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Aptitude::Area

Home > Quantitative Aptitude > Area > Subjective Solved Examples

Example 1 / 16

The sides of a rectangle are in the ratio of 6:5 and its area is 1331 sq. m. Find the perimeter of the rectangle.
Let 6x and 5x be the sides of the rectangle.
So, 6x*5x= 1331
11 \({x}^{2}\)=1331
\({x}^{2}\)=\(\frac{1331}{11}\)= 121
X=11
Length= 6x= 6*11=66
Breadth= 5x= 5*11=55
Perimeter= 2(l+b)= 2(66+55)= 121m Ans.

Area of rectangle= length*breadth
Perimeter of rectangle= 2*(length+breadth)
NA
If the sides are in the ratio, then take it with respect to x and find x, then find length and breadth by equating to the formula of the area. Once, length and breadth are obtained, we can find the perimeter by using the formula.
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Example 2 / 16

A rectangle measures 15 cm on the breadth and its diagonal measures 17 cm. What is the perimeter of the rectangle?
Length= \(\sqrt{diagonal^2-breadth^2}\)
So, \(\sqrt{17^2-15^2}\)
\(\sqrt{8^2}\)
Length= 8 cm
Perimeter= 2(l+b)= 2(8+15)= 46cm Ans.

diagonal of a rectangle=\(\sqrt{length^2+breadth^2}\)
Perimeter of rectangle= 2*(length+breadth)
NA
The diagonal formula of the rectangle is based on the Pythagoras theorem, where the hypotenuse is diagonal, the base is the breadth, the length is perpendicular.
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Example 3 / 16

A rectangular courtyard 3.78 m long and 5.25 m broad is to be paved exactly with square tiles, all of the same sizes. The minimum number of such tiles is:
length= 378 cm
Breadth= 525 cm
Maximum length of the square tile= HCF(378, 525)= 21 cm
No. of tiles= \(\frac{378*525}{21*21}\)=(18*25)= 450. Ans.

Maximum length of the square tile= HCF(length,breadth)
NA
To find the maximum length, we have to find HCF always.
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Example 4 / 16

The length of a room is 6 m and the width is 5 m. What is the cost of painting the floor at the rate of Rs.900 per sq. metre.
Area= 6*5 sq. m= 30 sq. m
Cost of 1 sq. m= Rs. 900
So, total cost= area*cost per sq. m= 30*900= Rs. 27000 Ans.

area of rectangle= length * breadth
Cost of painting floor= total cost= area*cost per sq. m
NA
to find the cost of painting the floor, calculate the area and multiply by cost of 1 sq. m.
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Example 5 / 16

The area of a square field is 3200 sq. m. The length of its diagonals will be?
Let the diagonal be d meters.
So, \(\frac{1}{2}\)* \({d}^{2}\)= 3200
\({d}^{2}\)= 3200*2= 6400
D = 80 m. Ans.

Area of the square=\(\frac{1}{2}\)* \({d}^{2}\)
NA
The two diagonals of the square are of the same length and each diagonal bisects each other and divide the square into a congruent isosceles right triangle.
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Example 6 / 16

The area of the largest triangle that can be inscribed in a semi-circle of radius R cm is:
Area of the largest triangle= \(\frac{1}{2}\) * 2R *R sq. cm.
= \({r}^{2}\) Ans.

Area of the largest triangle that can be inscribed in a semi-circle of radius R= \(\frac{1}{2}\) * 2R *R sq. cm.
NA
It is based on a direct formula.
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Example 7 / 16

The diagonal of a rhombus is 82 m and 90 m. Its area is:
Area of the rhombus = \(\frac{1}{2}\) * d1* d2.
= \(\frac{1}{2}\) * 82* 90
= 3690 sq. m. Ans.

Area of the rhombus = \(\frac{1}{2}\) * d1* d2
NA
The rhombus has two diagonals and both are perpendicular, which means a rhombus is an orthodiagonal quadrilateral.
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Example 8 / 16

The inner circumference of a circular race track 18 m wide is 880 m. Find the radius of the outer circle.
Let the inner radius be 'r' meters.
Then 2 \(\pi\) r = 880
2* \(\frac{22}{7}\)* r = 880
\(\frac{44}{7}\)= 880
R = 140 m Ans.

