Sum of the areas of two squares is 468 m2.3. If the difference of their perimeters is 24 m, find the sides of the two squares. (A) 40 cm, 16 cm (B) 48 cm, 24 cm (C) 6 cm, 78 cm (D) 18 cm, 12 cm
Answers
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Sum of the areas of two squares is 468 m2.3. If the difference of their perimeters is 24 m, find the sides of the two squares. (A) 40 cm, 16 cm (B) 48 cm, 24 cm (C) 6 cm, 78 cm (D) 18 cm, 12 cm
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here is the solution
area of 1 st square be a²
area of 2 st square be b²
ATQ,
a² + b² = 468............[1]
a² = 468 - b²
the difference of their perimeters is 24 m
so
4a - 4b = 24
let us square on both sides
16a² -16b² = 576
a² - b² = 36............[2]
so
a² + b² = 468............[1]
a² - b² = 36............[2]
now substitute and find the value of a and b by adding this up
thank you
Step-by-step explanation:
Answer:
→ 18m and 12 m .
Step-by-step explanation:
Let the sides of two squares be x m and y m respectively .
Case 1 .
→ Sum of the areas of two squares is 468 m² .
A/Q,
∵ x² + y² = 468 . ...........(1) .
[ ∵ area of square = side² . ]
Case 2 .
→ The difference of their perimeters is 24 m .
A/Q,
∵ 4x - 4y = 24 .
[ ∵ Perimeter of square = 4 × side . ]
⇒ 4( x - y ) = 24 .
⇒ x - y = 24/4.
⇒ x - y = 6 .
∴ y = x - 6 ..........(2) .
From equation (1) and (2) , we get
∵ x² + ( x - 6 )² = 468 .
⇒ x² + x² - 12x + 36 = 468 .
⇒ 2x² - 12x + 36 - 468 = 0 .
⇒ 2x² - 12x - 432 = 0 .
⇒ 2( x² - 6x - 216 ) = 0 .
⇒ x² - 6x - 216 = 0 .
⇒ x² - 18x + 12x - 216 = 0 .
⇒ x( x - 18 ) + 12( x - 18 ) = 0 .
⇒ ( x + 12 ) ( x - 18 ) = 0 .
⇒ x + 12 = 0 and x - 18 = 0 .
⇒ x = - 12m [ rejected ] . and x = 18m .
∴ x = 18 m .
Put the value of 'x' in equation (2), we get
∵ y = x - 6 .
⇒ y = 18 - 6 .
∴ y = 12 m ...............
Hence, sides of two squares are 18m and 12m respectively .