Question 13 Marks
Find the area ofa recta[gular plot, one side of whtch is $48m$ and lts diagonal is so $m.$
AnswerLength of rectangular plot $(l) = 48m$ and its diagonal $= 50m$
$\therefore$ Second side $(b) =\sqrt{50^2-48^2}$
$=\sqrt{2500-2304}$
$=\sqrt{196}=14\text{m}$
$\therefore\text{Area l}\times\text{b}$
$=48\times14=672\text{m}^2$ View full question & answer→Question 23 Marks
Find the area of a right triangle whose base is $1.2m$ and hypotenuse $3.7m.$
AnswerIn right angled $\triangle\text{ABC},$ Base $BC = 1.2m$
and hypotenuse $AC =3.7 m$
But $A C^2=A B^2+B C^2$ (Pythagoras Theorem)
$\Rightarrow(3.7)^2=A B^2+(1.2)^2$
$\Rightarrow 13.69=A B^2+1.44$
$\Rightarrow A B^2=13.69-1.44$
$\Rightarrow A B^2=12.25=(3.5)^2$
$\Rightarrow A B=3.5 m$
Now, area of $\triangle ABC =\frac{1}{2} \times$ Base $\times$ Altitude
$=\frac{1}{2} \times 1.2 \times 3.5 m^2=2.1 m^2$ View full question & answer→Question 33 Marks
A diagonal of a quadrilateral is $26\ cm$ and the perpendiculars drawn to it from the opposite vertices are $12.8\ cm$ and $11.2\ cm$. Find the area of the quadrilateral.
AnswerIn quadrilateral $ABCD$, diagonal $AC = 26cm$
and perpendiculars $DL = 12.8\ cm, BM = 11.2\ cm$
Area of quadrilateral $ABCD =\frac{1}{2}(\text{Sum of perpendicular)}\times\text{diagonal})$
$=\frac{1}{2}(12.8+11.2)\times26\text{cm}^2$
$=\frac{1}{2}\times24\times26=312\text{cm}^2$ View full question & answer→Question 43 Marks
A room is $9m$ by $8m$ by $6.5m$. It has one door of dimensions $(2m \times 1.5m)$ and four windows each of dimensions $(1.5m \times 1m)$. Find the cost of painting the walls at $Rs 50$ per $m^2$.
Answer$\text { Length }=9 \mathrm{~m} ; \text { Breadth }=8 \mathrm{~m}$
$\text { Height }=6.5 \mathrm{~m}$
Area of the four walls $=\{2(1+b) \times h\}$ sq. units
$=\{2(9+8) \times 6.5\} \mathrm{m}^2=\{34 \times 6.5) \mathrm{m}^2=221 \mathrm{~m}^2$
Area of one door $=(2 \times 1.5) \mathrm{m}^2=3 \mathrm{~m}^2$
Area of one window $=(1.5 \times 1) \mathrm{m}^2=1.5 \mathrm{~m}^2$
Area of four windows $=(4 \times 1.5) \mathrm{m}^2=6 \mathrm{~m}^2$
Total area of one door and four windows $=(3+6) \mathrm{m}^2=9 \mathrm{~m}^2$
Area to be painted $=(221-9) \mathrm{m}^2=212 \mathrm{~m}^2$
Rate of painting $=$ Rs 50 per $\mathrm{m}^2$
Total cost of painting $=\operatorname{Rs}(212 \times 50)=\operatorname{Rs} 10,600$
View full question & answer→Question 53 Marks
A $115$-m-long and $64-m$-broad lawn has two roads at right angles, one $2m$ wide, running parallel to its length, and the other $2.5m$ wide, running parallel to its breadth. Find the cost of gravelling the roads at $Rs 60$ per $m^2$.
AnswerLength of lawn $(\mathrm{l})=115 \mathrm{~m}$
and breadth $(b)=64 \mathrm{~m}$
Width of road parallel to length $=2 \mathrm{~m}$
and width of road parallel to breadth $=2.5 \mathrm{~m}$
$\text { Area of roads }=(115 \times 2+64 \times 2.5-2 \times 2.5) \mathrm{m}^2$
$=(230+160-5) \mathrm{m}^2=(390-5) \mathrm{m}^2=385 \mathrm{~m}^2$
$\text { Cost of gravelling = Rs. } 60 \mathrm{~m}^2$
$\text { Total cost }=\text { Rs. } 60 \times 385=\text { Rs. } 23100$ View full question & answer→Question 63 Marks
The area of a rhombus is equal to the area of a triangle whose base and the corresponding height are $24.8\ cm$ and $16.5\ cm$ respectively. If one of the diagonals of the rhombus is $22\ cm$, find the lenght of the other diagonal.
AnswerArea of a triangle $=\frac{1}2{}\times\text{Base}\times\text{Height}$
$=\Big(\frac{1}{2}\times24.8\times16.5\Big)\text{cm}^2=204.6\text{cm}^2$
Given: Area of the rhombus = Area of the triangle Area of the rhombus = $204.6\text{cm}^2$
Area of the rhombus $=\frac{1}2{}\times\text{(Product of the diagonals)}$
Given: Length of one diagonal $= 22\ cm$
$\therefore$ Length of the other diagonal $=\Big(\frac{204.6\times4}{22}\Big)\text{cm}$
$=18.6\text{cm}$
View full question & answer→Question 73 Marks
The area of a parallelogram is $3385m^2$. If its altitude is twice the corresponding base, find the base and the altitude.
AnswerLet the base of the parallelogram be $x \mathrm{~m}$.
The, the altitude of the parallelogram will be $2 \times \mathrm{m}$.
It is given that the area of the parallelogram is $338 \mathrm{~m}^2$.
Area of a parallelogram $=$ Base $\times$ Altitude
$\Rightarrow 338=x \times 2 x$
$\Rightarrow 338=2 x^2$
$\Rightarrow x^2=169 \mathrm{~m}^2$
$\Rightarrow x=13 \mathrm{~m}$
Base $=x m=13 m$
Altitude $=2 \times \mathrm{m}=(2 \times 13) \mathrm{m}=26 \mathrm{~m}$
View full question & answer→Question 83 Marks
In the following figures, find the area of the shaded region. 
Answer Side of square = 20cm
Area of square $=\text{a}^2=(20)^2=400\text{cm}^2$
Area of right $\triangle\text{LPM}=\frac{1}{2}\times10\times10\text{cm}^2=50\text{cm}^2$
Area of right $\triangle\text{RMQ}=\frac{1}{2}\times10\times20=100\text{cm}^2$
and area of right $=\triangle\text{RSL}=\frac{1}{2}\times20\times10=100\text{cm}^2$
Area of shaded region $=400-(50+100+100)\text{cm}^2$
$=400-250=150\text{cm}^2$
View full question & answer→Question 93 Marks
The adjacent sides of a parallelogram are $15\ cm$ and $8\ cm$. If the distance between the longer sides is $4\ cm$, find the distance between the shorter sides.
