|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
12 Oct 2021, 03:15
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
35% (03:07) correct
65%
(02:37)
wrong
based on 34
sessions
History
Date
Time
Result
Not Attempted Yet
There are 6 concentric squares and a circle is inscribed in the smallest square. If the area of the circle is 77m^2 and the distance between any two consecutive corners of the square is 2 meters, find the approximate sum of areas of the outermost and innermost square.
A. 376 m2
B. 476 m2
C. 576 m2
D. 676 m2
E. 776 m2
A. 376 m2
B. 476 m2
C. 576 m2
D. 676 m2
E. 776 m2
02 Dec 2021, 02:01
Kudos
Bookmarks
answer is D
Step 1:
diameter of the inscribed circle is 7(root 2) hence one side of the smallest square is 7(root 2) using a^2+b^2 = c^2 you can find the diagonal of the smallest square = 14 and the area of the square = 98
Step 2:
diagonal of the largest square = (14 + 20)^2 = 34^2 (20 is used here as the distance between any two consecutive corners of the square is 2 meters hence 10 corners times 2 = 20)
hence a^2+b^2 = 1156
since this is a square 1156/2 will give you both a^2 and b^2 = 578
Hence side of largest square = (root 578) but since this is a square area = side x side therefore area = 578
Step 3:
Add both areas = 578 + 98 = 676m^2
Step 1:
diameter of the inscribed circle is 7(root 2) hence one side of the smallest square is 7(root 2) using a^2+b^2 = c^2 you can find the diagonal of the smallest square = 14 and the area of the square = 98
Step 2:
diagonal of the largest square = (14 + 20)^2 = 34^2 (20 is used here as the distance between any two consecutive corners of the square is 2 meters hence 10 corners times 2 = 20)
hence a^2+b^2 = 1156
since this is a square 1156/2 will give you both a^2 and b^2 = 578
Hence side of largest square = (root 578) but since this is a square area = side x side therefore area = 578
Step 3:
Add both areas = 578 + 98 = 676m^2
07 Dec 2021, 00:08
Kudos
Bookmarks
Let the squares be numbered from 1 to 6, 1 being the innermost square containing the circle and 6 the outermost.
Area of the innermost square 1:
The diameter of the circle would be the side of the smallest square and hence the area of the circle can be expressed as
A = (pi) *r2 = 77
=>r= 7/√2 m =>Diameter =7√2
Side of the square = diameter of the circle = 7√2
Thus area of this square(1) =side*side= 7√2 * 7√2
= 98 square units
Area of the outermost square 6:
Diagonal of the smallest square(Square 1) = √2*side = 14 units
The distance between the top-right corners of square 1 and square 2 is 2 units; b/w square 2 and square 3 is 2 units and so on till we reach at the distance b/w square 5 and square 6 corners which is also 2 units.
So the total distance covered is 2*5=10 units on top right corner.
Similarly the total distance covered in bottom right = 2*5=10 units.
Hence the diagonal of the outermost square that is square 6 =10(top right) + diagonal of the innermost square + 10(bottom right)
=10 + 14 + 10 = 34 units
Thus the side of the outermost square(6) = 34 / √2
Thus the area of the square = side*side
= (34 / √2 ) * (34 / √2 )
=578 m^2
Sum of the areas = 98 + 578 =676 m^2
(option d)
D.S
GMAT SME
Area of the innermost square 1:
The diameter of the circle would be the side of the smallest square and hence the area of the circle can be expressed as
A = (pi) *r2 = 77
=>r= 7/√2 m =>Diameter =7√2
Side of the square = diameter of the circle = 7√2
Thus area of this square(1) =side*side= 7√2 * 7√2
= 98 square units
Area of the outermost square 6:
Diagonal of the smallest square(Square 1) = √2*side = 14 units
The distance between the top-right corners of square 1 and square 2 is 2 units; b/w square 2 and square 3 is 2 units and so on till we reach at the distance b/w square 5 and square 6 corners which is also 2 units.
So the total distance covered is 2*5=10 units on top right corner.
Similarly the total distance covered in bottom right = 2*5=10 units.
Hence the diagonal of the outermost square that is square 6 =10(top right) + diagonal of the innermost square + 10(bottom right)
=10 + 14 + 10 = 34 units
Thus the side of the outermost square(6) = 34 / √2
Thus the area of the square = side*side
= (34 / √2 ) * (34 / √2 )
=578 m^2
Sum of the areas = 98 + 578 =676 m^2
(option d)
D.S
GMAT SME
