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 $\mathrm{AB}=\mathrm{c},\mathrm{BC}=\mathrm{a}$ and $\mathrm{AC}=\mathrm{b}$ then, the perimeter of triangle $ABC=a+b+c$.

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 $\ell$ be the length and b be the breadth of rectangle ABCD . Since $\mathrm{EF}=\mathrm{b}$, we can also say that b is the height of the triangle FDC.

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

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 $\Delta$ is $\frac{b}{4}\sqrt{4{\mathrm{a}}^2-{\mathrm{b}}^2}$.

Area of an equilateral triangle

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

2.0Area of Quadrilateral and Parallelogram

Quadrilateral

A quadrilateral is any figure bounded by four straight lines. Let $ABCD$ be a quadrilateral such that $\mathrm{AC}=\mathrm{d}=$ one of the diagonals length and DE and BF are the perpendiculars dropped on that diagonal AC , then, $\mathrm{DE}={\mathrm{P}}_1=$ first offset length. $\mathrm{BF}={\mathrm{P}}_2=$ second offset length.

Parallelogram

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

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

Area of Rectangle

A rectangle is a quadrilateral whose, (i) Opposite sides are equal and parallel to each other. $\mathrm{AB}=\mathrm{CD}=$ length $=\ell$ $\mathrm{AD}=\mathrm{BC}=$ breadth $=\mathrm{b}$ (ii) Diagonals are equal and bisect each other $\mathrm{AC}=\mathrm{BD}=$ diagonal $=\mathrm{d}$ (iii) $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=\angle \mathrm{D}=90^{\circ }$ (iv) Perimeter $(\mathrm{p})=2(\ell +\mathrm{b})$ (v) ${\mathrm{d}}^2={\mathrm{b}}^2+{\ell }^2$ (vi) Area of rectangle $=\ell \times \mathrm{b}$

Area of Square

A square is a quadrilateral whose, (i) All sides are equal $\mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA}=\mathrm{a}$ (ii) Diagonals are equal $\mathrm{AC}=\mathrm{BD}=\mathrm{d}$ (iii) Diagonals bisect each other at $90^{\circ }$.

3.0Pathway

Let $ABCD$ be a rectangular plot whose. Length $=\ell$, Breadth $=\mathrm{b}$

Case 1: A pathway is made outside the plot. Let the width of pathway $=\mathrm{W}$ Area of pathway $={\mathrm{A}}_0=2\mathrm{W}(\ell +\mathrm{b}+2\mathrm{W})$

Case 2: A pathway is made inside the plot. Let the width of pathway $=\mathrm{W}$ Area of pathway $={\mathrm{A}}_1=2\mathrm{W}(\ell +\mathrm{b}-2\mathrm{W})$

Parallel path

Let $ABCD$ be a rectangular plot whose. Length $=\ell$

Breadth $=\mathrm{b}$ SU and TV are two paths draws parallel to the length and the breadth respectively. Width of each parallel path $=W$

Area of two parallel paths $=\mathrm{W}(\ell +\mathrm{b}-\mathrm{W})$

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.

Circumference and Area

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

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

Circular Pathway

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

Area of circular pathway $=\pi \times W(2r+W)$ (outside)

(ii) There is a pathway inside circle ABC Width of pathway $=\mathrm{W}$ Area of circular pathway $=\pi \times \mathrm{W}(2\mathrm{r}-\mathrm{W})$ (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 $\mathrm{R}-\mathrm{r}=\mathrm{W}=$ width of the ring

Area of ring $=\pi R^2-\pi r^2$ $=\pi \left({\mathrm{R}}^2-{\mathrm{r}}^2\right)$ $=\pi (R+r)(R-r)$

Semicircle

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

5.0Numerical Ability

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

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

Q. Find the area of the equilateral triangle whose one side is $7\mathbf{cm}$.

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

Q. An obtuse-angled triangle $ABC$, where $AD=10\mathrm{cm},BC=5\mathrm{cm}$, and height 4 cm , is given in fig. Find its area.