Answer$ABCD$ is a parallelogram with side $AB$ of length $15\ cm$ and the corresponding altitude $AE$ of length $4\ cm$.
The adjacent side $AD$ is of length $8\ cm$ and the corresponding altitude is $CF.$
Area of a parallelogram = Base $\times $ Height
We have two altitudes and two corresponding bases.
$\therefore AD \times CF = AB \times AE$
$ \Rightarrow 8cm \times CF = 15cm \times 4cm$
$\Rightarrow\text{CF}=\Big(\frac{15\times4}{8}\big)\text{cm}=7.5\text{cm}$
Hence, the distance between the shorter sides is $7.5cm.$ View full question & answer→Question 103 Marks
Calculate the area of the shaded region in each of the figures given below:
AnswerOuter length $=43 \mathrm{~m}$
and breadth $=27 \mathrm{~m}$
Area $=43 \times 27=1161 \mathrm{~m}^2$
Inner length $=43-2 \times 1.5=43-3=40 \mathrm{~m}$
and breadth $=27-2 \times 1=27-2=25 \mathrm{~m}$
Inner area $=40 \times 25=1000 \mathrm{~m}^2$
Area of shaded portion $=1161-1000=161 \mathrm{~m}^2$
View full question & answer→Question 113 Marks
A square lawn has a $2-m$-wide path surrounding it. If the area of the path is $136m^2$, find the area of the lawn.
AnswerLet $A B C D$ be the square lawn
and PQRS be the outer boundary of the square path
Let one side of the lawn $(A B)$ be $\times m$
Area of the square lawn $=\mathrm{x}^2$
Length $P Q=(x m+2 m+2 m)=(x+4) m$
Area of PQRS $=(x+4)^2=\left(x^2+8 x+16\right) m^2$
Now, Area of the path $=$ Area of PQRS - Area of the square lawn
$\Rightarrow 136=x^2+8 x+16 x-x^2$
$\Rightarrow 136=8 x+16$
$\Rightarrow 136-16=8 x$
$\Rightarrow 120=8 x$
$\Rightarrow x=15$
Side of the laws $=15 \mathrm{~m}$
Area of the lawn $=(\text { Side })^2=(15 m)^2=225 m^2$
View full question & answer→Question 123 Marks
In the given figure, $ABCD$ is a rectangle with length $= 36m$ and breadth $= 24m$. In $\triangle\text{ADE},\text{EF}\perp\text{AD}$ and $EF = 15\ m$. Calculate the area of the shaded region.
Answer$A B C D$ is a rectangle in which $A B=36 m$
and $B C=24 m$
In $\triangle AED$,
$EF =15 m$
$A D=B C=24 m$
Now area of rectangle $A B C D=l \times b$
$=36 \times 24 cm^2=864 cm^2$
Area of $\triangle AED =\frac{1}{2} \times AD \times EF$
$=\frac{1}{2} \times 24 \times 15 cm^2=180 cm^2$
Area of shaded portion $=864-180=684 m^2$
View full question & answer→Question 133 Marks
In the given figure, all steps are $0.5m$ high. Find the area of the shaded region.
AnswerDividing the figure an shown
Area of rectangle I$=3.5 \times 0.5 \mathrm{~m}^2=1.75 \mathrm{~m}^2$
Area of rectangle II $=(3.5-2 \times 0.5) \times 0.5=(3.5-1) \times 0.5=2.5 \times 0.5=1.25 \mathrm{~m}^2$
Area of rectangle III $=(2.5-1) \times 0.5=1.5 \times 0.5=0.75 \mathrm{~m}^2$
Area of rectangle IV $=(1.5-1.0) \times 0.5 \times 0.5=0.25 \mathrm{~m}^2$
Total area of shaded portion $=(1.75+1.25+0.75+0.25) \mathrm{m}^2=4 \mathrm{~m}^2$ View full question & answer→Question 143 Marks
The legs of a right triangle are in the ratio $3 : 4$ and its area is $1014cm^2$. Find the lengths of its legs.
AnswerLegs of a right angled triangle $=3: 4$
Let one leg (base) $=3 x$
Then second leg (altitude) $=4 x$
$\text { Area }=\frac{1}{2} \times \text { Base } \times \text { Altitude }$
$=\frac{1}{2} \times 3 \mathrm{x} \times 4 \mathrm{x}=6 \mathrm{x}^2$
$6 \mathrm{x}^2=1014$
$=\mathrm{x}^2=\frac{1014}{6}=169=(13)^2$
$\mathrm{x}=13$
one leg $^{\prime}($ Base $)=3 \mathrm{x}=3 \times 13=39 \mathrm{~cm}$
and second leg (altitude) $=4 \mathrm{x}=4 \times 13=52 \mathrm{~cm}$ View full question & answer→Question 153 Marks
Find the area of the triangle in which $a = 91m, b = 98m, c = 105m.$
Answer$a = 91m, b = 98m, c = 105m$
$\therefore\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{91+98+105}{2}=\frac{294}{2}=147$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{147(147-91)(147-98)(147-105)}$
$=\sqrt{147\times56\times49\times42}$
$=\sqrt{7\times7\times3\times7\times2\times2\times2\times7\times7\times7\times2\times3}$
$=2\times2\times3\times7\times7\times7\times=4116\text{m}^2$
View full question & answer→Question 163 Marks
A wire is looped in the form of a circle of radius $35\ cm$. If it is rebent in the form of a square, what will be the length of each side of the square?
AnswerIt is given that the radius of the circle is $35\ cm.$
Length of the wire = Circumference of the circle
$\Rightarrow $ Circumference of the circle $=2\pi\text{r}=\Big(2\times\frac{22}{7}\times35\Big)\text{cm}=220\text{cm}$
Let the wire be bent into the form of a square of side a cm.
Perimeter of the square $= 220cm$
$\Rightarrow 4a = 220$
$\Rightarrow\text{a}=\Big(\frac{220}{4}\Big)\text{cm}=55\text{cm}$
Hence, each side of the square will be 55cm.
View full question & answer→Question 173 Marks
The diameter of the wheel of a cycle is $70\ cm.$ How far will it go in $250$ revolutions?
AnswerIt may be noted that in one revolution, the cycle covers a distance equal to the circumference of the wheel.
Diameter of the wheel $= 70cm$
$\therefore$ Circumference of the wheel $=\pi\text{d}=\Big(\frac{22}{7}\times70\Big)\text{cm}=220\text{cm}$
Thus, the cycle covers $220cm$ in one revolution.
$\therefore$ Distance covered by the cycle in $250$ revolutions $= (220 \times 250)cm = 55000cm = 550m$ [since $1m = 100cm]$
Hence, the cycle will cover $550m$ in $250$ revolutions.
View full question & answer→Question 183 Marks
The circumference of a circle exceeds its diameter by $30\ cm$. Find the radius of the circle.
