Questions on Area and Perimeter are asked in IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC CGL, SSC CHSL and other competitive exams that you can think of. So read on as we discuss the different kind of questions asked on area and perimeter.
Area and Perimeter is a formula oriented topic where most mensuration problems can be solved by using the appropriate formulas for IBPS PO, IBPS Clerk, SBI PO, SBI Clerk, SSC CGL, SSC CHSL and other competitive exams. After discussing the introduction to mensuration with a list of mensuration formulas, it’s time to move to the second part where we discuss mensuration problems, some simple and some tricky. In the second part of this series on mensuration, we will be discussing area and perimeter based mensuration problems on 2D shapes. So revise the list of formulas and get ready!
We have divided problems from area and perimeter on 2D shapes in 3 parts- Basic Problems, Problems on Fencing and Carpeting and Problems on Paths around a Plot.
Set I: Basic Mensuration Problems on Area and Perimeter
Problems discussed in this part involve basic usage of the area and perimeter formula with a little bit of tweaking.
Problem 1: The ratio between the perimeter and length of a rectangle is 5:1. If the area of the rectangle is 216cm2, then what is the length of the rectangle?
We know the formulas for the area and perimeter of a rectangle-
Ratio between the length and the perimeter is given to us, so we can for the proportionality equation-
P : B :: 5 : 1
P/B= 5/1 (i)
We will replace the perimeter formula in the equation (i)-
2(l+b)/b = 5/1
Solve the equation by cross multiplication-
2(l+b) = 5b
2l + 2b = 5b
2l = 3b
b = 2/3 l (ii)
Replacing equation (ii) in the area formula-
Area = l x b
Area = l x 2/3l (iii)
Area = 216 (iv – given)
We have two area equations and two variables, so equating (iii) and (iv) we get-
216 = l x 2/3l
l2 = 216 x 3/2
l2 = 108 x 3
l2 = 324
l = 18
Therefore the length of the rectangle is 18cm
Set II: Basic Area and Perimeter Problems about Fencing and Carpeting a given Plot
Mensuration problems discussed in this part questions on area and perimeter deal with drawing fences around a given plot or covering a plot with carpet, tiles etc. To solve such mensuration problems you need to use the area perimeter formulas indirectly.
Problem 1: A fence is to be drawn around a circular ground of radius 7m. What will be the total expenditure, if the cost of fencing it is Rs 120/m?
On observing the diagram above you will realize that you need to know the length of the boundary to know the cost of fencing it. This boundary is equal to the perimeter of the plot. So the length of the fence needs to be equal to the circumference (the perimeter of a circle is called a circumference).
We know that the circumference/ perimeter of a circle is-
Substituting the values in the circumference formula-
Circumference = 2 x 22/7 x 7
Circumference = 2 x 22 = 44m
We know the cost of fencing 1m is Rs 120, therefore using unitary method you can find the cost for fencing 44m –
Total expenditure = Circumference x Cost/m
Total expenditure = 44 x 120
Total expenditure = 5280
Therefore the total expenditure of fencing the circular ground is Rs 5240.
For such mensuration problems in area and perimeter, the ground can be of any 2D shape, square, rectangle or even triangle and the activity can be that of building a compound wall or even just putting a rope at the boundary.
Problem 2: The floor of a rectangular auditorium of length 40 feet and breadth 28 feet is to be covered with a wall to wall carpet. Find the total cost, if the carpet costs Rs 72/sq. feet and the pasting charges are Rs6/sq. feet.
The figure represents a rectangular auditorium of length 40 feet and breadth 28 feet, whose floor has to be, covered a wall to wall carpet. So to get the cost of carpeting we need to first find out the area to be carpeted which is actually equal to the area of the floor of the rectangular auditorium.
We know the formula for the area of a rectangle is-
Substituting the values in the area formula we get-
Area = 28 x 40
Area = 1120 sq. feet
So the area of the auditorium to be carpeted is 1120 sq. feet
Now to get the cost carpeting the floor, we need the multiply per sq. feet cost of carpeting the floor with the area of the floor. So we first we need the total per sq. feet cost, which is a sum of cost of carpeting the floor and the pasting charges.
Total Cost of Carpeting/ Sq. Feet = Cost of the Carpet + Cost of Pasting
Total Cost = 72 + 6 = 78
The total cost of carpeting the floor will be the product of the total area and the total per sq. feet cost.
Total Cost = Total Area x Total Cost of Carpeting/ Sq. Feet
Total Cost = 1120 x 78
Total Cost = 87360
Therefore the total cost of carpeting the rectangular auditorium floor is Rs 87360
When solving mensuration problems like this on area and perimeter we need to always first take into account the shape of the area in consideration, find its total area and the n find the cost of the action to be done. The action can be of laying tiles, polishing or painting a certain area etc.
Set III: Basic Area and Perimeter Problems about Paths Around a given Plot
The mensuration problems discussed in this part, deal with settings when a path or garden is drawn around a 2D shape. The approach to such area and perimeter problems is by finding the area of the actual garden and plot and subtracting it from the new area which includes the path.
Problem 1: A rectangular path of length 20m and breadth 18m is to be bounded by a 2m wide jogging track from outside. What will be the area of the track?