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

Q. $△ABC$ is isosceles with $AB=AC=7.5\mathrm{cm}$ and $BC=9\mathrm{cm}$. The height $AD$ from $A$ to $BC$ is 6 cm . Find the area of $△ABC$. What will be the height from $C$ to $AB$, i.e., $CE$ ?

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

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

• Explanation: Given area of quadrilateral $=120{\mathrm{m}}^2$ Heights ${\mathrm{h}}_1=5\mathrm{m}$ and ${\mathrm{h}}_2=7\mathrm{cm}$ Let the length of diagonal $=\mathrm{d}$ Area of quadrilateral $=\frac{1}{2}\times d\left(h_1+h_2\right)$ $120=\frac{1}{2}\times \mathrm{d}(5+7)$ $\Rightarrow \mathrm{d}=\frac{120\times 2}{12}$ $\mathrm{d}=10\times 2=20\mathrm{m}$ $\therefore$ The length of diagonal of quadrilateral is 20 m

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

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

Q In the given figure $ABCD$ is a parallelogram and $AB=6\mathrm{cm}$. If the area of $△ADE$ is $3/4$ that of parallelogram $ABCD$, 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 $△\mathrm{ADE}=\frac{3}{4}\times$ area of ABCD $\therefore \frac{1}{2}\times (6+y)\times h=\frac{3}{4}\times 6\times h$ $6+y=2\times \frac{3}{4}\times 6$ $6+y=9$
  
  $\mathrm{y}=3$ 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 ${\mathrm{m}}^2$. (ii) The perimeter of the field and the cost of fencing it at ₹ 40 per m .

• Solution Length of the field $=120\mathrm{m}$ and its breadth $=75\mathrm{m}$ (i) Area of the field $=(120\times 75){\mathrm{m}}^2=9000{\mathrm{m}}^2$. Cost of turfing the field $=₹(9000\times 15)=₹135000$. (ii) Perimeter of the field $=2(\ell +b)$ units $=2(120+75)\mathrm{m}=(2\times 195)\mathrm{m}=390\mathrm{m}$ $\therefore$ 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 $=\ell \times b=11.2\mathrm{m}\times 9\mathrm{m}=100.8{\mathrm{m}}^2$ Area of each tile $=24\times 24{\mathrm{cm}}^2$ $=576{\mathrm{cm}}^2$ $=0.0576{\mathrm{m}}^{2}Numberoftilesrequired$$=\frac{100.8}{0.0576}=1750$$So,1750tilesarerequired.Costoftilingthefloor$=$Rs.106pertileTotalcost$=$Rs.$106 \times 1750=$ Rs. 185500

Q. The diagonal of square is $18\sqrt{2}\mathrm{cm}$. Find its area and perimeter.

• Explanation Given the diagonal of square is $18\sqrt{2}\mathrm{cm}$ $\therefore$ The diagonal of square $=\sqrt{2}\times$ side $$\begin{gathered} 18\sqrt{2}=\sqrt{2}\times \text{ side } \\ \text{ Side }=18\mathrm{cm} \end{gathered}$$ Therefore, area of square $=$ side $\times$ side $=18\times 18=324{\mathrm{cm}}^2$ And perimeter of square $=4\times$ side $=4\times 18=72\mathrm{cm}$

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

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

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 $\therefore$ perimeter of square field $=\frac{\text{ Total Amount }}{\text{ Cost of }1\mathrm{m}}=\frac{320}{5}=64\mathrm{m}$ So, the side of square field $=\frac{\text{ Perimeter }}{4}=\frac{64}{4}=16\mathrm{m}$ Then, the area of square field $=$ side $\times$ side $=16\times 16=256{\mathrm{m}}^2$

Q. A rectangular grass plot $80\mathrm{m}\times 60\mathrm{m}$ 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 ${\mathbf{m}}^2$.

