Areas and mensuration is a topic depends entirely on application of formulas. If you remember the formula, solving problems in this area is a cakewalk. For easy learning and to remember the formulas we provided them into a simple table. Before you move on to have a look at the solved examples, study the tables and try to understand the relevant formula

Two dimensional figures have only any two of length, breadth, and height. They have only areas but not volumes. Perimeter is a uni-dimensional measure. If you observe carefully, the power of the terms in the formulas of perimeter is 1, and of the terms in the areas is 2.

Square:

Area: $a^2$

Perimeter: $4a$

Rectanlge:

Area: $lb$

Perimeter: $2(l+b)$

Triangle:

Area: $\dfrac{1}{2}bh$

Area: $\sqrt {s(s - a)(s - b)(s - c)} $

Here, $s = \dfrac{1}{2}\left( {a + b + c} \right)$

Equilateral Triangle:

Area: $\dfrac{{\sqrt 3 }}{4}{a^2}$

Perimeter: $3a $

Height = $\dfrac{{\sqrt 3 }}{2}a$

Rightangle Triangle:

Area: $\dfrac{1}{2}ab$

Perimeter: $a + b + c $

Also, = ${a^2} + {b^2} = {c^2}$

Isosceles Triangle:

Area: $\dfrac{b}{4}\sqrt {{a^2} - \dfrac{{{b^2}}}{4}} $

Height: $\sqrt {{a^2} - \dfrac{{{b^2}}}{4}} $

Perimeter = $2a + b$

Circle:

Area: $\pi {r^2}$

Circumference: $2\pi r$

Parallelogram:

Area: $ah$

Circumference: $2(a+b)$

Rhombus:

Area: $\dfrac{1}{2}{d_1}{d_2}$

Circumference: $4a$

Ring:

Area: $\pi \left( {{R^2} - {r^2}} \right)$

Sector:

Area: $\dfrac{\theta }{{360}} \times \pi {r^2}$

Circumference: $\dfrac{\theta }{{360}} \times 2\pi r$

Semi-circle:

Area: $\dfrac{{\pi {r^2}}}{2}$

Circumference: $\pi r + 2r$

Semi-circle:

Area: $\dfrac{1}{2}h\left( {a + b} \right)$

1. If sides of a triangle are 8 cm, 15 cm and 17 cm respectively. Find its area.

Area of the triangle when all the three sides are given = $\sqrt {s(s - a)(s - b)(s - c)} $ where $s = \displaystyle\frac{{a + b + c}}{2}$

s = $\displaystyle\frac{{8 + 15 + 17}}{2}$ = 20 cm

Therefore, Area = $\sqrt {20 \times \left( {20 - 8} \right) \times \left( {20 - 15} \right) \times \left( {20 - 17} \right)} $

= $\sqrt {20 \times 12 \times 5 \times 3} $

= $\sqrt {4 \times 5 \times 3 \times 4 \times 5 \times 3} $ = 4 x 5 x 3 = 60 ${\rm{cm}}^{\rm{2}} $

Trick:

The triangle is right angle triangle as $17^2 = 15^2 + 8^2 $

Therefore, Area of right angle triangle = $\displaystyle\frac{1}{2}$ x 8 x 15 = 60 ${\rm{cm}}^{\rm{2}} $

2. Two parallel sides of a trapezium are 4 cm and 5 cm respectively. The perpendicular distance between the parallel sides is 6 cm. Find the area of the trapezium.

Area of trapezium when height and two parallel sides are given = $\displaystyle\frac{1}{2} \times h \times (a + b)$ = $\displaystyle\frac{1}{2}$ x 6 x (4 + 5) = 27${\rm{cm}}^{\rm{2}} $

3. If perimeter and area of a square are equal. Side of the square (in cm) is:

Given, $\left( {{\rm{Side}}} \right)^{\rm{2}} $ = 4 x (Side)

Therefore, Side = 4 cm

4. The diameter of the wheel of a vehicle is 5 metre. It makes 7 revolutions per 9 seconds. What is speed of the vehicle in km/h?

Radius of the wheel = $\displaystyle\frac{5}{2}$ metre

Distance covered in 1 revolution = Circumference of the wheel = ${\rm{2\pi r}}$ = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2}$ metre

Therefore, Distance covered in one second = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2} \times \displaystyle\frac{7}{9}$ metre

Therefore, Speed per hour = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2} \times \displaystyle\frac{7}{9} \times \displaystyle\frac{{18}}{5}$ = 44 km/h

5. The perimeter of a rhombus is 60 cm and one of its diagonal is 24 cm. Find the other diagonal of the rhombus.

Let ABCD is the rhombus whose diagonals BD and AC intersecting at point 'O'.

Side of rhombus = $\displaystyle\frac{1}{4}$ x Perimeter = $\displaystyle\frac{1}{4}$ x 60 = 15 cm.

Let BD = 24 cm

Then, BO = $\displaystyle\frac{1}{2}$BD = 12 cm

Now, AO = $\sqrt {{\rm{AB}}^{\rm{2}} {\rm{ - BO}}^{\rm{2}} } {\rm{ = }}\sqrt {{\rm{15}}^{\rm{2}} {\rm{ - 12}}^{\rm{2}} } $ = 9

Therefore, AC = 2 AO = 2 x 9 = 18 cm

6. A field is 40 metre long and 35 metre wide. The field is surrounded by a path of uniform width of 2.5 metre runs round it on the outside. Find the area of the path.

Remember the formula for the Area of path

= 2 x Width x [Length + Breadth + (2 x Width)]

= 2 x 25 x (40 + 35 + 2 x 2.5)

= 5 x (75 + 5) = 400 ${\rm{m}}^{\rm{2}} $

7. Find area of uniform path of width 2 metre running from centre of each side of the opposite side of a rectangle field measuring 17 metre by 12 metre.

Remember the formula for the Area of path

= Width of path x (Length of field + Breadth of field) - $\left( {{\rm\text{Width of path}}} \right)^{\rm{2}} $

= 2 x (17 + 12) - $\left( 2 \right)^2 $

= 58 - 4 = 54 ${\rm{m}}^{\rm{2}} $

8. A square and a rectangle each has perimeter 60 metre. Difference between the areas of the two figures is 16 square metre. Find length of the rectangle.