In the figure above the smaller rectangle represents the plot that has to bounded by a jogging track. The rectangle outside represents the plot along with the jogging track and the shaded area is the jogging track whose area we have to find. So we find the area of both the rectangles and subtract them from each to get the answer.
The area of the Inner Rectangle is easy to find, it is the direct application of the formula.
Area of Inner Rectangle = 20 x 18
Area of Inner Rectangle = 360 m2
Now come to the area of the outer rectangle. We need to find the new length and breadth to find its area. For the new length and breadth we have to add the 2m extensions form both th ends.
Length of the Outer Rectangle = Length of the Inner Rectangle + Width of the Jogging Track
Length of the Outer Rectangle = 20 + 2 + 2 (since the width has extended by 2m on both left and right side)
Length of the Outer Rectangle = 24m
Similarly, Breadth of the Outer Rectangle = Breadth of the Inner Rectangle + Width of the Jogging Track
Breadth of the Outer Rectangle = 18 + 2 + 2 (since the width has extended by 2m on both left and right side)
Breadth of the Outer Rectangle = 22m
Area of the Jogging Track = Area of Outer Rectangle – Area of Inner Rectangle
Area of the Jogging Track = ( 24 x 22 ) – ( 20 x 18 )
Area of the Jogging Track = 528 – 360
Area of the Jogging Track = 168m2
Therefore the area of the jogging path around the plot is 168m2.
So every time you solve such questions on area and perimeter, remember that you have to take the width into consideration from both the ends and not just one end. Another variation of such mensuration problems is when the path is made inside and not outside, so when the path is made inside you have to subtract the width from both the ends and not add it.
Problem 2: The outer circumference of a circular track is 220m. Find the cost of leveling the track at the rate of 50p/m2, if the track is 7m wide everywhere.
In the diagram we have a circular plot with a circumference of 220m with a track of 7m all around it. So the shaded area actually represents the track and that is the area which has to be leveled. So, the cost of leveling will be the product of product of the area of the track and the cost of leveling per sq. unit. For this we need to find the area of the track.
To find the area of the circle we should know the radius of the circle, assume the radius of the outer circle to be ‘R’ and the inner circle to be ‘r’
We know the formula of the circumference of a circle-
Also we know that circumference of the outer circle is 220m in this case.
2pR = 220
2 x 22/7 x R = 220
R = 220 x 7 / (22 x 2)
R = 35m
Now that we know R= 35m, we can also find the inner radius ‘r’
r = 35 – 7
r = 28m
Now that we know both the radii we can easily find area of both the circles.
Area of the Track = Area of Outer Circle – Area of Inner Circle
Area of the Track = pR2 – pr2 = p ( R2 – r2 )
Area of the Track = 22/7 (352 – 282)
Area of the Track = (22/7) x 72 (52 – 42) (Taking 72 as common)
Area of the Track = 22 x 7 x 9 m2
The cost of leveling the track, will the product of the area of the track and the cost of leveling per sq. unit.
Cost of Leveling the Track = Area of the Track x Cost of Leveling per sq. unit
Cost of Leveling the Track = 22 x 7 x 9 x 0.50
Cost of Leveling the Track = 22 x 7 x 9 x ½ (50p = ½ of a rupee)
Cost of Leveling the Track = 693
Therefore cost of leveling the path is Rs 693
Problems like these on the area are perimeter often need you find out a parameter from a given quantity and then use that parameter to make calculations.
Mensuration Problems on Area and Perimeter
Do try these mensuration problems and leave your answers in the comments below.
Question 1: Length and breadth of a rectangle are in the ratio 5: 3. If its perimeter is 840 m, what is the area of the rectangle?
1) 2400 sqm 2) 2000 sqm 3) 24,000 sqm 4) 16,000 sqm 5) None of these
Question 2: Perimeter of an equilateral triangle is 54 cm. What is the length of its side?
1) 18 cm 2) 27 cm 3) 13.5 cm 4) 9 cm 5) None of these
Question 2: The floor of a rectangular auditorium of length 40 feet and breadth 28 feet is to be covered with a wall to wall carpet. Find the total cost, if the carpet costs ` 72 per sq ft and the pasting charge is ` 6 per sq ft?
1) ` 65640 2) ` 85920 3) ` 87360 4) ` 56440 5) None of these
Question 3: What will be the cost of gardening 1 m broad boundary around a rectangular plot having a perimeter of 340 m at the rate of ` 10 per sqm?
1) ` 1700 2) ` 3400 3) ` 3440 4) ` 3000 5) None of these
Question 4: A rectangular area having length and breadth equals to 12 m and 8 m respectively is to be bounded by 50 cm broad garden from outside. What is the total area of the garden? 1) 25 m 2) 21 m 3) 10 m 4) 15 m 5) None of these
Question 5: The outer circumference of a circular track is 220 m. Find the cost of leveling the track at the rate of 50 paise/sqm, if the track is 7m wide everywhere?
1) ` 564 2) ` 693 3) ` 612 4) ` 564 5) None of these
For more practice questions download our e-books and do read the next post where we discuss how to solve mensuration problems on 3D shapes.