Answer(Circumference) - (Diameter) $= 30\ cm$
$\therefore(2\pi\text{r}-2\text{r})=30$
$\Rightarrow2\text{r}(\pi-1)=30$
$\Rightarrow2\text{r}\Big(\frac{22}{7}-1\Big)=30$
$\Rightarrow2\text{r}\times\frac{15}{7}=30$
$\Rightarrow\text{r}=\Big(30\times\frac{7}{30}\Big)=7$
$\therefore$ Radius of the given circle $= 7\ cm$
View full question & answer→Question 193 Marks
A verandah is $40m$ long and $15m$ broad. It is to be paved with stones, each measuring $6$ dm by $5$ dm. Find the number of stones required.
AnswerLength of verandah $(l) = 40m$ Breadth $(b) = 15m$
Area $= l \times b = 40 \times 15 = 600m^2$
Length of one stone $=6\text{dm}=\frac{6}{10}\text{m}$
and breadth $=5\text{dm}=\frac{5}{10}\text{m}$
Area of one stone $=\frac{6}{10}\times\frac{5}{10}$
$=\frac{30}{100}=\frac{3}{10}\text{m}^2$
$\therefore$ Number of stones
$=\frac{\text{Total area of verandah}}{\text{area of one stone}}$
$=\frac{600}{\frac{3}{10}}=\frac{600\times10}{3}$
$=2000$
View full question & answer→Question 203 Marks
A rectangular sheet of acrylic is $34\ cm$ by $24\ cm$. From it, $64$ circular buttons, each of diameter $3.5\ cm$, have been cut out. Find the area of the remaining sheet.
AnswerLength of rectangular sheet $(1)=34 \mathrm{~cm}$
and breadth $(b)=24 \mathrm{~cm}$
Area $=l \times b=34 \times 24 \mathrm{~cm}^2=816 \mathrm{~cm}^2$
Diameter of one button $=3.5 \mathrm{~cm}$
Radius $(r)=\frac{3.5}{2} \mathrm{~cm}$
and area of one button $=\pi \mathrm{r}^2$
$=\frac{22}{7} \times \frac{3.5}{2} \times \frac{3.5}{2} \mathrm{~cm}^2$
$=9.625 \mathrm{~cm}$
Area of 64 buttons $=9.625 \times 64 \mathrm{~cm}^2=616 \mathrm{~cm}^2$
Area of remaining sheet $=816-616=200 \mathrm{~cm}^2$
View full question & answer→Question 213 Marks
Find the area of a rhombus, the lengths of whose diagonals are: $8\ dm \ 5\ cm$ and $5\ dm \ 6\ cm.$
AnswerLength of one diagonal $= 8dm 5cm = (8 \times 10 + 5)cm = 85cm$ [since $1dm = 10cm]$
Length of the other diagonal $= 5dm 6cm = (5 \times 10 + 6)cm = 56cm$
$\therefore$ Area of the rhombus $=\frac{1}{2}\times\text{(Product of the diagonals)}$
$=\Big(\frac{1}{2}\times85\times56\Big)\text{cm}^2$$=2380\text{cm}^2$
View full question & answer→Question 223 Marks
Calculate the area of the shaded region in each of the figures given below. Fig. $(ii)$ has uniform width of $3cm$ and it is given that $AB = CD.$
Answeri. Outer length $=24 \mathrm{~m}$
and breadth $=19 \mathrm{~m}$
Area $=24 \times 19=456 \mathrm{~m}^2$
Length of unshaded portion $=24-4=20 \mathrm{~m}$
and breadth $=16.5 \mathrm{~m}$
Area of unshaded portion $=20 \times 16.5 \mathrm{~m}^2=330.0 \mathrm{~m}^2$
Area of shaded portion $=456-330=126 \mathrm{~m}^2$
ii. Dividing the figure an shown
Area of rectangle I$=15 \times 3 \mathrm{~cm}^2=45 \mathrm{~cm}^2$
Area of rectangle II $=(12-3) \times 3=9 \times 3=27 \mathrm{~cm}^2$
Area of rectangle III $=5 \times 3=15 \mathrm{~cm}^2$
and area of rectangle IV $=(12-3) \times 3=9 \times 3=27 \mathrm{~cm}^2$
Total area of shaded portion $=45+27+15+27=114 \mathrm{~cm}^2$ View full question & answer→Question 233 Marks
A horse is tethered to one corner of a rectangular field, $60m$ by $40m$, by a rope $14m$ long. On how much area can the horse graze?
AnswerLength of field $= 60m$ and
breadth $= 40m$
Length of rope $= 14m$
Area covered by the horse $=\frac{1}{4}$ of
area of circle $=\frac{1}{4}\times\pi\text{r}^2$
$=\frac{1}{4}\times\frac{22}{7}\times14\times14=154\text{cm}^2$
View full question & answer→Question 243 Marks
Calculate the area of the shaded region in each of the figures given below:
AnswerSide of square $(a)=40 \mathrm{~m}$
Area $=(a)^2=40 \times 40=1600 \mathrm{~m}^2$
Area of larger road $=40 \times 3=120 \mathrm{~m}^2$
and area of shorter road $=40 \times 2=80 \mathrm{~m}^2$
Area of roads $=(120+80)-3 \times 2=200-6=194 m^2$
Area of shaded portion $=(1600-194) \mathrm{m}^2=1406 \mathrm{~m}^2$
View full question & answer→Question 253 Marks
The lengths of the sides of a triangle are $33\ cm, 44\ cm$ and $55$ respectively. Find the area of the triangle and hence find the height corresponding to the side measuring $44\ cm.$
AnswerLet $a =33\ cm, b = 44\ cm, c = 55\ cm$
$\therefore\text{s}=\frac{\text{a + b + c}}{2}$ $=\frac{33+44+55}{2}$ $=\frac{132}{2}=66$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})(\text{s}-\text{b})(\text{s}-\text{c})}$
$=\sqrt{66(66-33)(66-44)(66-55)}$
$=\sqrt{66\times33\times22\times11}$
$=\sqrt{2\times3\times11\times3\times11\times2\times11\times11}$
$=2\times3\times11\times11=726\text{cm}^2$
Let base $=44\text{cm}$ then height $=\frac{\text{Area}\times2}{\text{Base}}$
$=\frac{726\times2}{44}$
$=33\text{cm}$ View full question & answer→Question 263 Marks
The, area of a rectangular field is $3584m^2$ and its length is $64m$. A boy runs around the fleld at the rate of $6km/h$. How long will he take to go $5$ times around it?
AnswerArea of rectangular field = $3584m^2$ Length $= 64m$
Area $= 3584$
Breadth $=\frac{\text{Area}}{\text{Length}}$
$=\frac{3584}{64}=56\text{m}$
Now perimeter $= 2(l + b) = 2(64 + 56)m = 2 \times 120 = 240m$
Distance covered in $5$ rounds $= 240 \times 5 = 1200m$
Speed $= 6km/h$
Time take $=\frac{1200}{1000}\times\frac{60}{6}=12\text{ minutes}$ ($1$ hour $= 60$ minutes)
View full question & answer→Question 273 Marks
The sides of a triangle are $42\ cm, 34\ cm$ and $20\ cm$. Calculate its area and the length of the height on the longest side.
