Question

Find the arae of triangle A(3, 0) (7, 0) and (8, 4)

Answer 1

Given :

• A ( 3 , 0 )
• B ( 7 , 0 )
• C ( 8 , 4 )

To Find :

• Area of the triangle

Solution :

$\large \sf \orange{Area_{triangle} = \dfrac{1}{2} \bigg[x_1(y_2 -y_2) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \bigg]} \\ \\ \sf x_1=3\: \: \:\: \:\:\:\:x_2=7\: \:\:\:\: \: \:\:\:x_3=8\\ \\\sf y_1 = 0\: \:\:\:\: \: \:\:\: y_2=0\: \:\:\:\: \: \:\:\:y_3=4 \\ \\ \sf \implies \dfrac{1}{2} \bigg[3(0 -4) + 7(4 - 0) + 8(0 - 0) \bigg] \\ \\ \sf \implies \dfrac{1}{2} \bigg[ 3( - 4) + 7(4 ) + 8(0) \bigg] \\ \\ \sf \implies \dfrac{1}{2} \bigg[ - 12 + 28 + 0 \bigg] \\ \\ \sf \implies \dfrac{1}{2} \times 16 \\ \\ \sf \implies8 \\ \\ \Large \underline{\bf \green{ Area \: of \: triange \: is \: 8 \: {unit}^{2}}}$

Answer 2

Question :

Find the area of triangle A(3, 0) (7, 0) and (8, 4).

Given :

A ( 3 , 0 )

B ( 7 , 0 )

C ( 8 , 4 )

To Find :

• Area of the triangle

Solution :

Area of Triangle = $\frac{1}{2}( x_{1}(y_{2} -y_{2}) +x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})$

–––––––––––––

$x_{1} = 3$

$x_{2} = 7$

$x_{3} = 8$

$y_{1} = 0$

$y_{2} = 0$

$y_{3} = 4$

–––––––––––––

$= > \frac{1}{2} \times 3(0 - 4) + 7(4 - 0) + 8(0 - 0)$

$= > \frac{1}{2} \times 3( - 4) + 7(4) + 8(0)$

$= > \frac{1}{2} \times - 12 + 28 + 0$

$= > \frac{1}{ \cancel2} \times \cancel16$

$\huge \boxed {= > 8}$

Area of triangle is $8 \: {unit}^{2}$