Perimeter and Area: Definition, Formulas and Solved Examples

Perimeter and Area

Mensuration : The process, art or the act of measuring is called mensuration. Anything that can be measured is said to be mensurable.

A plane figure is called rectilinear figure if it is made up of only line segments. E.g., triangles, square, rectangle and pentagons etc.

Perimeter: We can define perimeter as the total boundary length of a closed figure.

Area : Area is the amount of surface covered by any shape. A planner region has two dimensions length and breadth hence, its side is measured in term of its area.

1.0Triangle

Perimeter of a triangle

Suppose a triangle whose sides $AB = c, BC = a$ and $AC = b$ then, the perimeter of triangle $A BC = a + b + c$ .

Area of a triangle

Let us first draw a rectangle ABCD and mark a point F anywhere on AB . If we join F to D and C , then FDC is a triangle. Draw EF perpendicular to DC. Let $ℓ$ be the length and b be the breadth of rectangle ABCD . Since $EF = b$ , we can also say that b is the height of the triangle FDC.

Now let EC be x units and, therefore, DE will be $ℓ$ - x units. Now, from the formula of area of a right-angled triangle, we have area of $△ FDE = \frac{1}{2} \times b \times (ℓ - x)$ area of $△ FCE = \frac{1}{2} \times b \times x$ So, area of $△ FDC =$ area of $△ FDE +$ area of $△ FEC$ $= \frac{1}{2} \times b \times (ℓ - x) + \frac{1}{2} \times b \times x$

= \frac{1}{2} \times b ℓ - \frac{1}{2} bx + \frac{1}{2} bx = \frac{1}{2} b ℓ

Thus, the area of any triangle is half the product of its base and height.

Area of an isosceles triangle

Suppose $a$ is length of each equal side and $b$ is length of third unequal side of an isosceles triangle then area of an isosceles $Δ$ is $\frac{b}{4} 4 a^{2} - b^{2}$ .

Area of an equilateral triangle

Let $A BC$ be an equilateral triangle with each side equal 'a' units Draw $AD ⊥ BC$ , then $BD = DC = \frac{a}{2}$ In $△ A B D, A D = A B^{2} - B D^{2}$ (Pythagoras theorem)

= a^{2} - (\frac{a}{2})^{2} = a^{2} - \frac{a ^{2}}{4} = \frac{3 a ^{2}}{4} = \frac{3}{2} a

∴

Area of

△ ABC = \frac{1}{2}

base

\times

altitude

= (\frac{1}{2} \times a \times \frac{3}{2} a)

sq units

= \frac{3}{4} a^{2}

sq. units

∴

Area of an equilateral triangle

= \frac{3}{4} a^{2} = \frac{3}{4} \times (side)^{2}

2.0Area of Quadrilateral and Parallelogram

Quadrilateral

A quadrilateral is any figure bounded by four straight lines. Let $A BC D$ be a quadrilateral such that $AC = d =$ one of the diagonals length and DE and BF are the perpendiculars dropped on that diagonal AC , then, $DE = P_{1} =$ first offset length. $BF = P_{2} =$ second offset length.

(i) Area of any quadrilateral

= \frac{1}{2} \times

diagonal

\times (

Sum of two offsets

) = \frac{1}{2} d (P_{1} + P_{2})

(ii) If the diagonal falls outside the figure, then Area of quadrilateral

= \frac{1}{2} \times

diagonal

\times (

difference of offsets

) = \frac{1}{2} d (P_{1} - P_{2}) (

where

P_{1} > P_{2})

Parallelogram

A parallelogram is a quadrilateral whose (i) Opposite sides are equal and parallel to each other

AB = base = b, CD = b and AD = BC = a

(ii) Diagonal $AC = d$ and its offsets $DE = BF = P$ (iii) Height $= h =$ distance between base and top

(iv) Area of parallelogram

=

base

\times

height

= bh

(v) Area of parallelogram

= (

any diagonal

) \times

(its offsets)