AnswerLet $a = 42\ cm, b = 34\ cm, c = 20\ cm$
$\therefore\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{42+34+20}{2}$
$=\frac{96}{2}=48$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{48(48-42)(48-34)(48-20)}$
$=\sqrt{48\times6\times14\times28}$
$=\sqrt{2\times2\times2\times2\times3\times3\times2\times2\times7\times7\times2\times2}$
$=2\times2\times2\times2\times3\times7=336\text{cm}^2$ Longest side (Base) $=42\text{cm}$
$\therefore$ Height on longest side $=\frac{\text{Area}\times2}{\text{Base}}$
$=\frac{336\times2}{42}\text{cm}=16\text{cm}$ View full question & answer→Question 283 Marks
Find the area of quadrilateral $ABCD$ in which diagonal $BD = 24\ cm$. $\text{AL}\perp\text{BD}$ and $\text{\ cm}\perp\text{BD}$ such that $AL = 5\ cm$ and $\ cm = 8\ cm.$
AnswerIn the quadrilateral $ABCD BD = 24\ cm$
$\text{AL}\perp\text{BD}$ and $\text{CM}\perp\text{BD}$
$AL = 5cm$ and $CM = 8cm$
Now area of $\triangle\text{ABD}=\frac{1}{2}\text{bh}$
$=\frac{1}{2}\times\text{BD}\times\text{AL}=\frac{1}{2}\times24\times5\text{cm}^2$
$=60\text{cm}^2$ and area of $\triangle\text{CBD}=\frac{1}{2}\text{BD}\times\text{CM}$
$=\frac{1}{2}\times24\times8\text{cm}^2=96\text{cm}^2$
$\therefore$ Total area of quadrilateral $ABCD =60+96=156\text{cm}^2$ View full question & answer→Question 293 Marks
A rectangular field is $50m$ by $40m$. It has two roads through its centre, running parallel to its sides. The width of the longer and the shorter roads are $2m$ and $2.5-m$-respectively. Find the area of the roads and the area of the remaining portion of the field.
AnswerLength of field $(l) = 50m$ and breadth $(b) = 40m$
Width of road parallel to length $=2 \mathrm{~m}$
and width of road parallel to breadth $=2.5 \mathrm{~m}$
Area of roads $=50 \times 2+40 \times 2.5-2.5 \times 2=(100+100-5) \mathrm{m}^2=195 \mathrm{~m}^2$
and area of remaining portion $=50 \times 40-195=2000-195=1805 \mathrm{~m}^2$ View full question & answer→Question 303 Marks
A piece of wire is bent in the shape of an equilateral triangle each of whose sides measures $8.8\ cm$. This wire is rebent to form a circular ring. What is the diameter of the ring?
AnswerLength of the wire = Perimeter of the equilateral triangle $= 3 \times $ Side of the equilateral triangle $= (3 \times 8.8)cm = 26.4cm$
Let the wire be bent into the form of a circle of radius $r cm.$
Circumference of the circle $= 26.4cm$
$\Rightarrow2\pi\text{r}=26.4$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=26.4$
$\Rightarrow\text{r}\Big(\frac{26.4\times7}{2\times22}\Big)\text{cm}=4.2\text{cm}$
$\therefore$ Diameter $= 2r = (2 \times 4.2)cm = 8.4cm$
Hence, the diameter of the ring is $8.4cm.$
View full question & answer→Question 313 Marks
The area of four walls of a room is $120m^2$. If the length of the room is twice its breadth and the height is $4m$, find the area of the floor.
Answer$\text { Area of } 4 \text { walls }=120 \mathrm{~m}^2$
$\text { Height }(\mathrm{h})=4 \mathrm{~m}$
$\text { Let breadth }(b)=x$
$\text { and length }(\mathrm{l})=2 \mathrm{x}$
$\text { Area of } 4 \text { walls }=2(l+\mathrm{h}) \times \mathrm{h}$
$=2(2 \mathrm{x}+\mathrm{x}) \times 4=8 \times 3 \mathrm{x}=24 \mathrm{x}$
$24 \mathrm{x}=120$
$\mathrm{x}=\frac{120}{24}=5$
$\text { Length of room }=2 \mathrm{x}=2 \times 5=10 \mathrm{~m}$
$\text { and breadth }=x=5 \mathrm{~m}$
$\text { Area of floor }=l \times \mathrm{b}=10 \times 5=50 \mathrm{~m}^2$
View full question & answer→Question 323 Marks
The sides of a rectangular park are in the ratio $4 : 3$. If its area is $1728m^2$, find the cost of fencing it at $₹ 30$ per metre.
Answer$\text { Ratio in the sides of a rectangle }=4: 3$
$\text { Area }=1728 \mathrm{~cm}^2$
$\text { Let length }=4 \mathrm{x}$
$\text { then breadth }=3 \mathrm{x}$
$\text { Area }=1 \times \mathrm{b}$
$1728=4 \mathrm{x} \times 3 \mathrm{x}$
$\Rightarrow 12 \mathrm{x}^2=1728$
$\Rightarrow \mathrm{x}^2=144=(12)^2$
$\Rightarrow \mathrm{x}=12$
$\text { Length }=4 \mathrm{x}=4 \times 12=48 \mathrm{~m}$
$\text { and breadth }=3 \mathrm{~m}=3 \times 12=36 \mathrm{~m}$
$\text { Now perimeter }=2(\mathrm{l}+\mathrm{b})=2(48+36) \mathrm{m}=2 \times 84=168 \mathrm{~m}$
$\text { Rate of fencing }=\text { Rs. } 30 \text { per metre }$
$\text { Total cost }=168 \times 30=\text { Rs. } 5040$
View full question & answer→Question 333 Marks
A rhombus has the same perimeter as the circumference of a circle. If each side of the rhombus measures $33\ cm$, find the radius of the circle.
AnswerCircumference of the circle = Perimeter of the rhombus $= 4 \times $
Side of the rhombus $= (4 \times 33)cm = 132cm$
$\therefore$ Circumference of the circle $= 132cm$
$\Rightarrow2\pi\text{r}=132$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=132$
$\Rightarrow\text{r}=\Big(\frac{132\times7}{2\times22}\Big)\text{cm}=21\text{cm}$
Hence, the radius of the circle is $21cm.$
View full question & answer→Question 343 Marks
A wire is in the shape of a square of side $10\ cm$. If the wire is rebent into a rectangle of length $12\ cm$, find its breadth. Which figure encloses more area and by how much?
