Math, asked by sparkleb0000, 1 month ago

find the lenghts of the parallel sides of a trapezuim, if the area is 720 cm^2 and the ratio of its parallel sides is 2:3 where the distance between them is 72 cm

Answers

Answered by MrAnonymous412
5

 \\  \\  \color{blue} \underline{ \large\sf \: Question :-} \\  \\

find the lenghts of the parallel sides of a trapezuim, if the area is 720 cm^2 and the ratio of its parallel sides is 2:3 where the distance between them is 72 cm.

 \\  \\  \color{blue} \underline{ \large\sf \: Solution :-} \\  \\

 \\  \\  \:  \:  \:  \:  \:  \:  \:  \sf \: Let  \: the \:  length \:  of \:  parallel  \: sides  \: be \:  2x  \: and \:  3x  \: respectively. \\  \\

 \\  \\  \sf \: We \:  know  \: that, \\  \\

 \\  \\  \sf \: Area  \: of  \: trapezium \:  =  \:  \frac{1}{2}  \times h(a + b) \\  \\

 \\  \\  \sf \: Here,  \: h  \: is  \: height  \: and \:  a  \: is  \: a \:  length  \: o f  \: side \:  \\  \sf \:  which  \: are  \: parallel  \: to  \: another \:  side  \: which  \: i s \:  b . \\  \\

 \\  \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \sf \: Now,  \: put  \: the  \: given  \: values  \: in  \: f ormula \\  \\

 \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies720 \:  =  \:  \frac{1}{2}  \times 72(2x + 3x) \\  \\

 \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies720 \:  =  \:  \frac{1}{2}  \times 72 \times 5x \\  \\

 \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies720 \:  =  \:  36 \times 5x \\  \\

 \\  \\  \sf \:   \therefore\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies \: 5x \:  =  \:  \frac {720}{36}\\  \\

 \\  \\  \sf \:   \therefore\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies \: 5x \:  =  \:  20\\  \\

 \\  \\  \sf \:   \therefore\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies \: x \:  =  \frac{20}{5}\\  \\

 \\  \\  \sf \:   \therefore\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:    : \implies \: \underline {\boxed{  \orange{ \tt x \:  =  4}}}  \\ \\

 \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: Now , \:  put \:  the  \: value  \: o f  \: x  \: in  \:given \: sides  \\  \\

 \\  \\  \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \therefore  \: (1): \:  2x = 2 \times 4 = 6cm \\

 \\    \sf \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \therefore  \: (2): \:  3x = 3 \times 4 = 12cm \\   \\

 \\  \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \underline{ \sf \: hence ,\: The \:  lenghts  \: of  \: the  \: parallel \:  sides \:  of \:  a \:  trapezuim \:  is \:  6  \: cm \:  and  \: 12  \: cm \:  respectively. }\\  \\

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