Side of the square = $\displaystyle\frac{{60}}{4}$ = 15 metre

Difference in areas of square and rectangle = 16${\rm{m}}^{\rm{2}} $

Therefore, Increase and decrease in dimensions = $\sqrt {16} $ = 4 metre

Therefore, Sides of rectangle are (15 + 4) and (15 - 4) metre

Therefore, Length of rectangle = 19 metre

9. Find the radius of a circle whose area is equal to the sum of areas of three circles with radii 8 cm, 9 cm, 12 cm respectively.

Let radius of new circle is 'R' cm.

Then ${\rm{\pi R}}^{\rm{2}} $ = $\pi 8^2 + \pi 9^2 + \pi 12^2 = 64\pi + 81\pi + 144\pi = 289\pi $

Therefore, ${\rm{R}}^{\rm{2}} $ = 289 => R = 17 cm.

10. The are of the ring between two concentric circles, whose circumferences are 88 cm and 132 cm is:

Area of Ring = $\displaystyle\frac{1}{{4\pi }} \times \left( {132^2 - 88^2 } \right) = \displaystyle\frac{1}{{4\pi }}$ x (132 + 88) x (132 - 88)

= $\displaystyle\frac{1}{4} \times \displaystyle\frac{7}{{22}}$ x 220 x 44 = 770 ${\rm{cm}}^{\rm{2}} $

11. Two poles whose heights are 11 metre and 5 metre stand upright in a field. If the distance between their feet is 8 metre, what is the distance between their tops?

Distance between feet of big pole and that of small pole = 8 metre

Difference between heights of two poles = 11 - 5 = 6 metre

Distance between the tops of poles = $\sqrt {6^2 + 8^2 } $ = 10 metre

12. Find the are of the largest circle that can be inscribed in a square of 14 cm, a side.

Side of the square = Diameter of the circle = 14 cm

Therefore, Radius of the circle = 7 cm.

Area of the circle = $\pi r^2 = \displaystyle\frac{{22}}{7}$ x 7 x 7 = 154 ${\rm{cm}}^{\rm{2}} $.

13. A horse is tied to one of the corners of a square field whose side is 20 metre. If length of the rope is 14 metre, find the area of ungrazed field.

Area of square field = $\left( {{\rm{20 m}}} \right)^{\rm{2}} $ = 400 ${\rm{m}}^{\rm{2}} $

Area of the field grazed by the horse = $\displaystyle\frac{1}{4}\pi r^2 $

= $\displaystyle\frac{1}{4} \times \displaystyle\frac{{22}}{7}$ x 14 x 14 = 154 ${\rm{m}}^{\rm{2}} $

Therefore, Field ungrazed = (400 - 154) = 246 ${\rm{m}}^{\rm{2}} $

14. A horses are tied one to each corner of a square with 14 m a side, and length of the rope is 7 m. Find the area of ungrazed field.

Ungrazed area = Area of square - Area of circle

= $14^2 - \displaystyle\frac{{22}}{7}$ x 7 x 7 = 196 - 154 = 42 ${\rm{m}}^{\rm{2}} $

Short-Cut Method:

Radius of circle = 7 m.

Therefore, Ungraged field = $\displaystyle\frac{6}{7} \times 7^2 $ = 42 ${\rm{m}}^{\rm{2}} $

1. A rectangular carpet has an area of 60 sq.m. Its diagonal and longer side together equal 5 times the shorter side. The length of the carpet is :

a. 5 m

b. 12 m

c. 13 m

d. 14.5 m

Correct Option: B

Explanation:

Let the length = p metres and breadth = q metres

Then pq = 60

$\sqrt {{p^2} + {q^2}} + p = 5q$

$ \Rightarrow {p^2} + {q^2} = {(5q - p)^2}$ =$25{q^2} + {p^2} - 10pq$

As pq = 60,

$ \Rightarrow 25{q^2} - 10 \times 60 = 0$

$ \Rightarrow {q^2} = 25$ or q = 5

p = 60/5 = 12m

Length of the carpet = 12m

2. A rectangular carpet has an area of 120 sq.m and perimeter of 46 m. The length of its diagonal is :

a. 15 m

b. 16 m

c. 17 m

d. 20 m

Correct Option: C

Explanation:

Let length = a metres and breadth = b metres

Then, 2(a+b)=46 or (a+b) = 23

and ab = 120

Diagonal = $\sqrt {{a^2} + {b^2}} $ = $\sqrt {{{(a + b)}^2} - ab} = \sqrt {{{(23)}^2} - 2 \times 120} $ = $\sqrt {289} = 17m$

3. A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Then, its area is :

a. 480 sq.m

b. 320 sq.m

c. 600$\sqrt {15} $ sq.m

d. 450 $\sqrt {15} $ sq.m

Correct Option: C

Explanation:

AB = 60m, BC=40m or AC = 80m

s=$\displaystyle\frac{1}{2}(60 + 40 + 80)m = 90m$

(s-a) = 30m, (s-b) = 50m and (s-c)=10m

Area of $\Delta ABC$ = $\sqrt {s(s - a)(s - b)(s - c)} $

=$\sqrt {90 \times 30 \times 50 \times 10} {\rm{ }}{{\rm{m}}^2} = 300\sqrt {15} {\rm{ }}{{\rm{m}}^2}$

Area of //gm ABCD = 600$\sqrt {15} {\rm{ }}{{\rm{m}}^2}$

4. If a square and a parallelogram stand on the same base, then the ratio of the area of the square and the parallelogram is :

a. greater than 1

b. equal to 1

c. equal to $\displaystyle\frac{1}{2}$

d. equal to $\displaystyle\frac{1}{4}$

Correct Option: B

Explanation:

Let ABCD be the square and ABEF be the parallelogram.