AnswerSide of a square wire $=10 \mathrm{~cm}$
Perimeter $=4 \mathrm{a}=4 \times 10 \mathrm{~cm}=40 \mathrm{~cm}$
or perimeter of rectangle $=40 \mathrm{~cm}$
Length of rectangle $=12 \mathrm{~cm}$
Breadth $=\frac{40}{2}-12=20-12=8 \mathrm{~cm}$
Now area of square $=a^2=(10)^2=100 \mathrm{~cm}^2$
and area of rectangle $=\mathrm{I} \times \mathrm{b}=12 \times 8=96 \mathrm{~cm}^2$
Difference in areas $=100-96=4 \mathrm{~cm}^2$
Square has $4 \mathrm{~cm}^2$ more area.
View full question & answer→Question 353 Marks
The area of a circle is $1381m^2$. Find its circumference.
AnswerArea of a circle $=1386\text{m}^2$
$\therefore\text{Radius (r)}=\sqrt{\frac{\text{Area}}{\pi}}=\sqrt{\frac{1386\times7}{22}}\text{m}$
$=\sqrt{441}=21\text{m}$
$\therefore\text{Circumference}=2\pi\text{r}=2\times\frac{22}{7}\times21\text{m}$
$=132\text{m}$
View full question & answer→Question 363 Marks
The sides of a triangle are in the ratio $13 : 14 : 15$ and its perimeter is $84\ cm$. Find the area of the triangle.
AnswerPerimeter of the triangle $= 84\ cm$
Ratio in side $= 13 : 14 : 15$
Sum of ratios $= 13 + 14 + 15 = 42$
Let then first side $=\frac{84\times13}{42}=26\text{cm}$
Second side $=\frac{84\times14}{42}=28\text{cm}$
Third side $=\frac{84\times15}{42}=30\text{cm}$
Now $\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{84}{2}=42$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{42(42-26)(42-28)(42-30)}$
$=\sqrt{42\times16\times14\times12}$
$=\sqrt{2\times3\times7\times2\times2\times2\times2\times2\times7\times2\times2\times3}$
$=2\times2\times2\times2\times3\times7=336\text{cm}^2$
View full question & answer→Question 373 Marks
A steel wire when bent in the form of a square encloses an area of $121cm^2$. The same wire is bent in the form of a circle. find the area of the circle.
AnswerLet a be one side of the square.
$\text { Area of the square }=121 \mathrm{~cm}^2 \text { (given) }$
$\Rightarrow a^2=121$
$\Rightarrow a=11 \mathrm{~cm}(\text { since } 11 \times 11=121)$
$\text { Perimeter of the square }=4 \times \text { side }=4 \mathrm{a}=(4 \times 11) \mathrm{cm}=44 \mathrm{~cm}$
Length of the wire $=$ Perimeter of the square
$=44 \mathrm{~cm}$
The wire is bent in the form of a circle.
Circumference of a circle $=$ Length of the wire
$\therefore$ Circumference of a circle $=44 \mathrm{~cm}$
$\Rightarrow 2 \pi \mathrm{r}=44$
$\Rightarrow\left(2 \times \frac{22}{7} \times \mathrm{r}\right)=44$
$\Rightarrow \mathrm{r}=\left(\frac{44 \times 7}{2 \times 22}\right)=7 \mathrm{~cm}$
$\therefore$ Area of the circle $=\pi r^2$
$=\left(\frac{22}{7} \times 7 \times 7\right) \mathrm{cm}^2$
$=154 \mathrm{~cm}^2$
View full question & answer→Question 383 Marks
In the given figure, four equal circles are described about the four corners of a square so that each circle touches two of the circle as shown in the figure. find the area of the shaded region, each side of the square measuring $14\ cm.$
AnswerEach side of square $=14 \mathrm{~cm}$
Area of square $=a^2=14 \times 14=196 \mathrm{~cm}^2$
Radius of each circle at each corner of square $=\frac{14}{2}=7 \mathrm{~cm}$
$\therefore$ Area of the quadrant $=\frac{1}{4} \times \pi \mathrm{r}^2$
$=\frac{1}{4} \times \frac{22}{7} \times 7 \times 7=\frac{77}{2} \mathrm{~cm}^2$
and Area of $4$ quadrant $=\frac{77}{2} \times 4$
$=154 \mathrm{~cm}^2$
Area of shaded portion $=$ Area of square - area of $4$ quadrants
$=196-154=42 \mathrm{~cm}^2$
View full question & answer→Question 393 Marks
A room is $8.5m$ long, $6.5m$ broad and $3.4m$ high. It has two doors, each measuring ($1.5m$ by $1m$) and two windows, each measuring $(2m$ by $1m).$ Find the cost of painting its four walls at $Rs 160$ per $m^2$.
Answer$\text { Length of a room }(\mathrm{l})=8.5 \mathrm{~m}$
$\text { Breadth }(\mathrm{b})=6.5 \mathrm{~m}$
$\text { and height }(\mathrm{h})=3.4 \mathrm{~m}$
$\text { Area of four walls }=2(\mathrm{l}+\mathrm{b}) \times \mathrm{h}$
$=2(8.5+6.5) \times 3.4 \mathrm{~m}^2$
$=2 \times 15 \times 3.4 \mathrm{~m}^2$
$=30 \times 3.4=102.0 \mathrm{~m}^2$
Area of two doors of size $1.5 \mathrm{~m} \times 1 \mathrm{~m}=2 \times 1.5 \times 1 \mathrm{~m}=3 \mathrm{~m}^2$
and area of two windows of size $2 \mathrm{~m} \times 1 \mathrm{~m}=2 \times 2 \times 1=4 \mathrm{~m}^2$
Area of remaining portion $=102-(3+4)=102-7 \mathrm{~m}^2=95 \mathrm{~m}^2$
Rate of painting $=$ Rs. 160 per $\mathrm{m}^2$
Total cost $=$ Rs. $160 \times 95=$ Rs. 15200
View full question & answer→Question 403 Marks
Find the area of a rhombus, the lengths of whose diagonals are: $16\ cm$ and $28\ cm.$
AnswerLength of one diagonal $= 16\ cm$
Length of the other diagonal $= 28\ cm$
$\therefore$ Area of the rhombus $=\frac{1}{2}\times(\text{Product of the diagonals)}$ $=\Big(\frac{1}{2}\times16\times28\Big)\text{cm}^2=224\text{cm}^2$
View full question & answer→Question 413 Marks
one slde of a parallelogram ts $l8 \ cm$ long and lts area ls $153 \mathrm{~cm}^2$. Ftnd the distance of the glven slde from lts opposlte side.
AnswerBase of the parallelogram $= 18\ cm$
Area of the parallelogram = $153 \mathrm{~cm}^2$
$\therefore$ Area of the parallelogram = Base $\times $ Height
$\Rightarrow\text{Height}=\frac{\text{Area of the parallelogram}}{\text{Base}}$
$=\Big(\frac{153}{18}\Big)\text{cm}=8.5\text{cm}$
Hence, the distance of the given side from its opposite side is $8.5\ cm.$
View full question & answer→Question 423 Marks
Find the radius of a circle whose area is $616cm^2$.
