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The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, then the sides of the two squares are:
- a)18 m and 24 m
- b)6 m and 12 m
- c)12 m and 18 m
- d)24 m and 28 m
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The sum of the areas of two squares is 468 m2. If the difference of th...
Let sides of two squares be s1 nd s2...such that s1>s2.. now.. 4(s1-s2)=24m.. so.. (s1-s2)=6m... (1).. also.. s1^2+s2^2=468m^2... (s1-s2)^2+2s1s2=468... (6)^2+2s1s2=468...(using 1)... 2s1s2=432m^2..(2).. from(1), s1=s2+6.. on putting value of s1 in (2)... 2(s2+6)s2=432.. 2s2^2+12s2-432=0... s2^2+6s2-216=0.. s2^2+18s2-12s2-216=0.. s2(s2+18)-12(s2+18)=0.. (s2+18) (s2-12)=0.. so; s2=12m.. s2 can’t be 18m becoz Distance can't be negative. .. now, s1=(s2+6)=(12+6)m=18m...
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The sum of the areas of two squares is 468 m2. If the difference of th...
Solution:
Let the sides of the two squares be a and b respectively.
Given, Sum of the areas of two squares = 468 m²
=> a² + b² = 468
Also, difference of their perimeters = 24 m
=> 4a - 4b = 24
=> a - b = 6
We can solve these equations simultaneously to find the values of a and b.
From the second equation, we get a = b + 6
Substituting this in the first equation, we get:
(b + 6)² + b² = 468
Simplifying this equation, we get:
2b² + 12b - 396 = 0
Dividing both sides by 2, we get:
b² + 6b - 198 = 0
Using the quadratic formula, we get:
b = (-6 ± √(6² + 4*198)) / 2
b = (-6 ± √(1800)) / 2
b = (-6 ± 30) / 2 (ignoring the negative root)
b = 12
Therefore, a = b + 6 = 18
Hence, the sides of the two squares are 12 m and 18 m.
Therefore, option (c) is correct.
Let the sides of the two squares be a and b respectively.
Given, Sum of the areas of two squares = 468 m²
=> a² + b² = 468
Also, difference of their perimeters = 24 m
=> 4a - 4b = 24
=> a - b = 6
We can solve these equations simultaneously to find the values of a and b.
From the second equation, we get a = b + 6
Substituting this in the first equation, we get:
(b + 6)² + b² = 468
Simplifying this equation, we get:
2b² + 12b - 396 = 0
Dividing both sides by 2, we get:
b² + 6b - 198 = 0
Using the quadratic formula, we get:
b = (-6 ± √(6² + 4*198)) / 2
b = (-6 ± √(1800)) / 2
b = (-6 ± 30) / 2 (ignoring the negative root)
b = 12
Therefore, a = b + 6 = 18
Hence, the sides of the two squares are 12 m and 18 m.
Therefore, option (c) is correct.
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The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, then the sides of the two squares are:a)18 m and 24 mb)6 m and 12 mc)12 m and 18 md)24 m and 28 mCorrect answer is option 'C'. Can you explain this answer?
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The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, then the sides of the two squares are:a)18 m and 24 mb)6 m and 12 mc)12 m and 18 md)24 m and 28 mCorrect answer is option 'C'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, then the sides of the two squares are:a)18 m and 24 mb)6 m and 12 mc)12 m and 18 md)24 m and 28 mCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, then the sides of the two squares are:a)18 m and 24 mb)6 m and 12 mc)12 m and 18 md)24 m and 28 mCorrect answer is option 'C'. Can you explain this answer?.
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