= dP

Area of Rectangle

A rectangle is a quadrilateral whose, (i) Opposite sides are equal and parallel to each other. $AB = CD =$ length $= ℓ$ $AD = BC =$ breadth $= b$ (ii) Diagonals are equal and bisect each other $AC = BD =$ diagonal $= d$ (iii) $∠ A = ∠ B = ∠ C = ∠ D = 9 0^{\circ}$ (iv) Perimeter $(p) = 2 (ℓ + b)$ (v) $d^{2} = b^{2} + ℓ^{2}$ (vi) Area of rectangle $= ℓ \times b$

Area of Square

A square is a quadrilateral whose, (i) All sides are equal $AB = BC = CD = DA = a$ (ii) Diagonals are equal $AC = BD = d$ (iii) Diagonals bisect each other at $9 0^{\circ}$ .

a^{2} = \frac{d ^{2}}{4} + \frac{d ^{2}}{4} = \frac{d ^{2}}{2}

(iv) Area of square

= (Side)^{2} = a^{2}

Or

\frac{(diagonal) ^{2}}{2} = \frac{d ^{2}}{2}

Or

\frac{( Perimeter ) ^{2}}{16} = \frac{p ^{2}}{16}

Diagonal of a square

= 2 (

sides

)

3.0Pathway

Let $A BC D$ be a rectangular plot whose. Length $= ℓ$ , Breadth $= b$

Case 1: A pathway is made outside the plot. Let the width of pathway $= W$ Area of pathway $AC = BD = d$

Case 2: A pathway is made inside the plot. Let the width of pathway $= W$ Area of pathway $= A_{1} = 2 W (ℓ + b - 2 W)$

Parallel path

Let $A BC D$ be a rectangular plot whose. Length $= ℓ$

Breadth $= b$ SU and TV are two paths draws parallel to the length and the breadth respectively. Width of each parallel path $= (Side)^{2} = a^{2}$

Area of two parallel paths $= W (ℓ + b - W)$

4.0Circle

A circle is a figure consisting of all points in a plane in such a way that the distance of all these points from a given point (i.e. centre) is same.

E.g., In a given figure

OB = OC^{'} = OA =

constant Line

AC^{'}

is known as the diameter of a circle. The fixed point ( 0 ) is the centre of a circle. The constant distance from the centre to the circumference is known as the radius of the circle.

\frac{( Perimeter ) ^{2}}{16} = \frac{p ^{2}}{16}

radius ( r ) Any line drawn through the centre and terminated both ways by the circumference is called the diameter of a circle (

AC^{'} =

d iam e t er

)

Circumference and Area

Circumference (C) $\to$ the perimeter of the circle. Area $(A) \to$ the space enclosed by the circle. $OB =$ radius $= r$ $A C^{'} =$ diameter $= d$ $d = 2 r$ Length $ABC^{'} A =$ Circumference $= C$ Measurement for area $= A$ Formula : $C = 2 π r = π d, A = π r^{2}$ Correlation formula $A = \frac{C ^{2}}{4 π}$

The region enclosed between two concentric circles is known as a ring.

Circular Pathway

Let ABC be a circle whose radius $= r$ . (i) There is a pathway PLK outside this circle ABC whose width $= W$ .

Area of circular pathway $= b$ (outside)

(ii) There is a pathway inside circle ABC Width of pathway $= W$ Area of circular pathway $= π \times W (2 r - W)$ (inside)

If $R$ and $r$ denote the radii of the outer and inner circles respectively, which bound a plane circular ring it is evident that $R - r = W =$ width of the ring

Area of ring $= π R^{2} - π r^{2}$ $= π (R^{2} - r^{2})$ $= π (R + r) (R - r)$

= π W (R + r)

[Since W = R - r]

Semicircle

Area of semicircle $= \frac{π r ^{2}}{2}$ Perimeter of semicircle $= π r + 2 r$

5.0Numerical Ability

Q. Find the area of a triangle in which base $= 25 cm$ and height $= 14 cm$ .

Explanation: Here, base $= 25 cm$ and height $= 14 cm$ . $∴$ area of the triangle $= (\frac{1}{2} \times$ base $\times$ height $)$

