Question

the area of a triangle whose vertices are (0,0) , (4,0) and (0,6) is ​

Answer 1

Solution :-

Let the vertices of triangle be :-

A ( 0 , 0 )

B ( 4 , 0 )

C ( 0 , 6 )

$\underline{\boxed{\bf Area \ of \ triangle = \dfrac{1}{2} \times [ \ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2) \ ] } }$

Here :-

$\bullet \sf \ x_1 = 0 \\\\\bullet \sf \ x_2 = 4 \\\\\bullet \sf \ x_3 = 0 \\\\\bullet \sf \ y_1 = 0 \\\\\bullet \sf \ y_2 = 0 \\\\\bullet \sf \ y_3 = 6$

$\longrightarrow \sf \dfrac{1}{2} \times [ \ 0(0-6)+4(6-0)+0(0-0) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ ( 0 \times -6 )+( 4 \times 6 )+( 0 \times 0 ) \ ] \\\\\longrightarrow \dfrac{1}{2} \times [ \ 0+24+0 \ ] \\\\\longrightarrow \dfrac{1}{2} \times 24 \\\\\longrightarrow\bf 12 \ sq.units$

Area of triangle ABC = 12 sq.units

Answer 2

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Given:-

Vertices of a triangle :-

• A =  (0,0)

• B = (4,0)

• C = (0,6)

To Find:-

• Area of the triangle.

Solution:-

‣ Formula to find the Area of the triangle:-

$\pink\leadsto\large\boxed{\tt\purple{\dfrac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)] }}$

$\rightarrow\: \sf {x_1 =0}\\\rightarrow\: \sf {y_2 =0}\\\rightarrow\: \sf {y_3=6}\\\rightarrow\: \sf{x_2 =4}\\\rightarrow\: \sf {y_1 =0}\\\rightarrow\: \sf {x_3 =0}$

‣ Putting the values together :-

$\pink\leadsto$$\sf{\dfrac{1}{2}[0(0-6)+4(6-0)+(0-0)]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[(0)-6)+4(6-0)+0-0]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[0+(4)(6)+0-0]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[0+24+0-0]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[24+0-0]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[24+0]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}[24]}$

$\pink\leadsto$$\sf{\dfrac{1}{2}\times\dfrac{24}{1} }$

$\pink\leadsto$$\sf{\dfrac{1\times24}{2\times1} }$

$\pink\leadsto$$\sf{\dfrac{24}{2} }$

$\sf\leadsto \underline{\boxed{\pink{\mathfrak{12 \:unit^{2} }}}}$

∴ Area of the triangle = $\sf{ 12\:unit^{2} }$

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