Then, in right triangles ADF and BCE, we have AD = BC (sides of a square ) and AF = BE (sides of parallelogram). Therefore, DF = CE

$[D{F^2} = A{F^2} - A{D^2} = B{E^2} - B{C^2} = C{E^2}]$

Thus, $\Delta ADF$ = $\Delta BCE$

$\Rightarrow \Delta ADF + \Box{\rm{ }}ABCF$ = $\triangle BCE + \Box ABCF$

$ \Rightarrow $ Area of parallelogram ABEF = Area of ABCD

5. In a rhombus, whose area is 144 sq.cim, one of its diagonals is twice as long as the other. The length of its diagonals are :

a. 24 cm, 48 cm

b. 12 cm, 24 cm

c. 6$\sqrt 2 $ cm, 12$\sqrt 2 $cm

d. 6 cm, 12 cm

Correct Option: B

Explanation:

$\displaystyle\frac{1}{2} \times x \times 2x = 144 \Rightarrow {x^2} = 144$ or x = 12

Length of diagonals = 12cm, 24cm

6. The length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m.is:

a. 56 m

b. 64 m

c. 88 m

d. 168 m

Correct Option: A

Explanation:

Grazing area is equal to the area of a circle with radius r.

$\displaystyle\frac{{22}}{7} \times {r^2} = 9856$

Then ${r^2} = \left( {9856 \times \displaystyle\frac{7}{{22}}} \right)$

r = 56 m

7. The circumferences of two concentric circles are 176 m and 132 m respectively. What is the difference between their radii?

a. 5 metres

b. 7 metres

c. 8 metres

d. 44 metres

Correct Option: B

Explanation:

$2\pi R - 2\pi r = (176 - 132)$

$ \Rightarrow 2\pi (R - r) = 44$

$ \Rightarrow (R - r) = \displaystyle\frac{{44 \times 7}}{{2 \times 22}} = 7m$

8. The diameter of a circle is 105 cm less than the circumference. What is the diameter of the circle ?

a. 44 cm

b. 46 cm

c. 48 cm

d. 49 cm

Correct Option: D

Explanation:

$\pi d - d = 105 \Rightarrow (\pi - 1)d = 105$

$ \Rightarrow \left( {\displaystyle\frac{{22}}{7} - 1} \right)$d = 105

d = $\left( {\displaystyle\frac{7}{{15}} \times 105} \right)$cm=49cm

9. A circle and a square have same area. The ratio of the side of the square and the radious of the circle is :

a. $\sqrt \pi $:1

b. 1 : $\sqrt r $

c. 1 : r

d. r : 1

Correct Option: B

Explanation:

${x^2} = \pi {r^2} \Rightarrow \displaystyle\frac{x}{r} = \sqrt \pi = \sqrt \pi :1$

10. The number of rounds that a wheel of diameter $\displaystyle\frac{7}{{11}}$m will make in going 4 km, is :

a. 1000

b. 1500

c. 1700

d. 2000

Correct Option: D

Explanation:

Number of rounds = $\displaystyle\frac{{4 \times 1000}}{{\displaystyle\frac{{22}}{7} \times \displaystyle\frac{7}{{11}}}} = 2000$

11. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6:5. The smaller side of the rectangle is :

a. 30 cm

b. 60 cm

c. 72 cm

d. 132 cm

Correct Option: B

Explanation:

Circumference = $\left( {2 \times \displaystyle\frac{{22}}{7} \times 42} \right)$ cm = 264 cm

Let the rectangle sides are 6x, 5x. Then circumference is

$2 \times (6x + 5x) = 264$ or x = 12

Smaller side of recatngle = 5x = 60 cm

12. If the diameter of a circle is increased by 100% , its area is increased by :

a. 100%

b. 200%

c. 300%

d. 400%

Correct Option: C

Explanation:

Original area = $\pi \times {\left( {\displaystyle\frac{d}{2}} \right)^2} = \displaystyle\frac{{\pi {d^2}}}{4}$

New area = $\pi \times {\left( {\displaystyle\frac{{2d}}{2}} \right)^2} = \pi {d^2}$

Increase in area = $\left( {\pi {d^2} - \displaystyle\frac{{\pi {d^2}}}{4}} \right) = \displaystyle\frac{{3\pi {d^2}}}{4}$

Increase percent = $\left( {\displaystyle\frac{{3\pi {d^2}}}{4} \times \displaystyle\frac{4}{{\pi {d^2}}} \times 100} \right)\% $ = 300%

You can use $\left( {A + B + \displaystyle\frac{{AB}}{{100}}} \right)\% $ formula. Substitute A = B = 100

13.If the radius of a circle is reduced by 50%, its area is reduced by :

a. 25%

b. 50%

c. 75%

d. 100%

Correct Option: C

Explanation:

Original area = $\pi \times {r^2}$

New are = $\pi \times {\left( {\displaystyle\frac{r}{2}} \right)^2} = \displaystyle\frac{{\pi {r^2}}}{4}$

Reduction in area = $\left( {\pi {r^2} - \displaystyle\frac{{\pi {r^2}}}{4}} \right) = \displaystyle\frac{{3\pi {r^2}}}{4}$

Reduction percent = $\left( {\displaystyle\frac{{3\pi {r^2}}}{4} \times \displaystyle\frac{1}{{\pi {r^2}}} \times 100} \right)\% $=75%

You can use $\left( {A + B + \displaystyle\frac{{AB}}{{100}}} \right)\% $ formula. Substitute A = B = - 50.

14. If the circumference of a circle is 352 metres, then its area in ${m^2}$ is :

a. 9856

b. 8956

c. 6589

d. 5986

Correct Option: A

Explanation:

$2 \times \displaystyle\frac{{22}}{7} \times r = 352 \Rightarrow r = \left( {352 \times \displaystyle\frac{7}{{22}} \times \displaystyle\frac{1}{2}} \right) = 56m$

Area = $\left( {\displaystyle\frac{{22}}{7} \times 56 \times 56} \right){m^2} = 9856{\rm{ }}{{\rm{m}}^2}$

15. Area of square with side x is equal to the area of a triangle with base x . The altitude of the triangle is :

a. $\displaystyle\frac{x}{2}$

b. x

c. 2x

d. 4x

Correct Option: C

Explanation:

${x^2} = \displaystyle\frac{1}{2} \times x \times h$ or h = $\displaystyle\frac{{2{x^2}}}{x} = 2x$

16. The perimeter of an isosceles triangle is equal to 14 cm; the lateral side is to the base in the ratio 5:4. The area of the triangle is :

a. $\displaystyle\frac{1}{2}\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

b. $\displaystyle\frac{3}{2}\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

c. $\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

d. $2\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

Correct Option: D

Explanation:

Let lateral side = (5x)cm and base = (4x) cm

5x+5x+4x=14 or x = 1

So, the sides are 5 cm, 5 cm and 4 cm

Semi perimeter, S = $\displaystyle\frac{1}{2}$(5+5+4)cm = 7cm.

