The area of the ΔAOB having vertices A(0, 6), O(0,0), and B(6, 0) isa

1.	The area of the ΔAOB having vertices A(0, 6), O(0,0), and B(6, 0) isa) 12 sq. unitsb) 36 sq. unitsc) 18 sq. unitsd) 24 sq. units
Answer» $\huge\red{\tt{\mathbb{HEYA \: MATE}}}$ Given:- VERTICES, A(0,6) , O(0,0) , B(6,0) To find:- Area of triangle. Solution:- Area Of triangle = ${\frac{1}{2}}$ \| X1(y2-y3) + x2(y3-y1) + x3(y1-y2) \| Where, x1=0 , y1=6 x2=0 , y2=0 x3=6 , y3=0 Area => ${\frac{1}{2}}$ \| 0(0-0) + 0(0-6) + 6(6-0) \| => ${\frac{1}{2}}$ \| 0+0+36 \| => ${\frac{1}{2}}$ × 36 => 18 Which is the REQUIRED answer. ${\huge{\underline{\small{\mathbb{\blue{HOPE\:HELP\:U\:BUDDY :)}}}}}}$ ${\huge{\underline{\small{\mathbb{\pink{~cαndчflσѕѕ♡ :)}}}}}}$

The area of the ΔAOB having vertices A(0, 6), O(0,0), and B(6, 0) isa) 12 sq. unitsb) 36 sq. unitsc) 18 sq. unitsd) 24 sq. units

Answer»

$\huge\red{\tt{\mathbb{HEYA \: MATE}}}$

Given:-

VERTICES, A(0,6) , O(0,0) , B(6,0)

To find:-

Area of triangle.

Solution:-

Area Of triangle = ${\frac{1}{2}}$ | X1(y2-y3) + x2(y3-y1) + x3(y1-y2) |

Where,

x1=0 , y1=6

x2=0 , y2=0

x3=6 , y3=0

Area => ${\frac{1}{2}}$ | 0(0-0) + 0(0-6) + 6(6-0) |

=> ${\frac{1}{2}}$ | 0+0+36 |

=> ${\frac{1}{2}}$ × 36

=> 18

Which is the REQUIRED answer.

${\huge{\underline{\small{\mathbb{\blue{HOPE\:HELP\:U\:BUDDY :)}}}}}}$

${\huge{\underline{\small{\mathbb{\pink{~cαndчflσѕѕ♡ :)}}}}}}$