= (\frac{1}{2} \times 25 \times 14) cm^{2} = 175 cm^{2}

Q. Find the area of the equilateral triangle whose one side is $7 cm$ .

Explanation: Area of equilateral $Δ$ $= \frac{3}{4} \times (side)^{2} = \frac{3}{4} \times (7)^{2} = \frac{3}{4} \times 49$ $= \frac{1.732 \times 49}{4} = \frac{84.868}{4} cm^{2}$ $= 21.217 cm^{2} = 21.22 cm^{2}$ (up to 2 decimal places)

Q. An obtuse-angled triangle $A BC$ , where $= (\frac{1}{2} \times$ , and height 4 cm , is given in fig. Find its area.

Solution: In the obtuse-angled triangle $ABC, CD$ represents the height. Hence, $CD = 4 cm$ . We know that, Area $= \frac{1}{2} \times$ base $\times$ height $= \frac{1}{2} \times AB \times CD$ Thus, we need to find the length of base $A B$ of the triangle. Since $C D$ is the height, so $C D ⊥ B D$ . In right-angled triangle BCD , $BC^{2} = BD^{2} + CD^{2}$ (Pythagoras theorem) $(5)^{2} = BD^{2} + (4)^{2}$ $BD^{2} = 25 - 16 = 9$ $∴ B D = 3 cm$ Now, $AB + BD = AD$ $∴ AB = AD - BD = 10 - 3 = 7 cm$ Hence area of $△ ABC = \frac{1}{2} \times AB \times CD = \frac{1}{2} \times 7 \times 4 = 14 cm^{2}$

Q. $△ A BC$ is isosceles with $A B = A C = 7.5 cm$ and $BC = 9 cm$ . The height $A D$ from $A$ to $BC$ is 6 cm . Find the area of $△ A BC$ . What will be the height from $C$ to $A B$ , i.e., $CE$ ?

Solution:
Area of $△ ABC = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times 9 \times 6 = 27 cm^{2}$ Now for $△ A BC$ , if $A B$ is the base and $CE$ is its corresponding height. Area of $△ ABC = \frac{1}{2} \times AB \times CE$ $\Rightarrow 27 cm^{2} = \frac{1}{2} \times 7.5 \times CE (∵$ area of $△ ABC = 27 cm^{2})$ $\Rightarrow CE = \frac{27 \times 2}{7.5} = \frac{54}{7.5} = \frac{540}{75} = 7.2 cm$

Q. Find the length of the diagonal of a quadrilateral whose area is $120 m^{2}$ and perpendicular height are 5 m and 7 m .

Explanation: Given area of quadrilateral $= 120 m^{2}$ Heights $h_{1} = 5 m$ and $h_{2} = 7 cm$ Let the length of diagonal $= d$ Area of quadrilateral $= \frac{1}{2} \times d (h_{1} + h_{2})$ $120 = \frac{1}{2} \times d (5 + 7)$ $\Rightarrow d = \frac{120 \times 2}{12}$ $d = 10 \times 2 = 20 m$ $∴$ The length of diagonal of quadrilateral is 20 m

Q. In the given figure, ABCD is a parallelogram, $AB = 8 cm, BC = 6.4 cm$ and $DE = 6 cm$ . Find (i) the area of the parallelogram, (ii) the length of $BF$ .

Explanation (i) Area of the parallelogram $=$ base $\times$ height $= AB \times DE$ $= 8 \times 6 = 48 cm^{2}$ (ii) If we take AD as the base, then BF is the corresponding height. Then $AD \times BF = 48$ $6.4 \times BF = 48$ $BF = \frac{48}{6.4} = 7.5 cm$ The length of BF is 7.5 cm .

Q In the given figure $A BC D$ is a parallelogram and $A B = 6 cm$ . If the area of $△ A D E$ is $3/4$ that of parallelogram $A BC D$ , find the length of $BE$ .

