In this page area of triangle question7 we are going to see solution of first question some
practice questions. In the lower classes we have studied about how to
find the area for any triangle. We will use the formula is (1/2 ) x
base x height if the base and perpendicular height is given.But here we
are going to see the another method that is if we are given the three
vertices of a triangle we can find the area by using this formula. Here
you can practice problems in the form of quiz. You can verify the
solution by clicking the link.

Question 1 :

Find the area of triangle whose vertices are (6,7) (2,-9) and (-4,1).

Question 2 :

Find the area of triangle whose vertices are (3,4) (2,-1) and (4,-6).

Question 3 : area of triangle question7

Find the area of triangle whose vertices are (5,6) (2,4) and (1,-3).

Question 4 :

Find the area of triangle whose vertices are (1,3) (-7,6) and (5,-1).

Question 5 :

Find the area of triangle whose vertices are (1,1) (3,4) and (5,-2).

Question 6 :

Find the area of triangle whose vertices are (-3,-9) (-1,6) and (3,9).

Question 7 :

Find the area of triangle whose vertices are (-3,-9) (3,9) and (5,-8).

Solution:

First we have to plot the point in the graph sheet as below.

Now we have to take anticlockwise direction. So we have to take the points in the order B (3,9) A (-3,-9) and C (5,-8)

Area of the triangle = ^{1}⁄_{2} {x_{1}(y_{2}-y_{3}) + x_{2}(y_{3}-y_{1}) +
x_{3}(y_{1}-y_{2})}

x₁ = 3 x₂ = -3 x₃ = 5

y₁ = 9 y₂ = -9 y₃ = -8

Area of the triangle BAC = 1/2 {3(-9-(-8)) + (-3) (-8-9) + 5 (9-(-9)}

= 1/2 {3 (-9+8) + (-3) (-17) + 5 (9+9)}

= 1/2 {3 (-1) + 51 + 5 (18)}

= 1/2 {-3 + 51 + 90}

= 1/2 {-3 + 141}

= 1/2 {138}

= 138/2

= 69 Square Units.

Therefore the area of BAC = 69 square units.

Question 8 :

Find the area of triangle whose vertices are (4,5) (4,2) and (-2,2).

Question 9 :

Find the area of triangle whose vertices are (3,1) (2,2) and (2,0).

Question 10 :

Find the area of triangle whose vertices are (3,1) (0,4) and (-3,1).