Question
The parallel sides of a trapezium are 20cm and 10cm. Its nonparallel sides are both equal, each being 13cm. Find the area of the trapezium.
✓
Answer
In trapezium ABCD,
AB || DC and AD = BC
AB = 20cm, CD = 10cm.
AD = BC = 13cm,
Through C, draw CE || DA and $\text{CF}\perp\text{EB}$ or AB,
Then CE = CA = 13cm.
EB = AB - AE = AB - CD
= 20 - 10 = 10cm.
Now side of $\triangle\text{ECB}$ are 13, 13, 10cm,
$\text{s}=\frac{13+13+10}{2}=\frac{36}{2}=18$
$\therefore$ area of $\triangle\text{ECB},$
$=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ (Hero's formula)
$=\sqrt{18(18-13)(18-13)(18-10)}$
$=\sqrt{18\times5\times5\times8}=\sqrt{3600}=60\text{cm}^2$
But area of $\triangle\text{CEB}=\frac{1}{2}\times\text{EB}\times\text{CF}$
$\Rightarrow60=\frac{1}{2}\times10\times\text{CF}$
$\Rightarrow\text{CF}=\frac{60\times2}{10}=12\text{cm}.$
$\therefore$ Distance between two parallel lines (h) = 12cm.
$\therefore$ Area of trapezium $=\frac{1}{2}\text{h}(\text{l}_1+\text{l}_2)$
$=\frac{1}{2}\times12(20+10)\text{cm}^2$
$=6\times30=180\text{cm}^2$
AB || DC and AD = BC
AB = 20cm, CD = 10cm.
AD = BC = 13cm,
Through C, draw CE || DA and $\text{CF}\perp\text{EB}$ or AB,
Then CE = CA = 13cm.
EB = AB - AE = AB - CD
= 20 - 10 = 10cm.
Now side of $\triangle\text{ECB}$ are 13, 13, 10cm,
$\text{s}=\frac{13+13+10}{2}=\frac{36}{2}=18$
$\therefore$ area of $\triangle\text{ECB},$
$=\sqrt{\text{s}(\text{s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$ (Hero's formula)
$=\sqrt{18(18-13)(18-13)(18-10)}$
$=\sqrt{18\times5\times5\times8}=\sqrt{3600}=60\text{cm}^2$
But area of $\triangle\text{CEB}=\frac{1}{2}\times\text{EB}\times\text{CF}$
$\Rightarrow60=\frac{1}{2}\times10\times\text{CF}$
$\Rightarrow\text{CF}=\frac{60\times2}{10}=12\text{cm}.$
$\therefore$ Distance between two parallel lines (h) = 12cm.
$\therefore$ Area of trapezium $=\frac{1}{2}\text{h}(\text{l}_1+\text{l}_2)$
$=\frac{1}{2}\times12(20+10)\text{cm}^2$
$=6\times30=180\text{cm}^2$
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