(s-a)=2 cm, (s-b)=2cm and (s-c)=3 cm

Area = $\sqrt {7 \times 2 \times 2 \times 3} {\rm{ c}}{{\rm{m}}^2} = 2\sqrt {21{\rm{ }}} c{m^2}$

17. If the diagonal of a square is doubled, how does the area of the square change ?

a. Becomes four fold

b. Becomes three fold

c. Becomes two fold

d. None of these

Correct Option: A

Explanation:

$\displaystyle\frac{{\displaystyle\frac{1}{2} \times {d^2}}}{{\displaystyle\frac{1}{2} \times {{(2d)}^2}}} = \displaystyle\frac{1}{4}$

New area becomes 4 fold.

18. If the base of a rectangle is increased by 10% and the area is unchanged, then the corresponding altitude must be decreased by :

a. $9\displaystyle\frac{1}{{11}}\% $

b. 10%

c. 11%

d. $11\displaystyle\frac{1}{9}\% $

Correct Option: A

Explanation:

Let base = b and altitude = h.

then area = (bh)

New base = $\left( {\displaystyle\frac{{110}}{{100}}b} \right) = \left( {\displaystyle\frac{{11}}{{10}}b} \right)$

Let new altitude = H

Then, ${\displaystyle\frac{{11}}{{10}}b \times H = bh}$ or H = $\left( {\displaystyle\frac{{10}}{{11}}h} \right)$

Decrease = $\left( {h - \displaystyle\frac{{10}}{{11}}h} \right) = \displaystyle\frac{1}{{11}}h$

Decrease percent = $\left( {\displaystyle\frac{1}{{11}}h \times \displaystyle\frac{1}{h} \times 100} \right)\% $ = $9\displaystyle\frac{1}{{11}}\% $

Assume Length is 10 units and Altitude is 11 Units. Now Area = 110

New length is 11 units and New altitude is h units. Now Area = 11h

But given 110 = 11h $ \Rightarrow $ h = 10

So change in altitude = 1/11 x 100 = $9\displaystyle\frac{1}{{11}}\% $

19. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle is increased by 75 cm$^2$. Therefore, the length of the rectangle is :

a. 20 cm

b. 30 cm

c. 40 cm

d. 50 cm

Correct Option: C

Explanation:

Let breadth = x cm and length = (2x) cm

Then. (2x-5)(x+5) - $x \times 2x = 75$

$2{x^2} + 5x - 25 - 2{x^2} = 75$ or 5x = 100

or x = 20

Length = (2x)cm = 40 cm.

20. A rectangle has 15 cm as its length and 150 cm$^2$ as its area. Its area is increased to $1\displaystyle\frac{1}{3}$ times the original area by increasing only its length of its new perimeter is :

a. 50 cm

b. 60 cm

c. 70 cm

d. 80 cm

Correct Option: B

Explanation:

Breadth of the rectangle = $\left( {\displaystyle\frac{{150}}{{15}}} \right)$ cm=10 cm

New area = $\left( {\displaystyle\frac{4}{3} \times 150} \right)c{m^2} = 200{\rm{ c}}{{\rm{m}}^2}$

New length = $\left( {\displaystyle\frac{{200}}{{10}}} \right)cm = 20{\rm{ cm}}$

New perimeter = 2(20+10)cm = 60 cm

21. The length of a rectangular room is 4 metres. If it can be partitioned into two equal square rooms, what is the length of each partition in metres?

a. 1

b. 2

c. 4

d. Data inadequate.

Correct Option: B

Explanation:

Let the side of each new room = y metres.

Then, ${{\rm{y}}^2} = 2x$

Clearly, 2x is a complete square when x = 2

${{\rm{y}}^2} = 4$ or y = 2 m

22. The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by :

a. 42%

b. 62%

c. 82%

d. None of these

Correct Option: C

Explanation:

Let the side of the square = 100 m

New length = 140 m. new breadth = 130 m

Increase in area =$[(140 \times 130) - (100 \times 100)]{\rm{ }}{m^2}$ = 8200 ${m^2}$

Increase percent = $\left( {\displaystyle\frac{{8200}}{{100 \times 100}} \times 100} \right)\% $ = 82%

23. A hall 20m long and 15m broad is surrounded by a verandah of uniform width of 2.5m. The cost of flooring the verandah at the rate of Rs.3.50 per sq.metre is :

a. Rs.500

b. Rs.600

c. Rs.700

d. Rs.800

Correct Option: C

Explanation:

Area of verandah = $[(25 \times 20) - (20 \times 15)]{\rm{ }}{m^2}$=200 ${m^2}$

Cost of flooring = Rs.$(200 \times 3.50)$ = Rs.700

24. If the ratio of the areas of two squares is 9:1, the ratio of their perimeters is :

a. 9:1

b. 3:1

c. 3:4

d. 1:3

Correct Option: B

Explanation:

Let the areas of squares be : $(9{x^2}){\rm{ }}{{\rm{m}}^2}{\rm{ and (}}{{\rm{x}}^2}){\rm{ }}{{\rm{m}}^2}$

Then, their sides are $\sqrt {9{x^2}} ,\sqrt {{x^2}} $ or (3x) metres & x metres respectively.