AnswerLet the radius of the circle be $r cm$
Area $=(\pi\text{r}^2)\text{cm}^2$
$\pi\text{r}^2=616$
$\Rightarrow\frac{22}{7}\times\text{r}\times\text{r}=616$
$\Rightarrow\text{r}^2=\Big(\frac{616\times7}{22}\Big)=196$
$\Rightarrow\text{r}=\sqrt{196}=14\text{cm}$
Hence, the radius of he given circle is $14\ cm.$
View full question & answer→Question 433 Marks
Find the area of a rectangular plot on side of which is $48m$ and its diagonal $50m.$
AnswerWe know that all the angles of a rectangle are $90^{\circ}$ and the diagonal divides the rectangle into two right angled triangles.
So, one side of the triangle will be $48 m$ and the diagonal, which is $50 m$ , will be the hypotenuse.
According to Pythagoras theorem:
$(\text { Hypotenuse })^2=(\text { Base })^2+(\text { Perpendicular })^2$
Perpendicular $=\sqrt{(\text { Hypotenuse })^2-(\text { Base })^2}$
Perpendicular $=\sqrt{(50)^2-(48)^2}$
$\therefore$ Other side of the rectangular plot $=14 \mathrm{~m}$
$\therefore$ Area of the rectangular plot $=48 \mathrm{~m} \times 14 \mathrm{~m}=672 \mathrm{~m}^2$
Hence, the area of a rectangular plot is $672 \mathrm{~m}^2$.
View full question & answer→Question 443 Marks
The area of a rhombus is $119cm^2$ and its perimeter is $56\ cm$. Find its height.
AnswerPerimeter of the rhombus $=56 \mathrm{~cm}$
Area of the rhombus $=119 \mathrm{~cm}^2$
Side of the rhombus $=\frac{\text { perimeter }}{4}=\left(\frac{56}{4}\right) \mathrm{cm}=14 \mathrm{~cm}$
Area of a rhombus $=$ Base $\times$ Height
$\therefore$ Height of the rhombus $=\frac{\text { Area }}{\text { Base }}$
$=\left(\frac{119}{14}\right) \mathrm{cm}$
$=8.5 \mathrm{~cm}$
View full question & answer→Question 453 Marks
A godown is $50m$ long, $40m$ broad and $10m$ high. Find the cost of whitewashing its four walls and ceiling at $Rs 20$ per square metre.
Answer$\text { Length of go down }(I)=50 \mathrm{~m}$
$\text { Breadth }(\mathrm{b})=40 \mathrm{~m}$
$\text { and height }(\mathrm{h})=10 \mathrm{~m}$
$\text { Area of } 4 \text { walls }=2(\mathrm{l}+\mathrm{b}) \times \mathrm{h}$
$=2(50+40) \times 10 \mathrm{~m}$
$=2 \times 90 \times 10=1800 \mathrm{~m}^2$
$\text { and area of ceiling }=1 \times \mathrm{b}=50 \times 40=2000 \mathrm{~m}^2$
$\text { Total area of walls and ceiling }=1800+2000=3800 \mathrm{~m}^2$
$\text { Rate of whitewashing }=\text { Rs. } 20 \text { per } \mathrm{m}^2$
$\text { Total cost }=\text { Rs. } 20 \times 3800=\text { Rs. } 76000$
View full question & answer→Question 463 Marks
In the following figures, find the area of the shaded region.
AnswerLength of rectangle $(l) = 18\ cm$ and
breadth $(b) = 10\ cm$
Area $= l × b = 18 × 10 = 180cm^2$
Area of right $\triangle\text{EBC}=\frac{1}{2}\times10\times8=40\text{cm}^2$ and
area of right $\triangle\text{EDF}=\frac{1}{2}\times10\times6=30\text{cm}^2$
Area of shaded region $=180-(40+30)=180-70=110\text{cm}^2$ View full question & answer→Question 473 Marks
The area of a rhombus is $148.8cm^2$. If one of its diagonals is $19.2\ cm,$ find the length of the other diagonal.
AnswerArea of a rhombus $=\frac{1}{2} \times$ (Product of the diagonals)
Given:
Length of one diagonal $=19.2 \mathrm{~cm}$
Area of the rhombus $=148.8 \mathrm{~cm}^2$
$\therefore$ Length of the other diagonal $=\left(\frac{148.8 \times 8}{19.2}\right) \mathrm{cm}=15.5 \mathrm{~cm}$
View full question & answer→Question 483 Marks
A room $9.5m$ long and $6m$ wide is surrounded by a $1.25-m-$ long verandah. Calculate cost of cementing the floor of this verandah at $Rs 80$ per $m^2$.
Answer$\text { Length' of room }(\mathrm{l})=9.5 \mathrm{~m}$
$\text { Breadth }(\mathrm{b})=6 \mathrm{~m}$
$\text { Width of outer verandah }=1.25 \mathrm{~m}$
$\text { Outer length }(\mathrm{L})=9.5+2 \times 1.25=9.5+2.5=12.0 \mathrm{~m}$
$\text { and breadth }(B)=6+2 \times 1.25=6+2.5=8.5 \mathrm{~m}$
$\text { Area of verandah }=\text { Outer area }- \text { Inner area }=\mathrm{L} \times \mathrm{B}-\mathrm{l} \times \mathrm{b}$
$=(12.0 \times 8.5-9.5 \times 6) \mathrm{m}^2-(102.0-57.0) \mathrm{m}^2=45 \mathrm{~m}^2$
$\text { Rate of cementing }=\text { Rs. } 15 \text { per } \mathrm{m}^2$
$\text { Total cost }=\text { Rs. } 80 \times 45=\text { Rs. } 3600$
View full question & answer→Question 493 Marks
A square lawn is surrounded by a path $2.5m$ wide. If the area of the path is $165m^2$, find the area of the lawn.
AnswerArea of path $=165 \mathrm{~m}^2$
Width of path $=2.5 \mathrm{~m}$.
Let side of square lawn $=\mathrm{x} \mathrm{m}$.
Outer side $=x+2 \times 2.5=(x+5) m$
Area of path $=(x+5) 2-x^2$
$\Rightarrow \mathrm{x}^2+10 \mathrm{x}+25-\mathrm{x}^2=165$
$\Rightarrow 10 x=165-25=140$
$\Rightarrow \mathrm{x}=\frac{140}{10}=14 \mathrm{~m}$
Side of lawn $=14 \mathrm{~m}$
and area of lawn $=(14)^2 \mathrm{~m}^2=196 \mathrm{~m}^2$ View full question & answer→Question 503 Marks
A rectangular lawn $70m$ by $50m$ has two roads, each $5m$ wide, running through its middle, one parallel to its length and the other parallel to its breadth. Find the cost of constructing the roads at $Rs 120$ per $m^2$.