Solution Let the length of BE be y cm . Let the height from D to the base AB be hcm . Area of $△ ADE = \frac{3}{4} \times$ area of ABCD $∴ \frac{1}{2} \times (6 + y) \times h = \frac{3}{4} \times 6 \times h$ $6 + y = 2 \times \frac{3}{4} \times 6$ $DE = 6 cm$
$y = 3$ The length of BE is 3 cm .

Q. The length and breadth of a rectangular field are 120 m and 75 m respectively. Find (i) The area of the field and the cost of turfing it at $₹15$ per $m^{2}$ . (ii) The perimeter of the field and the cost of fencing it at ₹ 40 per m .

Solution Length of the field $= 120 m$ and its breadth $= 75 m$ (i) Area of the field $= (120 \times 75) m^{2} = 9000 m^{2}$ . Cost of turfing the field $= ₹ (9000 \times 15) = ₹135000$ . (ii) Perimeter of the field $= 2 (ℓ + b)$ units $= 2 (120 + 75) m = (2 \times 195) m = 390 m$ $∴$ the cost of fencing $= ₹ (390 \times 40) = ₹15600$

Q. The dimensions of a room are 11.2 m by 9 m . The floor of the room is to be covered by marble tiles, each measuring 24 cm by 24 cm . Find the total number of tiles required. What is the cost of tiling the floor at Rs. 106 per tile?

Solution Dimensions of a room are 11.2 m by 9 m Area of room $= ℓ \times b = 11.2 m \times 9 m = 100.8 m^{2}$ Area of each tile $= 24 \times 24 cm^{2}$ $= 576 cm^{2}$ $= 0.0576 m^{2} N u mb ero f t i l esre q u i re d$ $= \frac{100.8}{0.0576} = 1750$ $S o, 1750 t i l es a rere q u i re d . C os t o f t i l in g t h e f l oor$ = $R s .106 p er t i l e T o t a l cos t$ = $R s .$ $106 \times 1750 = R s .185500$

Q. The diagonal of square is $182 cm$ . Find its area and perimeter.

Explanation Given the diagonal of square is $182 cm$ $∴$ The diagonal of square $= 2 \times$ side $182 = 2 \times side Side = 18 cm$ Therefore, area of square $=$ side $\times$ side $= 18 \times 18 = 324 cm^{2}$ And perimeter of square $= 4 \times$ side $= 4 \times 18 = 72 cm$

Q. The cost of cementing a square courtyard and at 6.75 per $m^{2}$ is $₹ 6075$ . Find the cost of fencing it at the rate of ₹ 4 per meter.

Solution Given cost of cementing at $₹6.75$ per $m^{2}$ is $₹6075$ $∴$ Area of square courtyard $= \frac{Total Amount}{Cost of 1 m ^{2}}$ $= \frac{6075}{6.75} = 900 m^{2}$ Therefore, side of courtyard $R s .$ Then, the perimeter of courtyard $= 4 \times$ side $= 4 \times 30 = 120 m$ $= 2 \times$ The cost of fencing $=$ perimeter $\times$ rate of $1 m = 120 \times 4 = ₹2480$

Q. If it costs ₹ $320$ to fence a square field at the rate of ₹ $5$ per m . find the length of the side and the area of the field?

Solution Given Total cost is ₹ 320 to fence a square field at the rate of ₹ 5 per m $∴$ perimeter of square field $= \frac{Total Amount}{Cost of 1 m} = \frac{320}{5} = 64 m$ So, the side of square field $= \frac{Perimeter}{4} = \frac{64}{4} = 16 m$ Then, the area of square field $=$ side $\times$ side $= 16 \times 16 = 256 m^{2}$

Q. A rectangular grass plot $80 m \times 60 m$ has two roads, each 10 m wide, running in the middle of it, one parallel to length and the other parallel to breadth. Find the cost of gravelling the roads at ₹ 2 per $m^{2}$ .

