The sum of the areas of two squares is 400 sq.m. If the difference between their perimeters is 16m. find the sides of two squares.

Open in App

Solution

$Let the perimeter of first square be p m . Then the perimeter of second square will be (p - 16) m . Now, the side o f the first square is (\frac{p}{4}) m and side of the second square is (\frac{p - 16}{4}) m . By the given condition, we get, {(\frac{p}{4})}^{2} + {(\frac{p - 16}{4})}^{2} = 400 \Rightarrow \frac{p^{2}}{16} + \frac{p^{2} - 32 p + 256}{16} = 400 \Rightarrow \frac{2 p^{2} - 32 p + 256}{16} = 400 \Rightarrow 2 p^{2} - 32 p - 6144 = 0 \Rightarrow 2 (p^{2} - 16 p - 3072) = 0 \Rightarrow p^{2} - 16 p - 3072 = 0 On splitting the middle term - 16 p as 48 p - 64 p, we ge t : p^{2} + 48 p - 64 p - 3072 = 0 \Rightarrow p (p + 48) - 64 (p + 48) = 0 \Rightarrow (p + 48) (p - 64) = 0 \Rightarrow p + 48 = 0 o r p - 64 = 0 \Rightarrow p = - 48 o r p = 64 Since p is the perimeter of the first square, which cannot be negative, p = 64 Thus, the perimeter of the first square is p = 64 m and perimeter of the second square is p - 16 = 64 - 16 = 48 m Therefore, the side of first square is \frac{p}{4} = \frac{64}{4} = 16 m and side of the second square is \frac{p - 16}{4} = \frac{64 - 16}{4} = \frac{48}{4} = 12 m$

Suggest Corrections

Similar questions

400 m^{2}

16 m

m^{2}

400 c m^{2} .

c m^{2}

Join BYJU'S Learning Program