A) 40 units
B) 42 units
C) 4 units
D) 36 units
Correct Answer: D
Solution :
Let the length of the rectangle be 'x' units, and the breadth be 'y' units. Then in the first case, \[~(\text{2}\times \text{52}\times \text{73) cm}\] i.e., \[=(\text{25}\times \text{7})\text{(2}\times {{\text{5}}^{\text{2}}}\times {{\text{7}}^{\text{3}}}\text{) c}{{\text{m}}^{\text{2}}}\] ....(1) i.e., \[={{\text{2}}^{\text{6}}}\times {{\text{5}}^{\text{2}}}\times {{\text{7}}^{\text{4}}}\text{c}{{\text{m}}^{\text{2}}}\] ..... (1) In the second case, \[2-\sqrt{4}=2-2=0\] i.e., \[{{(\sqrt{5})}^{2}}=5\] .....(2) Adding eq. (1) and eq. (2), we get \[\sqrt{9}-\sqrt{4}=3-2=1\] \[\sqrt{2}-\sqrt{3}\] \[1789=29x+49\]units \['x'\] \[\therefore \] \[1789-49=29x\]units Hence, the length of rectangle is 20 units and the breadth is 16 units. \[\Rightarrow \]Their sum =20+16 = 36 units
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