25. The length of a rectangle is increased by 60% . By what percent would the width have to be decreased to maintain the same area ?

a. $37\displaystyle\frac{1}{2}$

b. 60%

c. 75%

d. 120%

Correct Option: A

Explanation:

Initially, let length = x and breadth = y

Let, new breadth = z. Then new length = $\left( {\displaystyle\frac{{160}}{{100}}x} \right) = \displaystyle\frac{8}{5}x$

$\displaystyle\frac{8}{5}x \times z$ =xy or z = $\displaystyle\frac{{5y}}{8}$

Decrease in breadth = $\left( {y - \displaystyle\frac{{5y}}{8}} \right) = \displaystyle\frac{3}{8}y$

Decrease percent = $\left( {\displaystyle\frac{3}{8}y \times \displaystyle\frac{1}{y} \times 100} \right)\% $ = $37\displaystyle\frac{1}{2}\% $

**2 Dimensional Figures:**Two dimensional figures have only any two of length, breadth, and height. They have only areas but not volumes. Perimeter is a uni-dimensional measure. If you observe carefully, the power of the terms in the formulas of perimeter is 1, and of the terms in the areas is 2.

Square:

Rectanlge:

Triangle:

Equilateral Triangle:

Rightangle Triangle:

Isosceles Triangle:

Circle:

Parallelogram:

Rhombus:

Ring:

Sector:

Semi-circle:

Semi-circle:

**Solved Examples**

1. If sides of a triangle are 8 cm, 15 cm and 17 cm respectively. Find its area.

Area of the triangle when all the three sides are given = $\sqrt {s(s - a)(s - b)(s - c)} $ where $s = \displaystyle\frac{{a + b + c}}{2}$

s = $\displaystyle\frac{{8 + 15 + 17}}{2}$ = 20 cm

Therefore, Area = $\sqrt {20 \times \left( {20 - 8} \right) \times \left( {20 - 15} \right) \times \left( {20 - 17} \right)} $

= $\sqrt {20 \times 12 \times 5 \times 3} $

= $\sqrt {4 \times 5 \times 3 \times 4 \times 5 \times 3} $ = 4 x 5 x 3 = 60 ${\rm{cm}}^{\rm{2}} $

Trick:

The triangle is right angle triangle as $17^2 = 15^2 + 8^2 $

Therefore, Area of right angle triangle = $\displaystyle\frac{1}{2}$ x 8 x 15 = 60 ${\rm{cm}}^{\rm{2}} $

2. Two parallel sides of a trapezium are 4 cm and 5 cm respectively. The perpendicular distance between the parallel sides is 6 cm. Find the area of the trapezium.

Area of trapezium when height and two parallel sides are given = $\displaystyle\frac{1}{2} \times h \times (a + b)$ = $\displaystyle\frac{1}{2}$ x 6 x (4 + 5) = 27${\rm{cm}}^{\rm{2}} $

3. If perimeter and area of a square are equal. Side of the square (in cm) is:

Given, $\left( {{\rm{Side}}} \right)^{\rm{2}} $ = 4 x (Side)

Therefore, Side = 4 cm

4. The diameter of the wheel of a vehicle is 5 metre. It makes 7 revolutions per 9 seconds. What is speed of the vehicle in km/h?

Radius of the wheel = $\displaystyle\frac{5}{2}$ metre

Distance covered in 1 revolution = Circumference of the wheel = ${\rm{2\pi r}}$ = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2}$ metre

Therefore, Distance covered in one second = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2} \times \displaystyle\frac{7}{9}$ metre

Therefore, Speed per hour = 2 x $\displaystyle\frac{{22}}{7} \times \displaystyle\frac{5}{2} \times \displaystyle\frac{7}{9} \times \displaystyle\frac{{18}}{5}$ = 44 km/h

5. The perimeter of a rhombus is 60 cm and one of its diagonal is 24 cm. Find the other diagonal of the rhombus.

Let ABCD is the rhombus whose diagonals BD and AC intersecting at point 'O'.

Side of rhombus = $\displaystyle\frac{1}{4}$ x Perimeter = $\displaystyle\frac{1}{4}$ x 60 = 15 cm.

Let BD = 24 cm

Then, BO = $\displaystyle\frac{1}{2}$BD = 12 cm

Now, AO = $\sqrt {{\rm{AB}}^{\rm{2}} {\rm{ - BO}}^{\rm{2}} } {\rm{ = }}\sqrt {{\rm{15}}^{\rm{2}} {\rm{ - 12}}^{\rm{2}} } $ = 9

Therefore, AC = 2 AO = 2 x 9 = 18 cm

6. A field is 40 metre long and 35 metre wide. The field is surrounded by a path of uniform width of 2.5 metre runs round it on the outside. Find the area of the path.

Remember the formula for the Area of path

= 2 x Width x [Length + Breadth + (2 x Width)]

= 2 x 25 x (40 + 35 + 2 x 2.5)

= 5 x (75 + 5) = 400 ${\rm{m}}^{\rm{2}} $

7. Find area of uniform path of width 2 metre running from centre of each side of the opposite side of a rectangle field measuring 17 metre by 12 metre.

Remember the formula for the Area of path

= Width of path x (Length of field + Breadth of field) - $\left( {{\rm\text{Width of path}}} \right)^{\rm{2}} $

= 2 x (17 + 12) - $\left( 2 \right)^2 $

= 58 - 4 = 54 ${\rm{m}}^{\rm{2}} $

8. A square and a rectangle each has perimeter 60 metre. Difference between the areas of the two figures is 16 square metre. Find length of the rectangle.

Side of the square = $\displaystyle\frac{{60}}{4}$ = 15 metre

Difference in areas of square and rectangle = 16${\rm{m}}^{\rm{2}} $

Therefore, Increase and decrease in dimensions = $\sqrt {16} $ = 4 metre

Therefore, Sides of rectangle are (15 + 4) and (15 - 4) metre

Therefore, Length of rectangle = 19 metre

9. Find the radius of a circle whose area is equal to the sum of areas of three circles with radii 8 cm, 9 cm, 12 cm respectively.

Let radius of new circle is 'R' cm.

Then ${\rm{\pi R}}^{\rm{2}} $ = $\pi 8^2 + \pi 9^2 + \pi 12^2 = 64\pi + 81\pi + 144\pi = 289\pi $

Therefore, ${\rm{R}}^{\rm{2}} $ = 289 => R = 17 cm.

