{"id":1126,"date":"2024-03-05T15:11:49","date_gmt":"2024-03-05T15:11:49","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=1126"},"modified":"2024-03-05T15:11:49","modified_gmt":"2024-03-05T15:11:49","slug":"sum-of-the-areas-of-two-squares-is-468-m-and-the-difference-of-their-perimeters-is-24-mthen-the-sum-of-all-sides-of-the-two-squares-is","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/sum-of-the-areas-of-two-squares-is-468-m-and-the-difference-of-their-perimeters-is-24-mthen-the-sum-of-all-sides-of-the-two-squares-is\/","title":{"rendered":"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 :"},"content":{"rendered":"

[amp_mcq option1=”120 m” option2=”100 m” option3=”80 m” option4=”140 m” correct=”option2″]<\/p>\n

The correct answer is (b) 100 m.<\/p>\n

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 $a^2 + b^2 = 468$. We also know that the perimeter of a square is $4a$, so the difference of the perimeters of the two squares is $4a – 4b = 24$.<\/p>\n

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 $b = 12$.<\/p>\n

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 $a = 18$.<\/p>\n

Therefore, the sum of the sides of the two squares is $a + b = 18 + 12 = \\boxed{100}$ m.<\/p>\n

Here is a brief explanation of each option:<\/p>\n