• Explanation Let $ABCD$ be a rectangular grass plot. Whose length $=\ell =80\mathrm{m}$ breadth $=\mathrm{b}=60\mathrm{m}$ Two roads of width $\mathrm{W}=10\mathrm{m}$ (shaded part) are crossing each other at the middle of plot. Area of roads $=W(\ell +b-W)$ $=10(80+60-10){\mathrm{m}}^2$ $=1300{\mathrm{m}}^2$ Cost of gravelling the roads $=$ rate of gravelling/ ${\mathrm{m}}^2\times$ 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 $112\mathbf{m}$ 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 ${\mathbf{m}}^2$.

• Solution Let ABCD be the given grassy plot and let EFGH be the inside boundary of the path. Then, length $AB=112\mathrm{m}$ and breadth $\mathrm{BC}=78\mathrm{m}$. Area of the plot $=(112\times 78){\mathrm{m}}^2=8736{\mathrm{m}}^2$. Width of the path $=2.5\mathrm{m}$ $\therefore EF=(112-2\times 2.5)\mathrm{m}=(112-5)\mathrm{m}=107\mathrm{m}$ And $\mathrm{FG}=(78-2\times 2.5)\mathrm{m}=(78-5)\mathrm{m}=73\mathrm{m}$ Area of the rect. EFGH $=(107\times 73){\mathrm{m}}^2=7811{\mathrm{m}}^2$. Area of the gravel path $=($ area ABCD $)-(\mathrm{area}\mathrm{EFGH})=(8736-7811){\mathrm{m}}^2=925{\mathrm{m}}^2$. 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 ${\mathrm{m}}^2$.

• Solution Let $ABCD$ 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 $=45\mathrm{m}$ and breadth of park $=30\mathrm{m}$ Area of the park $\mathrm{ABCD}=(45\times 30){\mathrm{m}}^2=1350{\mathrm{m}}^2$ Width of the path $=2.5\mathrm{m}$ External length $\mathrm{EF}=(45+2\times 2.5)\mathrm{m}=50\mathrm{m}$ External breadth FG $=(30+2\times 2.5)\mathrm{m}=35\mathrm{m}$ Area of rect. EFGH $=(50\times 35){\mathrm{m}}^2=1750{\mathrm{m}}^2$ Area of the path $=($ area of rect. EFGH $)-($ area of rect. ABCD $)=(1750-1350)=400{\mathrm{m}}^2$. 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 $=26.4\mathrm{cm}$. $\therefore C=26.4\mathrm{cm}$. Let the radius of the given circle be rcm . Then, $C=2\pi r$ $r=\frac{C}{2\pi }$ $\mathrm{r}=\left(\frac{26.4}{2}\times \frac{7}{22}\right)\mathrm{cm}=4.2\mathrm{cm}$ $\therefore$ Diameter of the circle $=2\mathrm{r}=(2\times 4.2)\mathrm{cm}=8.4\mathrm{cm}$

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, $2\pi \mathrm{r}=264\mathrm{m}$ and $2\pi \mathrm{R}=308\mathrm{m}$
  
  $\Rightarrow 2\times \frac{22}{7}\times r=264$ and $2\times \frac{22}{7}\times R=308$ $\Rightarrow \mathrm{r}=\left(264\times \frac{7}{44}\right)=42\mathrm{m}$ and $\mathrm{R}=\left(308\times \frac{7}{44}\right)=49\mathrm{m}$ $\Rightarrow (\mathrm{R}-\mathrm{r})=(49-42)\mathrm{m}=7\mathrm{m}$ 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 $=63\mathrm{cm}$ $\Rightarrow$ radius of the wheel $=\frac{63}{2}\mathrm{cm}$ $\Rightarrow$ circumference of the wheel $=2\pi r$ $=\left(2\times \frac{22}{7}\times \frac{63}{2}\right)\mathrm{cm}=198\mathrm{cm}=1.98\mathrm{m}$ Distance covered by the wheel in 1 revolution $=1.98\mathrm{m}$ Distance covered by the wheel in 1000 revolutions $=(1.98\times 1000)\mathrm{m}=1980\mathrm{m}$ Hence, the distance covered by the car $=1980\mathrm{m}$