10. The are of the ring between two concentric circles, whose circumferences are 88 cm and 132 cm is:

Area of Ring = $\displaystyle\frac{1}{{4\pi }} \times \left( {132^2 - 88^2 } \right) = \displaystyle\frac{1}{{4\pi }}$ x (132 + 88) x (132 - 88)

= $\displaystyle\frac{1}{4} \times \displaystyle\frac{7}{{22}}$ x 220 x 44 = 770 ${\rm{cm}}^{\rm{2}} $

11. Two poles whose heights are 11 metre and 5 metre stand upright in a field. If the distance between their feet is 8 metre, what is the distance between their tops?

Distance between feet of big pole and that of small pole = 8 metre

Difference between heights of two poles = 11 - 5 = 6 metre

Distance between the tops of poles = $\sqrt {6^2 + 8^2 } $ = 10 metre

12. Find the are of the largest circle that can be inscribed in a square of 14 cm, a side.

Side of the square = Diameter of the circle = 14 cm

Therefore, Radius of the circle = 7 cm.

Area of the circle = $\pi r^2 = \displaystyle\frac{{22}}{7}$ x 7 x 7 = 154 ${\rm{cm}}^{\rm{2}} $.

13. A horse is tied to one of the corners of a square field whose side is 20 metre. If length of the rope is 14 metre, find the area of ungrazed field.

Area of square field = $\left( {{\rm{20 m}}} \right)^{\rm{2}} $ = 400 ${\rm{m}}^{\rm{2}} $

Area of the field grazed by the horse = $\displaystyle\frac{1}{4}\pi r^2 $

= $\displaystyle\frac{1}{4} \times \displaystyle\frac{{22}}{7}$ x 14 x 14 = 154 ${\rm{m}}^{\rm{2}} $

Therefore, Field ungrazed = (400 - 154) = 246 ${\rm{m}}^{\rm{2}} $

14. A horses are tied one to each corner of a square with 14 m a side, and length of the rope is 7 m. Find the area of ungrazed field.

Ungrazed area = Area of square - Area of circle

= $14^2 - \displaystyle\frac{{22}}{7}$ x 7 x 7 = 196 - 154 = 42 ${\rm{m}}^{\rm{2}} $

Short-Cut Method:

Radius of circle = 7 m.

Therefore, Ungraged field = $\displaystyle\frac{6}{7} \times 7^2 $ = 42 ${\rm{m}}^{\rm{2}} $

**MCQ's**

1. A rectangular carpet has an area of 60 sq.m. Its diagonal and longer side together equal 5 times the shorter side. The length of the carpet is :

a. 5 m

b. 12 m

c. 13 m

d. 14.5 m

Correct Option: B

Explanation:

Let the length = p metres and breadth = q metres

Then pq = 60

$\sqrt {{p^2} + {q^2}} + p = 5q$

$ \Rightarrow {p^2} + {q^2} = {(5q - p)^2}$ =$25{q^2} + {p^2} - 10pq$

As pq = 60,

$ \Rightarrow 25{q^2} - 10 \times 60 = 0$

$ \Rightarrow {q^2} = 25$ or q = 5

p = 60/5 = 12m

Length of the carpet = 12m

2. A rectangular carpet has an area of 120 sq.m and perimeter of 46 m. The length of its diagonal is :

a. 15 m

b. 16 m

c. 17 m

d. 20 m

Correct Option: C

Explanation:

Let length = a metres and breadth = b metres

Then, 2(a+b)=46 or (a+b) = 23

and ab = 120

Diagonal = $\sqrt {{a^2} + {b^2}} $ = $\sqrt {{{(a + b)}^2} - ab} = \sqrt {{{(23)}^2} - 2 \times 120} $ = $\sqrt {289} = 17m$

3. A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m long. Then, its area is :

a. 480 sq.m

b. 320 sq.m

c. 600$\sqrt {15} $ sq.m

d. 450 $\sqrt {15} $ sq.m

Correct Option: C

Explanation:

AB = 60m, BC=40m or AC = 80m

s=$\displaystyle\frac{1}{2}(60 + 40 + 80)m = 90m$

(s-a) = 30m, (s-b) = 50m and (s-c)=10m

Area of $\Delta ABC$ = $\sqrt {s(s - a)(s - b)(s - c)} $

=$\sqrt {90 \times 30 \times 50 \times 10} {\rm{ }}{{\rm{m}}^2} = 300\sqrt {15} {\rm{ }}{{\rm{m}}^2}$

Area of //gm ABCD = 600$\sqrt {15} {\rm{ }}{{\rm{m}}^2}$

4. If a square and a parallelogram stand on the same base, then the ratio of the area of the square and the parallelogram is :

a. greater than 1

b. equal to 1

c. equal to $\displaystyle\frac{1}{2}$

d. equal to $\displaystyle\frac{1}{4}$

Correct Option: B

Explanation:

Let ABCD be the square and ABEF be the parallelogram.

Then, in right triangles ADF and BCE, we have AD = BC (sides of a square ) and AF = BE (sides of parallelogram). Therefore, DF = CE

$[D{F^2} = A{F^2} - A{D^2} = B{E^2} - B{C^2} = C{E^2}]$

Thus, $\Delta ADF$ = $\Delta BCE$

$\Rightarrow \Delta ADF + \Box{\rm{ }}ABCF$ = $\triangle BCE + \Box ABCF$

$ \Rightarrow $ Area of parallelogram ABEF = Area of ABCD

5. In a rhombus, whose area is 144 sq.cim, one of its diagonals is twice as long as the other. The length of its diagonals are :

a. 24 cm, 48 cm

b. 12 cm, 24 cm

c. 6$\sqrt 2 $ cm, 12$\sqrt 2 $cm

d. 6 cm, 12 cm

Correct Option: B

Explanation:

$\displaystyle\frac{1}{2} \times x \times 2x = 144 \Rightarrow {x^2} = 144$ or x = 12

Length of diagonals = 12cm, 24cm

6. The length of a rope by which a cow must be tethered in order that she may be able to graze an area of 9856 sq.m.is:

a. 56 m

b. 64 m

c. 88 m

d. 168 m

Correct Option: A

Explanation:

Grazing area is equal to the area of a circle with radius r.