AnswerLength of lawn $(l)=70 \mathrm{~m}$
Breadth (b) $=50 \mathrm{~m}$
Width of crossing roads $=5 \mathrm{~m}$
Area of roads $=70 \times 5+50 \times 5-(5)^2$
$=350+250-(5)^2$
$=600-25=575 \mathrm{~m}^2$
Cost of constructing $=$ Rs. 120 per $\mathrm{m}^2$
Total cost Rs. $120 \times 575= Rs. 69000$ View full question & answer→Question 513 Marks
A racetrack is in the form of a ring whose inner circumference is $528m$ and the outer circumference is $616m$. Find the width of the track.
AnswerLet the inner and outer radii of the track be $r$ metres and $R$ metres, respectively.
Then, $2\pi\text{r}=528$
$2\pi\text{R}=616$
$\Rightarrow2\times\frac{22}{7}\times\text{r}=528$
$2\times\frac{22}{7}\times\text{R}=616$
$\Rightarrow\text{r}=\Big(528\times\frac{7}{44}\Big)=84$
$\text{R}=\Big(616\times\frac{7}{44}\Big)=98$
$\Rightarrow (R - r) = (98 - 84)m = 14m$
Hence, the width of the track is $14m.$ View full question & answer→Question 523 Marks
How long will a man take to make a round of a circular field of radius $21m$, cycling at the speed of $8km/h?$
AnswerRadius of the circular field, r = 21m Circumference $=2\pi\text{r}$
$=2\times\frac{22}{7}\times21\text{m}=132\text{m}$
Speed of cyclist $= 8km/hr = 8000m/hr$
$\therefore$ Time taken for making one round $=\frac{132\times60}{8000}$
$[60$minutes $= 1hr]$
$=\frac{99}{100}\text{minutes}$
$=\frac{99}{100}\times60\sec.$
$=\frac{594}{10}=59.4\text{ seconds}$
View full question & answer→Question 533 Marks
The base of a parallelogram is twice its height. If the area of the parallelogram is $512cm^2$. find the base and the height.
AnswerLet the height of the parallelogram be $x cm .$
Then, the base of the parallelogram will be $2 xcm .$
It is given that the area of the parallelogram is $512 \mathrm{~cm}^2$.
Area of a parallelogram $=$ Base $\times$ Height
$\therefore 512 \mathrm{~cm}^2=2 \mathrm{x} \times \mathrm{x}$
$512 \mathrm{~cm}^2=2 \mathrm{x}^2$
$\Rightarrow \mathrm{x}^2=\left(\frac{512}{2}\right) \mathrm{cm}^2=256 \mathrm{~cm}^2$
$\Rightarrow \mathrm{x}^2=(16 \mathrm{~cm})^2$
$\Rightarrow \mathrm{x}=16 \mathrm{~cm}$
$\therefore \text { Base }=2 \mathrm{x}=2 \times 16$
$=32 \mathrm{~cm}$
Height $=x=16 \mathrm{~cm}$
View full question & answer→Question 543 Marks
In the given figure, $ABCD$ is a rectangle in which $AB = 40\ cm$ and $BC = 25\ cm$. If $P, Q, R, S$ be the midpoints of $AB, BC, CD$ and $DA$ respectively, find the area of the shaded region.
AnswerIn the fig. $ABCD$ is a rectangle in which $AB = 40cm, BC = 25cm.$
$P, Q, R$ and $S$ and the mid points of sides, $PQ, QR, RS$ and $SP$
respectively Then $PQRS$ is a rhombus.
Now, join $PR $and $QS$. $PR = BC = 25cm$ and $QS = AB = 40cm$
Area of PQRS $=\frac{1}{2}\times\text{PR}\times\text{QS}$
$=\frac{1}{2}\times25\times40=500\text{cm}^2$
View full question & answer→Question 553 Marks
The length and breadth of a rectangular piece of land are in the ratio of $5 : 3$. If the total cost of fencing it at $Rs 24$ per metre is $Rs 9600$, find its length and breadth.
AnswerRatio in length and breadth of a rectangular piece of land $= 5 : 3$
Cost of fencing $= Rs. 9600$ and rate $= Rs. 24$ per m
Perimeter $=\frac{9600}{24}=400\text{m}$
Let length $= 5x$
Then breadth $= 3x$
Perimeter $= 2(l + b) $
$\Rightarrow 400 = 2(5x + 3x) $
$\Rightarrow 400 = 2 \times 8x = 16x$
$ \Rightarrow 16x = 400 $
$\Rightarrow x = 25$
Length of the land $= 5x = 5 \times 25 = 125m$ and
width $= 3x = 3 \times 25 = 75m$
View full question & answer→Question 563 Marks
The area of a circle is $616cm^2$. Find its circumference.
AnswerArea of the circle $=616\text{cm}^2$
$\therefore\text{Radius (r)}=\sqrt{\frac{\text{Area}}{\pi}}=\frac{616\times7}{22}$
$=\sqrt{28\times7}=\sqrt{196}=14\text{cm}$
$\therefore$ Circumference $=2\pi\text{r}=2\times\frac{22}{7}\times14\text{cm}$
$=88\text{cm}$
View full question & answer→Question 573 Marks
A rectangular ground is $90m$ long and $32m$ broad. In the middle of the ground there is a circular tank of radius $14$ metres. Find the cost of turfing the remaining portion at the rate of $Rs 50$ per square metre.
AnswerArea of the rectangular ground $=90 \mathrm{~m} \times 32 \mathrm{~m}=(90 \times 32) \mathrm{m}^2=2880 \mathrm{~m}^2$
Given:
Radius of the circular tank $(r)=14 m$
$\therefore$ Area covered by the circular tank $=\pi \mathrm{r}^2=\left(\frac{22}{7} \times 14 \times 14\right) \mathrm{m}^2$
$=616 \mathrm{~m}^2$
$\therefore$ Remaining portion of the rectangular ground for turfing $=$ (Area of the rectangular ground - Area covered by the circular tank)
$=(2880-616) m^2=2264 m^2$
Rate of turfing $= Rs 50$ per sq. metre
$\therefore$ Total cost of turfing the remaining ground $=$ Rs $(50 \times 2264)=$ Rs $1,13,200$
View full question & answer→Question 583 Marks
Each side of a square flower bed is $2m$ $80\ cm$ long. It is extended by digging a strip $30\ cm$ wide all around it. Find the area of the enlarged flower bed and also the increase in the area of the flower bed.
AnswerEach side of square bed $(\mathrm{a})=2 \mathrm{~m} ~80 \mathrm{~cm}=2.8 \mathrm{~m}$
Width of strip $=30 \mathrm{~cm}$
Outer side $(A)=2.8 \mathrm{~m}+2 \times 30 \mathrm{~cm}=2.8+0.6=3.4 \mathrm{~m}$
Outer area $=(3.4 \mathrm{~m})^2=11.56 \mathrm{~m}^2$
Inner area $=(2.8)^2=7.84 \mathrm{~m}^2$
Area of increased bed flower $=11.56-7.84=3.72 \mathrm{~m}^2$ View full question & answer→Question 593 Marks
A wire in the form of a rectangle $18.7\ cm$ long and $14.3\ cm$ wide is reshaped and bent into the form of a circle. Find the radius of the circle so formed.
