Problems on Area - Solved Examples(Set 2)
6. A rectangular field has to be fenced on three sides leaving a side of feet uncovered. If the area of the field is sq. feet, how many feet of fencing will be required? | |
A. | B. |
C. | D. |
answer with explanation
Answer: Option C
Explanation:
Solution 1
Area of the field sq. feet.
Length of the adjacent sides are
feet and feet.
Required length of the fencing
feet
Solution 2
Area of the field sq. feet
sq. feet
Required length of the fencing
feet
Explanation:
Solution 1
Area of the field sq. feet.
Length of the adjacent sides are
feet and feet.
Required length of the fencing
feet
Solution 2
Area of the field sq. feet
sq. feet
Required length of the fencing
feet
7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is feet, and the sum of the lengths of the painted sides is feet, find out the area of the parking space in square feet? | |
A. sq. ft. | B. sq. ft. |
C. sq. ft. | D. sq. ft. |
answer with explanation
Answer: Option A
Explanation:
Solution 1
length feet
breadth feet
Area square feet
Solution 2
Let feet. Then,
Area square feet
Explanation:
Solution 1
length feet
breadth feet
Area square feet
Solution 2
Let feet. Then,
Area square feet
8. The area of a rectangular plot is square metres. If the length is more than the breadth, what is the breadth of the plot? | |
A. metres | B. metres |
C. metres | D. metres |
answer with explanation
Answer: Option B
Explanation:
Solution 1
Let breadth
Then length,
From and ,
Solution 2
length of breadth.
length×breadth
⇒ of breadth×breadth
⇒ breadth×breadth
⇒ breadth×breadth
⇒ breadth
Explanation:
Solution 1
Let breadth
Then length,
From and ,
Solution 2
length of breadth.
length×breadth
⇒ of breadth×breadth
⇒ breadth×breadth
⇒ breadth×breadth
⇒ breadth
9. A large field of hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares? | |
A. | B. |
C. | D. |
answer with explanation
Answer: Option D
Explanation:
Solution 1
Let area of the larger part hectares,
area of the smaller part hectares.
Difference of the areas of the two parts
one-fifth of the average of the two areas
(∵ total area is )
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
Hence, area of smaller part
hectares.
Solution 2
Average of the two areas
one-fifth of the average of the two areas
⇒ Difference of the two areas
Let area of the smaller part hectares.
Then, area of the larger part hectares.
Explanation:
Solution 1
Let area of the larger part hectares,
area of the smaller part hectares.
Difference of the areas of the two parts
one-fifth of the average of the two areas
(∵ total area is )
Given that difference of the areas of the two parts = one-fifth of the average of the two areas
Hence, area of smaller part
hectares.
Solution 2
Average of the two areas
one-fifth of the average of the two areas
⇒ Difference of the two areas
Let area of the smaller part hectares.
Then, area of the larger part hectares.
10. The length of a room is m and width is m. What is the cost of paying the floor by slabs at the rate of Rs. per sq. metre. | |
A. Rs. | B. Rs. |
C. Rs. | D. Rs. |
answer with explanation
Answer: Option D
Explanation:
Area sq. metre.
Cost for sq. metre. = Rs.
Hence, total cost
Explanation:
Area sq. metre.
Cost for sq. metre. = Rs.
Hence, total cost
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