$\displaystyle\frac{{22}}{7} \times {r^2} = 9856$

Then ${r^2} = \left( {9856 \times \displaystyle\frac{7}{{22}}} \right)$

r = 56 m

7. The circumferences of two concentric circles are 176 m and 132 m respectively. What is the difference between their radii?

a. 5 metres

b. 7 metres

c. 8 metres

d. 44 metres

Correct Option: B

Explanation:

$2\pi R - 2\pi r = (176 - 132)$

$ \Rightarrow 2\pi (R - r) = 44$

$ \Rightarrow (R - r) = \displaystyle\frac{{44 \times 7}}{{2 \times 22}} = 7m$

8. The diameter of a circle is 105 cm less than the circumference. What is the diameter of the circle ?

a. 44 cm

b. 46 cm

c. 48 cm

d. 49 cm

Correct Option: D

Explanation:

$\pi d - d = 105 \Rightarrow (\pi - 1)d = 105$

$ \Rightarrow \left( {\displaystyle\frac{{22}}{7} - 1} \right)$d = 105

d = $\left( {\displaystyle\frac{7}{{15}} \times 105} \right)$cm=49cm

9. A circle and a square have same area. The ratio of the side of the square and the radious of the circle is :

a. $\sqrt \pi $:1

b. 1 : $\sqrt r $

c. 1 : r

d. r : 1

Correct Option: B

Explanation:

${x^2} = \pi {r^2} \Rightarrow \displaystyle\frac{x}{r} = \sqrt \pi = \sqrt \pi :1$

10. The number of rounds that a wheel of diameter $\displaystyle\frac{7}{{11}}$m will make in going 4 km, is :

a. 1000

b. 1500

c. 1700

d. 2000

Correct Option: D

Explanation:

Number of rounds = $\displaystyle\frac{{4 \times 1000}}{{\displaystyle\frac{{22}}{7} \times \displaystyle\frac{7}{{11}}}} = 2000$

11. A circular wire of radius 42 cm is cut and bent in the form of a rectangle whose sides are in the ratio of 6:5. The smaller side of the rectangle is :

a. 30 cm

b. 60 cm

c. 72 cm

d. 132 cm

Correct Option: B

Explanation:

Circumference = $\left( {2 \times \displaystyle\frac{{22}}{7} \times 42} \right)$ cm = 264 cm

Let the rectangle sides are 6x, 5x. Then circumference is

$2 \times (6x + 5x) = 264$ or x = 12

Smaller side of recatngle = 5x = 60 cm

12. If the diameter of a circle is increased by 100% , its area is increased by :

a. 100%

b. 200%

c. 300%

d. 400%

Correct Option: C

Explanation:

Original area = $\pi \times {\left( {\displaystyle\frac{d}{2}} \right)^2} = \displaystyle\frac{{\pi {d^2}}}{4}$

New area = $\pi \times {\left( {\displaystyle\frac{{2d}}{2}} \right)^2} = \pi {d^2}$

Increase in area = $\left( {\pi {d^2} - \displaystyle\frac{{\pi {d^2}}}{4}} \right) = \displaystyle\frac{{3\pi {d^2}}}{4}$

Increase percent = $\left( {\displaystyle\frac{{3\pi {d^2}}}{4} \times \displaystyle\frac{4}{{\pi {d^2}}} \times 100} \right)\% $ = 300%

**Shortcut:**

You can use $\left( {A + B + \displaystyle\frac{{AB}}{{100}}} \right)\% $ formula. Substitute A = B = 100

13.If the radius of a circle is reduced by 50%, its area is reduced by :

a. 25%

b. 50%

c. 75%

d. 100%

Correct Option: C

Explanation:

Original area = $\pi \times {r^2}$

New are = $\pi \times {\left( {\displaystyle\frac{r}{2}} \right)^2} = \displaystyle\frac{{\pi {r^2}}}{4}$

Reduction in area = $\left( {\pi {r^2} - \displaystyle\frac{{\pi {r^2}}}{4}} \right) = \displaystyle\frac{{3\pi {r^2}}}{4}$

Reduction percent = $\left( {\displaystyle\frac{{3\pi {r^2}}}{4} \times \displaystyle\frac{1}{{\pi {r^2}}} \times 100} \right)\% $=75%

**Shortcut:**

You can use $\left( {A + B + \displaystyle\frac{{AB}}{{100}}} \right)\% $ formula. Substitute A = B = - 50.

a. 9856

b. 8956

c. 6589

d. 5986

Correct Option: A

Explanation:

$2 \times \displaystyle\frac{{22}}{7} \times r = 352 \Rightarrow r = \left( {352 \times \displaystyle\frac{7}{{22}} \times \displaystyle\frac{1}{2}} \right) = 56m$

Area = $\left( {\displaystyle\frac{{22}}{7} \times 56 \times 56} \right){m^2} = 9856{\rm{ }}{{\rm{m}}^2}$

15. Area of square with side x is equal to the area of a triangle with base x . The altitude of the triangle is :

a. $\displaystyle\frac{x}{2}$

b. x

c. 2x

d. 4x

Correct Option: C

Explanation:

${x^2} = \displaystyle\frac{1}{2} \times x \times h$ or h = $\displaystyle\frac{{2{x^2}}}{x} = 2x$

16. The perimeter of an isosceles triangle is equal to 14 cm; the lateral side is to the base in the ratio 5:4. The area of the triangle is :

a. $\displaystyle\frac{1}{2}\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

b. $\displaystyle\frac{3}{2}\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

c. $\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

d. $2\sqrt {21} {\rm{ c}}{{\rm{m}}^2}$

Correct Option: D

Explanation:

Let lateral side = (5x)cm and base = (4x) cm

5x+5x+4x=14 or x = 1

So, the sides are 5 cm, 5 cm and 4 cm

Semi perimeter, S = $\displaystyle\frac{1}{2}$(5+5+4)cm = 7cm.

