Question
In figure, $CD || AE$ and $CY || BA.$ Prove that $\text{ar}(\triangle\text{CBX}) =\text{ar}(\triangle\text{AXY})$
✓
Answer
$CD || AE$ and $CY || BA$. we have to prove that $\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY}).$
Since triangle on the same base and between the same parallel are equal in area,
so we have $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABY})$
$\Rightarrow\text{ar}(\triangle\text{CBX})+\text{ar}(\triangle\text{ABX})=\text{ar}(\text{ABX})+\text{ar}(\triangle\text{AXY})$
Hence, $\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY})$ [cancelling $\text{ar}(\triangle\text{ABX)}$ from both sides.
Since triangle on the same base and between the same parallel are equal in area,
so we have $\text{ar}(\triangle\text{ABC})=\text{ar}(\triangle\text{ABY})$
$\Rightarrow\text{ar}(\triangle\text{CBX})+\text{ar}(\triangle\text{ABX})=\text{ar}(\text{ABX})+\text{ar}(\triangle\text{AXY})$
Hence, $\text{ar}(\triangle\text{CBX})=\text{ar}(\triangle\text{AXY})$ [cancelling $\text{ar}(\triangle\text{ABX)}$ from both sides.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
In figure, $O$ is the centre of the circle, then prove that $\angle\text{x}=\angle\text{y}+\angle\text{z}.$
→Draw the graph of the equation $x + 2y - 3 = 0$ From your graph, find the value of $y$ when,
→$i. x = 5$
$ii. x = -5.$
If a hollow sphere of internal and external diameters $4\ cm$ and $8\ cm$ respectively melted into a cone of base diameter $8\ cm$, then find the height of the cone.
→In the adjoining figure, the point $D$ divides the side $BC$ of $\triangle\text{ABC}$ in the ratio $m : n$. Prove that $\text{ar}(\triangle\text{ABD}):\text{ar}(\triangle\text{ADC})=\text{m}:\text{n}$.
→$ABC$ is an isosceles triangle in which $AB = AC. BE$ and $CF$ are its two medians. Show that $BE = CF.$
→In Fig. $AB = AC $and $DB = DC$, find the ratio $\angle\text{ABD}:\angle\text{ACD}.$
→If $V$ is the volume of a cuboid of dimensions $a, b, c$ and $S$ is its surface area, then prove that:
$\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
→→$\frac{1}{\text{V}}=\frac{2}{\text{S}}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)$
Verify whether the indicated numbers are zeros of the polynomials corresponding to them in the following case: $p(x) = x^3 - 6x^2 + 11x - 6, x = 1, 2, 3$
→Simplify the following products: $\Big(\text{m}+\frac{\text{n}}{7}\Big)^3\Big(\text{m}-\frac{\text{n}}{7}\Big)$
→