Question

the area of ∆AOB having vertices A(0,0) O(1,0) B(0,1).

Answer 1

Question :-

the area of ∆AOB having vertices A(0,0) O(1,0) B(0,1).

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Solution :-

➲ Given Information :-

• Vertices of a Right Angled Triangle ∆AOB
• A ➢ ( 0,0 )
• ⇒ A ➢ ( X ₁,Y ₁ )
• O ➢ ( 1,0 )
• ⇒ O ➢ ( X ₂,Y ₂ )
• B ➢ ( 0,1 )
• ⇒ B ➢ ( X ₃,Y ₃ )

➲ To Find :-

• The area of Right Angled Triangle ∆SOB

➲ Concept :-

• Area And Perimeter of Plane Figures / Graphical Representation of Plane Figures

➲ Formula Used :-

• Area = $\boxed{ \boxed{ \sf \dfrac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]}}$

➲ Explanation :-

• Simply substituting the values in the formula above, We will get our answer, i.e., area of ∆AOB

➲ Diagram :-

• Attached in the image above.

➲ Calculation :-

Substituting the values in the formula stated above, We get,

$\sf: \implies \sf \dfrac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) ]$

After substitution of values, We get,

$\sf: \implies \sf \dfrac{1}{2} [0(0 - 1) + 1(1 - 0) + 0(0 - 0) ]$

Calculating further, We get,

$\sf: \implies \sf \dfrac{1}{2} [0 + 1 +0 ]$

Calculating further, We get,

$\sf: \implies \sf \dfrac{1}{2} [1] = \dfrac{1}{2}$

Cancelling & Calculating further, We get,

$\sf: \implies \red {\dfrac{1}{2} }$

$\sf \therefore \bf area \: of \: \triangle = \boxed{\red{\dfrac{1}{2} \: units^{2} }}$

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Final Answer :-

• Area of ∆AOB =$\dfrac{1}{2} \: units^{2}$

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