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Areas Of Parallelograms And Triangles — MATHS STD 9 — Question

Rajasthan BoardEnglish MediumSTD 9MATHSAreas Of Parallelograms And Triangles5 Marks
Question
If $\text{ABCD}$ is a parallelogram, then prove that
$\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD}) \ =\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar} (\|^{ gm} \text{ABCD})$
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Answer


Given, $\text{ABCD}$ is a parallelogram.
To prove: $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\text{ar}(\triangle\text{ABC})$
$\ =\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar} (\|^{ gm} \text{ABCD})$
Proof: We know that diagonal of a parallelogram divides it into two equal triangles.
Since,$AC$ is the diagonal
Then, $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\frac{1}{2}\text{ar} (\|^{ gm} \text{ABCD}) ...(1)$
Since, BD is the diagonal
Then $\text{ar}(\triangle\text{ABD})=\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar} (\|^{gm} \text{ABCD}) ...(2)$
Compare equation $(1)$ and $(2)$
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABD})$
$\ =\text{ar}(\triangle\text{BCD})=\frac{1}{2}\text{ar} (\|^{ gm} \text{ABCD})$

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