Circumference of circle= 2 \(\pi\) r
NA
Directly based on the formula.
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Example 9 / 16

Find the cost of carpeting a room 130 m long and 90 m broad with a carpet 1000 cm broad at the rate of Rs. 20 per meter.
Area of the carpet= Area of the room = 130*90= 11700 sq. m.
Breadth of the carpet = 1000 cm = 10 m
Length of the carpet = \(\frac{area}{breadth}\)
= \(\frac{11700}{10}\)
= 1170m Ans.

Area of the room= length*breadth
NA
Area of the carpet= Area of the room.
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Example 10 / 16

A rectangular field has to be fenced on the three sides leaving asides of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
Area of the field= 680 sq. feet.
L*b= 680 sq. feet.
Length= 20 feet.
20*breadth= 680
Breadth= 34 feet.
Required length of fencing= l+2b(since 1 side is uncovered)
= 20+(2*34)= 88 feet. Ans.

Area of the rectangular field= length*breadth
Length of fencing= perimeter of the rectangle= 2(l+b)

NA
if 1 length side is uncovered= length of fencing= l+2b
if 1 breadth side is uncovered= length of fencing= 2l+b

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Example 11 / 16

The area of the rectangle plot is 460 sq. m. If the length is 15% more than the breadth, what is the breadth of the plot?
Let the breadth be b m.
Lb = 460 sq. m-------(i)
Length= b+15% of b
= \(\frac{115b}{100}\)------(ii)
Putting (ii) in (i), we get,
\(\frac{115b}{100}\)*b= 460
\({b}^{2}\)= 400
B = \(\sqrt{400 }\) = 20m. Ans.

Area of the rectangular field= length*breadth
If x is increased by y%, then z= x+y% of x.
NA
make the relation of length and breadth and put in the given area.
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Example 12 / 16

A large field of 1000 hectares is divided into two parts. The difference of the areas of the two parts is one-tenth of the average of the two areas. What is the area of the smaller part in hectares?
Let the areas of the parts be x hectares and (1000-x) hectares.
Difference of the areas of the two parts= x-(1000-x)= 2x-1000
Given that, area difference between the two parts = one-tenth of the average of the two areas.
So,
2x-1000= \(\frac{1}{10}\)*\(\frac{[x+(1000-x)]}{2}\)
On solving, we get,
X= 525
Area of the smaller part= (1000-525)= 475 hectares. Ans.

average of x and y= \(\frac{x+y}{2}\)
NA
1 hectare= 10000 sq. m.
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Example 13 / 16

A girl walking at the rate of 12 km per hour crosses a square field diagonally in 18 seconds. The area of the field is:
Distance covered is \(\frac{12*1000}{3600}\)*18= 60m.
Diagonal of square field= 60 m
Area of square field= \(\frac{{d}^{2}}{2}\)
= \(\frac{{60}^{2}}{2}\)
= 1800 sq. m. Ans.

Area of the square=\(\frac{{d}^{2}}{2}\)
NA
1 kmph= \(\frac{5}{18}\) m/s
1 m/s= \(\frac{18}{5}\) kmph
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Example 14 / 16

The perimeter of the square circumscribed about a circle of radius r is:
Each side of square= 2r
Perimeter= 4*side= 4*2r= 8r Ans.

The perimeter of the square= 4*side
NA
if square is circumscribed then, side of square= 2r.
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Example 15 / 16

Find the area of the sector whose arc is making an angle of 90® at the center of a circle of radius 3.2 cm.
Area of the sector= \(\frac{90}{360}\)* \(\pi\)* \({r}^{2}\)
= \(\frac{90}{360}\)*\(\frac{22}{7}\)*\({3.2}^{2}\)
= \(\frac{11*10.24}{2}\)
= 56.32 sq. cm. Ans.

\(\frac{Angle Made By An Arc}{360}\)* \(\pi\)* \({r}^{2}\)
NA
Sector is the portion of a disk enclosed by two radii and an arc.
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Example 16 / 16

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and the rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Area of the park = (60 x 40) sq. m = 2400 sq. m

Area of the lawn = 2109 sq. m

Area of the crossroads = (2400 - 2109) sq. m = 291 sq. m

Let the width of the road be x metres. Then,

60x + 40x – \({x}^{2}\) = 291.

\({x}^{2}\) - 100x + 291 = 0.

(x - 97)(x - 3) = 0.

x = 3 Ans.

Area= length*breadth
NA
The road is in the shape of the plus mark. But in the middle, the square has been calculated twice above. So subtract that area \({x}^{2}\).
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