Aptitude::Area
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Example 1 / 16
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.
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)
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.
Example 2 / 16
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.
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)
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.
Example 3 / 16
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.
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.
Example 4 / 16
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.
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
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.
Example 5 / 16
Let the diagonal be d meters.
So, \(\frac{1}{2}\)* \({d}^{2}\)= 3200
\({d}^{2}\)= 3200*2= 6400
D = 80 m. Ans.
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.
Example 6 / 16
Area of the largest triangle= \(\frac{1}{2}\) * 2R *R sq. cm.
= \({r}^{2}\) Ans.
= \({r}^{2}\) 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.
Example 7 / 16
Area of the rhombus = \(\frac{1}{2}\) * d1* d2.
= \(\frac{1}{2}\) * 82* 90
= 3690 sq. m. Ans.
= \(\frac{1}{2}\) * 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.
Example 8 / 16
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.
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.
Example 9 / 16
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.
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.
Example 10 / 16
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.
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)
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
if 1 breadth side is uncovered= length of fencing= 2l+b
Example 11 / 16
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.
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.
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.
Example 12 / 16
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.
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.
Example 13 / 16
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.
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
1 m/s= \(\frac{18}{5}\) kmph
Example 14 / 16
Each side of square= 2r
Perimeter= 4*side= 4*2r= 8r Ans.
Perimeter= 4*side= 4*2r= 8r Ans.
The perimeter of the square= 4*side
NA
if square is circumscribed then, side of square= 2r.
Example 15 / 16
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.
= \(\frac{90}{360}\)*\(\frac{22}{7}\)*\({3.2}^{2}\)
= \(\frac{11*10.24}{2}\)
= 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.
Example 16 / 16
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 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 \({x}^{2}\).