]

= 35/2

 

Area of quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

= (21/2 + 35/2)

= 28 square units .

 

Q5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A(4,-6) B (3,-2) and C(5,2).

Solution:-

 

A(4,-6)  B(3,-2)  and C(5,2).

 

NCERT Solutions for Class 10 Chapter 7-28

 

 

 

 

 

 

 

Let D be the mid – point of side BC of ∆ABC. Therefore , AD is the median in ∆ABC.

Coordinates of point D = Midpoint of BC = (3+5)/2  , (-2+2)/2 = (4,0)

Area of ∆ABD = 1/2[(4) (-2) -(0) + 3(0 – (-6) + 4(-6 – (-2)]

= -3 square units.

 

Area of ∆ACD = 1/2 [ 4(0-2) +4 (2-(-6) + 5(-6) – (0)]

= -3 square units.

The area of both sides is the same Thus median AD has divided ∆ABC in two triangles of equal areas.