Question

26. Find the area of a triangle whose vertices are A(-3,6), B(5,0) and C(-6,2).​

Answer 1

Answer:

answer is given in above photo

Answer 2

Given:-

• Vertices of triangle are A(-3,6),B(5,0) and C(-6,2).

To find:-

• Area of triangle

Required Formula:-

• Area of ∆ ABC = ½ × [x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂) ]

Solution:-

• Area of triangle ABC is 25 sq.units

Step-by-step explanation:

$\\ :\implies\sf{Area \: of \: \triangle \: = \frac{1}{2} \times \bigg[ - 3(0 - 2) + 5(2 - 6) - 6(6 - 0) \bigg]} \\ \\ :\implies\sf{Area \: of \: \triangle \: = \frac{1}{2} \bigg[ - 3( - 2) + 5( - 4) - 6(6) \bigg] } \\ \\ :\implies\sf{Area \: of \: \triangle \: = \frac{1}{2} \bigg[6 + ( - 20) - 36 \bigg] } \\ \\ :\implies\sf{Area \: of \: \triangle \: = \frac{1}{ \cancel{2}}\times \: \cancel{- 50} \: } \\ \\ :\implies \underbrace{\boxed{\frak{Area \: of \: \triangle \: = 25}}} \: \gray{ \bigstar}$

• ∴ Hence,The area of triangle is 25 sq.units.