The lengths of the parallel sides of a trapezium are in the ratio 4:7 . If the height of trapezium is 14 m and it's area is 385 sq.m , fins the length of its parallel sides .
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The area of a trapezium is given by,
[math]\text{A}=\frac{1}{2}\cdot \text{h}\cdot(b_{1}+b_{2})[/math]
Given:
[math]b_{1}:b_{2}= 2:5[/math], [math]\text{h}=10\text{cm}[/math]
[math]\text{A}=350cm^{2}[/math]
[math]\therefore \frac{b_{1}}{b_{2}}=\frac{2}{5}[/math]
[math]\Rightarrow b_{1}=\frac{2}{5}\cdot b_{2}\cdots(1)[/math]
Now substituting in the formula of Area of trapezium we get,
[math]\frac{1}{2}\cdot 10\cdot \left ( \frac{2}{5}b_{2}+b_{2} \right )=350[/math]
Simplifying we get,
[math]\frac{7}{5}b_{2}=70[/math]
Hence,
[math]b_{2}=50cm[/math]
Back-substituting [math]b_{2}[/math] in equation [math](1)[/math]
We get, [math]b_{1}=20cm[/math]..
Example please
[math]\text{A}=\frac{1}{2}\cdot \text{h}\cdot(b_{1}+b_{2})[/math]
Given:
[math]b_{1}:b_{2}= 2:5[/math], [math]\text{h}=10\text{cm}[/math]
[math]\text{A}=350cm^{2}[/math]
[math]\therefore \frac{b_{1}}{b_{2}}=\frac{2}{5}[/math]
[math]\Rightarrow b_{1}=\frac{2}{5}\cdot b_{2}\cdots(1)[/math]
Now substituting in the formula of Area of trapezium we get,
[math]\frac{1}{2}\cdot 10\cdot \left ( \frac{2}{5}b_{2}+b_{2} \right )=350[/math]
Simplifying we get,
[math]\frac{7}{5}b_{2}=70[/math]
Hence,
[math]b_{2}=50cm[/math]
Back-substituting [math]b_{2}[/math] in equation [math](1)[/math]
We get, [math]b_{1}=20cm[/math]..
Example please
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