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Areas of Parallelograms and Triangles — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsAreas of Parallelograms and Triangles1 Mark
Question
Write True or False and justify your answer: $ABCD$ is a parallelogram and $X$ is the mid-point of $AB$. If ar $(AXCD) = 24\ cm^2,$ then ar $(ABC) = 24\ cm^2.$
✓

Answer

Given, $ABCD$ is a parallelogram and ar $(AXCD) = 24\ cm^2$ 
Let area of parallelogram $ABCD$ is $24\ cm$ and join $AC. $ 
We know that, diagonals divides the area of parallelogram in two equal areas.
$\therefore\text{ar}(\triangle\text{ABC})=\text{ar}(\text{ACD})=\text{y}$
Also, $X$ is the mid $-$ point of $Ab.$
So, $\text{ar}(\triangle\text{ACX})=\text{ar}(\text{BCX})$
$[$since, $X$ is the median in $\triangle\text{ABC}]$
$=\frac{1}{2}\text{ar}(\text{ABC})=\frac{1}{2}\text{y}$
Now, $\text{ar}(\text{AXCD})=\text{ar}(\text{ADC})+\text{ar}(\text{ACX})$
$24=\text{y}+\frac{\text{y}}{2}$
$\Rightarrow24=\frac{3\text{y}}{2}$
$\Rightarrow\text{y}=\frac{24\times2}{3}=16\text{cm}^2$
Hence, $\text{ar}\triangle(\text{ABC})=16\text{cm}^2$

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