(s-a)=2 cm, (s-b)=2cm and (s-c)=3 cm

Area = $\sqrt {7 \times 2 \times 2 \times 3} {\rm{ c}}{{\rm{m}}^2} = 2\sqrt {21{\rm{ }}} c{m^2}$

17. If the diagonal of a square is doubled, how does the area of the square change ?

a. Becomes four fold

b. Becomes three fold

c. Becomes two fold

d. None of these

Correct Option: A

Explanation:

$\displaystyle\frac{{\displaystyle\frac{1}{2} \times {d^2}}}{{\displaystyle\frac{1}{2} \times {{(2d)}^2}}} = \displaystyle\frac{1}{4}$

New area becomes 4 fold.

18. If the base of a rectangle is increased by 10% and the area is unchanged, then the corresponding altitude must be decreased by :

a. $9\displaystyle\frac{1}{{11}}\% $

b. 10%

c. 11%

d. $11\displaystyle\frac{1}{9}\% $

Correct Option: A

Explanation:

Let base = b and altitude = h.

then area = (bh)

New base = $\left( {\displaystyle\frac{{110}}{{100}}b} \right) = \left( {\displaystyle\frac{{11}}{{10}}b} \right)$

Let new altitude = H

Then, ${\displaystyle\frac{{11}}{{10}}b \times H = bh}$ or H = $\left( {\displaystyle\frac{{10}}{{11}}h} \right)$

Decrease = $\left( {h - \displaystyle\frac{{10}}{{11}}h} \right) = \displaystyle\frac{1}{{11}}h$

Decrease percent = $\left( {\displaystyle\frac{1}{{11}}h \times \displaystyle\frac{1}{h} \times 100} \right)\% $ = $9\displaystyle\frac{1}{{11}}\% $

**Short cut:**

Assume Length is 10 units and Altitude is 11 Units. Now Area = 110

New length is 11 units and New altitude is h units. Now Area = 11h

But given 110 = 11h $ \Rightarrow $ h = 10

So change in altitude = 1/11 x 100 = $9\displaystyle\frac{1}{{11}}\% $

19. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and the breadth is increased by 5 cm, the area of the rectangle is increased by 75 cm$^2$. Therefore, the length of the rectangle is :

a. 20 cm

b. 30 cm

c. 40 cm

d. 50 cm

Correct Option: C

Explanation:

Let breadth = x cm and length = (2x) cm

Then. (2x-5)(x+5) - $x \times 2x = 75$

$2{x^2} + 5x - 25 - 2{x^2} = 75$ or 5x = 100

or x = 20

Length = (2x)cm = 40 cm.

20. A rectangle has 15 cm as its length and 150 cm$^2$ as its area. Its area is increased to $1\displaystyle\frac{1}{3}$ times the original area by increasing only its length of its new perimeter is :

a. 50 cm

b. 60 cm

c. 70 cm

d. 80 cm

Correct Option: B

Explanation:

Breadth of the rectangle = $\left( {\displaystyle\frac{{150}}{{15}}} \right)$ cm=10 cm

New area = $\left( {\displaystyle\frac{4}{3} \times 150} \right)c{m^2} = 200{\rm{ c}}{{\rm{m}}^2}$

New length = $\left( {\displaystyle\frac{{200}}{{10}}} \right)cm = 20{\rm{ cm}}$

New perimeter = 2(20+10)cm = 60 cm

21. The length of a rectangular room is 4 metres. If it can be partitioned into two equal square rooms, what is the length of each partition in metres?

a. 1

b. 2

c. 4

d. Data inadequate.

Correct Option: B

Explanation:

Let the side of each new room = y metres.

Then, ${{\rm{y}}^2} = 2x$

Clearly, 2x is a complete square when x = 2

${{\rm{y}}^2} = 4$ or y = 2 m

22. The length and breadth of a square are increased by 40% and 30% respectively. The area of the resulting rectangle exceeds the area of the square by :

a. 42%

b. 62%

c. 82%

d. None of these

Correct Option: C

Explanation:

Let the side of the square = 100 m

New length = 140 m. new breadth = 130 m

Increase in area =$[(140 \times 130) - (100 \times 100)]{\rm{ }}{m^2}$ = 8200 ${m^2}$

Increase percent = $\left( {\displaystyle\frac{{8200}}{{100 \times 100}} \times 100} \right)\% $ = 82%

23. A hall 20m long and 15m broad is surrounded by a verandah of uniform width of 2.5m. The cost of flooring the verandah at the rate of Rs.3.50 per sq.metre is :

a. Rs.500

b. Rs.600

c. Rs.700

d. Rs.800

Correct Option: C

Explanation:

Area of verandah = $[(25 \times 20) - (20 \times 15)]{\rm{ }}{m^2}$=200 ${m^2}$

Cost of flooring = Rs.$(200 \times 3.50)$ = Rs.700

24. If the ratio of the areas of two squares is 9:1, the ratio of their perimeters is :

a. 9:1

b. 3:1

c. 3:4

d. 1:3

Correct Option: B

Explanation:

Let the areas of squares be : $(9{x^2}){\rm{ }}{{\rm{m}}^2}{\rm{ and (}}{{\rm{x}}^2}){\rm{ }}{{\rm{m}}^2}$

Then, their sides are $\sqrt {9{x^2}} ,\sqrt {{x^2}} $ or (3x) metres & x metres respectively.

25. The length of a rectangle is increased by 60% . By what percent would the width have to be decreased to maintain the same area ?

a. $37\displaystyle\frac{1}{2}$

b. 60%

c. 75%

d. 120%

Correct Option: A

Explanation:

Initially, let length = x and breadth = y

Let, new breadth = z. Then new length = $\left( {\displaystyle\frac{{160}}{{100}}x} \right) = \displaystyle\frac{8}{5}x$

$\displaystyle\frac{8}{5}x \times z$ =xy or z = $\displaystyle\frac{{5y}}{8}$

Decrease in breadth = $\left( {y - \displaystyle\frac{{5y}}{8}} \right) = \displaystyle\frac{3}{8}y$

Decrease percent = $\left( {\displaystyle\frac{3}{8}y \times \displaystyle\frac{1}{y} \times 100} \right)\% $ = $37\displaystyle\frac{1}{2}\% $