# Aptitude::Area

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*(20)*

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

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

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.

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

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.

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.

= \({r}^{2}\)

= \({r}^{2}\)

**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.

= \(\frac{1}{2}\) * 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

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

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.

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.

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.

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.

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

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.

= \(\frac{90}{360}\)*\(\frac{22}{7}\)*\({3.2}^{2}\)

= \(\frac{11*10.24}{2}\)

= 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

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 \({x}^{2}\).