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AreaAreas of Parallelograms and Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreaAreas of Parallelograms and Triangles3 Marks
Question
In the adjoining figure, $DE \| BC.$ Prove that:
$i. \text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABE})$
$ii. \text{ar}(\triangle\text{OCE})=\text{ar}(\triangle\text{OBD}).$
✓

Answer

$\triangle\text{DEC}$ and $\triangle\text{DEB}$ lies on the same base and between the same parallel lines.
So, $\text{ar}(\triangle\text{DEC})=\text{ar}(\triangle\text{DEB})\dots(1)$
$i.$ On adding $\text{ar}(\triangle\text{ADE})$ in both sides of equation $(1),$ we get:
$\text{ar}(\triangle\text{DEC})+\text{ar}(\triangle\text{ADE})=\text{ar}(\triangle\text{DEB})+\text{ar}(\triangle\text{ADE})$
$\Rightarrow\ \text{ar}(\triangle\text{ACD})=\text{ar}(\triangle\text{ABE})$
$ii.$ On adding $\text{ar}(\triangle\text{ODE})$ in both sides of equation $(1),$ we get:
$\text{ar}(\triangle\text{DEC})-\text{ar}(\triangle\text{ODE})=\text{ar}(\triangle\text{DEB})-\text{ar}(\triangle\text{ODE})$
$\Rightarrow\ \text{ar}(\triangle\text{OCE})=\text{ar}(\triangle\text{OBD}).$

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