SIDES be a and b as shown in figure Area\[=ab=120\]                                ...(1) Perimeter\[=2\left( a+b \RIGHT)=46\] \[\Rightarrow a+b=23\]                                  ?(2) We have \[{{(a-b)}^{2}}={{(a+b)}^{2}}-4AB\] \[\therefore \]\[a-b=\sqrt{{{(23)}^{2}}-4\times 120}=7\]   ?(3) SOLVING (2) and (3), we get \[a=15,b=8\] \[\therefore \] The length of diagonal of the carpet \[=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{15}^{2}}+{{8}^{2}}}=17\,m.\]

"> SIDES be a and b as shown in figure Area\[=ab=120\]                                ...(1) Perimeter\[=2\left( a+b \RIGHT)=46\] \[\Rightarrow a+b=23\]                                  ?(2) We have \[{{(a-b)}^{2}}={{(a+b)}^{2}}-4AB\] \[\therefore \]\[a-b=\sqrt{{{(23)}^{2}}-4\times 120}=7\]   ?(3) SOLVING (2) and (3), we get \[a=15,b=8\] \[\therefore \] The length of diagonal of the carpet \[=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{15}^{2}}+{{8}^{2}}}=17\,m.\]

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A rectangular carpet has an area of \[120\text{ }{{m}^{2}}\] and a perimeter of 46 m. The length of its diagonal is

7th Class Mathematics in 7th Class 11 months ago

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Let the SIDES be a and b as shown in figure Area\[=ab=120\]                                ...(1) Perimeter\[=2\left( a+b \RIGHT)=46\] \[\Rightarrow a+b=23\]                                  ?(2) We have \[{{(a-b)}^{2}}={{(a+b)}^{2}}-4AB\] \[\therefore \]\[a-b=\sqrt{{{(23)}^{2}}-4\times 120}=7\]   ?(3) SOLVING (2) and (3), we get \[a=15,b=8\] \[\therefore \] The length of diagonal of the carpet \[=\sqrt{{{a}^{2}}+{{b}^{2}}}=\sqrt{{{15}^{2}}+{{8}^{2}}}=17\,m.\]

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