27. The diagram shows a sequence of square wire frames. The lengths of a side of these frames
are x cm, (x+ 3) cm, (x+6) cm......respectively. The sum of the areas of the first three
squares is 525 sq cm.
1. Express the length of a side of the frame Square wire frames
in terms of x and n
2. Find the value of x
3. piece of wire is 99 cm long. It is cut and
bent into a frame in the sequence Find the
length of a side of the largest frame than can be formed.
Answers
Step-by-step explanation:
where is the diagram
plz provide the diagram as attachment
27. The diagram shows a sequence of square wire frames.
Given,
The lengths of a side of wire frames = x cm, x+3 cm, x+6 cm
1. Express the length of a side of the frame Square wire frames in terms of x and n
Length of side of frame in terms of x and n be given by,
x + 3n, where n = 0,1,2,3.....
2. Find the value of x
Areas of wire frames = (x)², (x+3)², (x+6)²
Sum of areas of wire frames = 525 sq cm
∴ (x)² + (x+3)² + (x+6)² = 525
⇒ x² + x² + 9 + 6x + x² + 36 + 12x = 525
3x² + 45 + 18x = 525
3x² + 18x - 480 = 0
x (x + 16) - 10 (x + 16) =0
x= 10
3. piece of wire is 99 cm long. It is cut and bent into a frame in the sequence Find the length of a side of the largest frame than can be formed.
Given, the piece of wire = 99cm
⇒ The perimeter of square after bending = 99cm
∴ 4a = 99, where a = side of square
Side = a = 99/4 = 24.75 cm
Side = 24.75 cm