The area of a triangle with vertices A(3,0),

Question:

The area of a triangle with vertices A(3,0), B(7, 0) and C(8, 4) is

(a) 14                    

(b) 28                        

(c) 8                          

(d) 6

Solution:

(c) Area of Δ ABC whose Vertices A≡(x1,y1),B≡(x2,y2) and C≡(x3, y3) are given by

$\Delta=\left|\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]\right|$

Here, $x_{1}=3, y_{1}=0, x_{2}=7, y_{2}=0, x_{3}=8$ and $y_{3}=4$

$\therefore \quad \Delta=\left|\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]\right|=\left|\frac{1}{2}(-12+28+0)\right|=\left|\frac{1}{2}(16)\right|=8$

Hence, the required area of AABC is 8.

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now