1998 AHSME Problem 10

p=2(l+b)=14 so l+b=7 now from the figure if we can see the bigger square's side can be expressed as the half of the perimeter so the area of the square is 7^2=49

let $a$ and $b$ be the measurements of 2 non congruent sides of the rectangle. Then, $P=2(a+b)=14$ $a+b=7$

From the figure above, the measure of a side of the bigger square can be expressed as $a+b$ ,which is equal to 7. Hence, the area of the bigger square is $(a+b)^{2}=7^{2}=\boxed{49}$

since perimeter of each rectangle is 2(length breadth)=14, by solving this we get length (l b=7 )of large square as 7 and area is 7*7=49

Each rectangle will have length 5 and width 2 cm. So total side vlength is 7 Cm so area of square is 7X7 =49 Ans K.K.GARG,India