###
**Questions on Area and Perimeter are
asked in SBI PO, IBPS, SSC CGL and any other competitive exam 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 SBI PO, IBPS and SSC CGL Exam.After discussing 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 216cm

^{2}, then what is the length of the rectangle?

**Solution 1:**

We know the
formulas for the area and perimeter of a rectangle-

**Step 1**

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)

**Step 2**

We will
replace the perimeter formula in the equation (i)-

2(l+b)/b =
5/1

**Step 3**

Solve the
equation by cross multiplication-

2(l+b) = 5b

2l + 2b = 5b

2l = 3b

b = 2/3
l (ii)

**Step 4**

Replacing
equation (ii) in the area formula-

Area = l x b

Area = l x
2/3l (iii)

Area =
216 (iv - given)

**Step 5**

We have two area
equations and two variables, so equating (iii) and (iv) we get-

216 = l x
2/3l

l

^{2 }= 216 x 3/2
l

^{2}= 108 x 3
l

^{2}= 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?

**Solution 1:**

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-

**Step 1**

Substituting
the values in the circumference formula-

Circumference
= 2 x 22/7 x 7

Circumference
= 2 x 22 = 44m

**Step 2**

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.

**Solution 2:**

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-

**Step 1**

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

**Step 2**

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

**Step 3**

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?

**Solution 1**

**:**

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.

**Step 1**

**Step 2**

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 m

^{2}

**Step 2**

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

**Step 3**

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 = 168m

^{2}
Therefore
the area of the jogging path around the plot is 168m

^{2}.
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/m

^{2}, if the track is 7m wide everywhere.

**Solution 2**

**:**

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.

**Step 1**

**Step 2**

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

**Step 3**

Now that we
know R= 35m, we can also find the inner radius ‘r’

r = 35 – 7

r = 28m

**Step 4**

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 = pR

^{2 }- pr^{2}= p ( R^{2 }- r^{2})
Area of the
Track = 22/7 (35

^{2}– 28^{2})
Area of the
Track = (22/7) x 7

^{2}(5^{2}– 4^{2}) (Taking 7^{2}as common)
Area of the
Track = 22 x 7 x 9 m

^{2}

**Step 5**

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 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 sq m
2) 2000 sq m 3) 24,000 sq m 4) 16,000 sq m 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 perimeter of 340 m at the rate of ` 10 per sq m?

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/sq m, 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.

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