Explanation Let $A BC D$ be a rectangular grass plot. Whose length $= ℓ = 80 m$ breadth $= b = 60 m$ Two roads of width $W = 10 m$ (shaded part) are crossing each other at the middle of plot. Area of roads $= W (ℓ + b - W)$ $= 10 (80 + 60 - 10) m^{2}$ $= 1300 m^{2}$ Cost of gravelling the roads $=$ rate of gravelling/ $m^{2} \times$ area of roads $=$ Rs. $2 \times 1300 =$ Rs. 2600 Hence, the cost of gravelling the roads is Rs. 2600.

Q. A rectangular grassy plot is $112 m$ long and 78 m broad. It has a 2.5 m wide gravel path all around it on the inside. Find the area of the path and the cost of constructing it at $₹120$ per $m^{2}$ .

Solution Let ABCD be the given grassy plot and let EFGH be the inside boundary of the path. Then, length $A B = 112 m$ and breadth $BC = 78 m$ . Area of the plot $= (112 \times 78) m^{2} = 8736 m^{2}$ . Width of the path $= 2.5 m$ $= 16 \times 16 = 256 m^{2}$ And $FG = (78 - 2 \times 2.5) m = (78 - 5) m = 73 m$ Area of the rect. EFGH $= (107 \times 73) m^{2} = 7811 m^{2}$ . Area of the gravel path $= ($ area ABCD $) - (area EFGH) = (8736 - 7811) m^{2} = 925 m^{2}$ . Cost of constructing the path $= ₹ (925 \times 120) = ₹111000$ .

Q. A rectangular park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path and the cost of constructing it at $₹125$ per $m^{2}$ .

Solution Let $A BC D$ be the given park surrounded externally by a 2.5 m wide path. Let EFGH be the external boundary of the path. Length of park $= 45 m$ and breadth of park $= 30 m$ Area of the park $ABCD = (45 \times 30) m^{2} = 1350 m^{2}$ Width of the path $= 2.5 m$ External length $EF = (45 + 2 \times 2.5) m = 50 m$ External breadth FG $= (30 + 2 \times 2.5) m = 35 m$ Area of rect. EFGH $= (50 \times 35) m^{2} = 1750 m^{2}$ Area of the path $= ($ area of rect. EFGH $) - ($ area of rect. ABCD $) = (1750 - 1350) = 400 m^{2}$ . Cost of constructing the path $= ₹ (400 \times 125) = ₹50000$ .

Q. Find the diameter of a circle whose circumference is 26.4 cm .

Explanation: Circumference of the given circle $= 26.4 cm$ . $∴ C = 26.4 cm$ . Let the radius of the given circle be rcm . Then, $C = 2 π r$ $r = \frac{C}{2 π}$ $r = (\frac{26.4}{2} \times \frac{7}{22}) cm = 4.2 cm$ $∴$ Diameter of the circle $= 2 r = (2 \times 4.2) cm = 8.4 cm$

Q. A racetrack is in the form of a ring whose inner circumference is 264 m and the outer circumference is 308 m . Find the width of the track.

Explanation Let the inner and outer radii of the track be $r$ metres and $R$ metres respectively. Then, $2 π r = 264 m$ and $2 π R = 308 m$
$\Rightarrow 2 \times \frac{22}{7} \times r = 264$ and $2 \times \frac{22}{7} \times R = 308$ $\Rightarrow r = (264 \times \frac{7}{44}) = 42 m$ and $R = (308 \times \frac{7}{44}) = 49 m$ $\Rightarrow (R - r) = (49 - 42) m = 7 m$ Hence, the width of the track is 7 m .

Q. The diameter of a wheel of a car is 63 cm . Find the distance travelled by the car during the period in which the wheel makes 1000 revolutions.

Solution: Diameter of the wheel $= 63 cm$ $\Rightarrow$ radius of the wheel $= \frac{63}{2} cm$ $\Rightarrow$ circumference of the wheel $= 2 π r$ $= (2 \times \frac{22}{7} \times \frac{63}{2}) cm = 198 cm = 1.98 m$ Distance covered by the wheel in 1 revolution $= 1.98 m$ Distance covered by the wheel in 1000 revolutions $= (1.98 \times 1000) m = 1980 m$ Hence, the distance covered by the car $= 1980 m$

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