Home » mcq » mathematics » Sum of the areas of two squares is 468 m and the difference of their perimeters is 24 m,then the sum of all sides of the two squares is :
120 m
100 m
80 m
140 m
Answer is Wrong!
Answer is Right!
The correct answer is (b) 100 m.
Let $a$ and $b$ be the sides of the two squares. We know that the area of a square is $a^2$, so the sum of the areas of the two squares is $a^2 +
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b^2 = 468$. We also know that the perimeter of a square is $4a$, so the difference of the perimeters of the two squares is $4a – 4b = 24$.
We can solve this system of equations to find $a$ and $b$. Subtracting the second equation from the first equation, we get $a^2 – b^2 = 444$. Adding $2b^2$ to both sides, we get $3b^2 = 444$. Dividing both sides by 3, we get $b^2 = 148$. Taking the square root of both sides, we get $b = \pm 12$. Since the sides of a square must be positive, we know that $b = 12$.
Substituting $b = 12$ into the equation $a^2 + b^2 = 468$, we get $a^2 + 144 = 468$. Solving for $a$, we get $a^2 = 324$. Taking the square root of both sides, we get $a = 18$.
Therefore, the sum of the sides of the two squares is $a + b = 18 + 12 = \boxed{100}$ m.
Here is a brief explanation of each option:
- Option (a): 120 m. This is not possible because the sum of the sides of a square must be a multiple of 4.
- Option (b): 100 m. This is the correct answer.
- Option (c): 80 m. This is not possible because the sum of the sides of a square must be a multiple of 4.
- Option (d): 140 m. This is not possible because the sum of the sides of a square must be a multiple of 4.