In this page area of triangle question8 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 question8

**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).**

**Question 8 :**

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

**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 A (4,5) C (-2,2) and B (4,2)

x₁ = 4 x₂ = -2 x₃ = 4

y₁ = 5 y₂ = 2 y₃ = 2

Area of the triangle ACB = 1/2 {4(2-2) + (-2) (2-5) + 4 (5-2)}

= 1/2 {4 (0) + (-2) (-3) + 4 (3)}

= 1/2 {0 + 6 + 12}

= 1/2 {18}

= 18/2

= 9 Square Units.

Therefore the area of ACB = 9 square units.

**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).**

HTML Comment Box is loading comments...