2}}\] \[\therefore Rs900\to \FRAC{100}{45}\times 900=2000{{m}^{2}}\] \[A=\frac{1}{2}\LEFT( 2b.\frac{h}{2} \right)=\frac{1}{2}.5h.\frac{h}{2}=\frac{5{{h}^{2}}}{4}\] Now \[\frac{5{{h}^{2}}}{4}=2000\Rightarrow {{h}^{2}}=1600\] \[=h=40\]

"> 2}}\] \[\therefore Rs900\to \FRAC{100}{45}\times 900=2000{{m}^{2}}\] \[A=\frac{1}{2}\LEFT( 2b.\frac{h}{2} \right)=\frac{1}{2}.5h.\frac{h}{2}=\frac{5{{h}^{2}}}{4}\] Now \[\frac{5{{h}^{2}}}{4}=2000\Rightarrow {{h}^{2}}=1600\] \[=h=40\]

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The cost of cutting grass from a triangular field at Rs. 45 per \[100\text{ }{{m}^{2}}\]is Rs. 900. Fine its height if twice the base of the triangle is 5 times the height.                        

7th Class Mathematics in 7th Class 10 months ago

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(b): Area of Field: \[Rs\text{ }45\to 100\,{{m}^{2}}\] \[\therefore Rs900\to \FRAC{100}{45}\times 900=2000{{m}^{2}}\] \[A=\frac{1}{2}\LEFT( 2b.\frac{h}{2} \right)=\frac{1}{2}.5h.\frac{h}{2}=\frac{5{{h}^{2}}}{4}\] Now \[\frac{5{{h}^{2}}}{4}=2000\Rightarrow {{h}^{2}}=1600\] \[=h=40\]

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