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let length of the rectangle be L and breadth be B
so by the problem perimeter 2(IL+B)=34
=> L+B =17 .......(1)
=> L2+B2+2LB=289 ..........(2)
and the square of the diagonal = L2+B2=132 =169 .....(.3)
By (3) and (2) we get 2LB = 120
So (L- B)2= L2+B2-2LB = 169-120=49
=> L-B =7 .........(4)
hence by (1)and(4) we get 2B = 10 and B = 5
So breadth B = 5 cm
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