AnswerLength of the wire = Perimeter of the rectangle $= 2(l + b) = 2 \times (18.7 + 14.3)cm = 66cm$
Let the wire be bent into the form of a circle of radius $r \ cm$.
Circumference of the circle $= 66cm$
$\Rightarrow2\pi\text{r}=66$
$\Rightarrow\Big(2\times\frac{22}7{}\times\text{r}\Big)=66$
$\Rightarrow\text{r}=\Big(\frac{66\times7}{2\times22}\Big)\text{cm}=10.5\text{cm}$
Hence, the radius of the circle formed is $10.5\ cm.$
View full question & answer→Question 603 Marks
The area of the $4$ walls of a room is $77m^2$. The length and breadth of the room are $7.5m$ and $3.5m$ respectively. Find the height of the room.
AnswerArea of 4 walls of a room $=77 \mathrm{~m}^2$
Length of room $(l)=7.5 \mathrm{~m}$
and breadth $(b)=3.5 \mathrm{~m}$
Let h be the height,
then area of four walls $=2(l+b) h$
$\Rightarrow 2(7.5+3.5) \mathrm{h}=77$
$\Rightarrow 2 \times 11 \times \mathrm{h}=77$
$\Rightarrow \mathrm{~h}=\frac{77}{2 \times 11}$
Height of room $=3.5 \mathrm{~m}$
View full question & answer→Question 613 Marks
The base of an isosceles triangle is $48\ cm$ and one of its equal sides is $30\ cm$. Find the area of the triangle.
AnswerIn isosceles $\triangle\text{ABC}$ Base BC $=48\text{cm}$ and $AB = AC =30\text{cm}$ Let $\text{AD}\perp\text{BC}$ 
Then $\text{BD = DC}=\frac{48}{2}=24\text{cm}$ In right $\triangle\text{ABD},$
$\text{AB}^2=\text{AD}^2+\text{BD}^2$ (Pythagoras Theorem)
$\Rightarrow(30)^2=(24)^2+(\text{AD})^2$
$\Rightarrow900=576+\text{AD}^2$
$\Rightarrow\text{AD}^2=900-576=324=(18)^2$
$\therefore\text{AD}=18\text{cm}$
Now, area of triangle $=\frac{\text{Base}\times\text{Altitude}}{2}$
$=\frac{48\times18}{2}=432\text{cm}^2$ View full question & answer→Question 623 Marks
The area of a square field is $\frac{1}{2}$ hectare. Find the length of its diagonal in metres. Hint: $1$ hectare = $10000m^2$.
AnswerArea of the square $=\Big\{\frac{1}{2}\times(\text{Diagonal})^2\Big\}\text{sq. units}$
Given:
Area of the square field $=\frac{1}{2}$ hectare
$=\Big(\frac{1}2{}\times10000\Big)\text{m}^2=5000\text{m}^2$ [since $1$ hectare = $10000m^2$]
Diagonal of the square $=\sqrt{2\times\text{Area of the square}}$
$=\big(\sqrt{2\times5000}\big)\text{m}=100\text{m}$
$\therefore$ Length of the diagonal of the square field $= 100m$
View full question & answer→Question 633 Marks
Find the cost of carpeting a room $13m$ by $9m$ with a carpet of width $75\ cm$ at the rate of $Rs 105$ per metre.
AnswerLength of a room $=13 \mathrm{~m}$
Breadth $=9 \mathrm{~m}$
Area of floor $=\mathrm{l} \times \mathrm{b}=13 \times 9 \mathrm{~m}^2=117 \mathrm{~m}^2$
or area of carpet $=117 \mathrm{~m}^2$
Width $=75 \mathrm{~cm}=\frac{75}{100}=\frac{3}{4} \mathrm{~m}$
Length of carpet $=$ Area $\div$ Width
$=117 \div \frac{3}{4}$
$=117 \times \frac{4}{3} \mathrm{~m}$
$=39 \times 4=156 \mathrm{~m}$
Rate $=$ Rs. 105 per m
Total cost $=$ Rs. $156 \times 105= Rs. 16380$
View full question & answer→Question 643 Marks
A rectangular lawn is $30m$ by $20m$. It has two roads each $2m$ wide running in the middle of it, one parallel to the length and the other parallel to the breadth. Find the area of the roads.
AnswerLet $A B C D$ be the rectangular park
EFGH and IJKL are the two rectangular roads with width $2 m .$
Length of the rectangular park $A D=30 \mathrm{~cm}$
Breadth of the rectangular park $C D=20 \mathrm{~cm}$
Area of the road $\mathrm{EFGH}=30 \mathrm{~m} \times 2 \mathrm{~m}=60 \mathrm{~m}^2$
Area of the road $\mathrm{IJKL}=20 \mathrm{~m} \times 2 \mathrm{~m}=40 \mathrm{~m}^2$
Clearly, area of MNOP is common to the two roads.
Area of MNOP $=2 \mathrm{~m} \times 2 \mathrm{~m}=4 \mathrm{~m}^2$
Area of the roads $=$ Area $($ EFGH $)+$ Area $($ IJKL $)-$ Area $(M N O P)$ $=(60+40) m^2-4 m^2=96 m^2$
View full question & answer→Question 653 Marks
In a parallelogram $ABCD, AB = 18\ cm, BC = 12\ cm$. $\text{AL}\perp\text{DC}$ and $\text{AM}\perp\text{BC}.$
If $AL = 6.4cm$, find the length of $AM.$ Answer$\text { Base, } A B=18 \mathrm{~cm}$
$\text { Height, } A L=6.4 \mathrm{~cm}$
$\therefore \text { Area of the parallelogram } A B C D=\text { Base } \times \text { Height }$
$=(18 \mathrm{~cm} \times 6.4 \mathrm{~cm})=115.2 \mathrm{~cm}^2 \ldots \text { (i) }$
Now, taking BC as the base:
Area of the parallelogram $A B C D=$ Base $\times$ Height
$=(12 \mathrm{~cm} \times \mathrm{AM}) \ldots \text { (ii) }$
From equation (i) and (ii):
$12 \mathrm{~cm} \times \mathrm{AM}=115.2 \mathrm{~cm}^2$
$\Rightarrow \mathrm{AM}=\left(\frac{115.2}{12}\right) \mathrm{cm}$
$=9.6 \mathrm{~cm}$
View full question & answer→Question 663 Marks
Find the area of a right triangle having base $= 24\ cm$ and hypotenuse $= 25\ cm.$
AnswerConsider $\triangle \mathrm{ABC}$ Here, $\angle \mathrm{B}=90^{\circ}$
$\mathrm{AB}=24 \mathrm{~cm}$
$\mathrm{AC}=25 \mathrm{~cm}$
Now, $\mathrm{AB}^2+\mathrm{BC}^2=\mathrm{AC}^2$
$\mathrm{BC}^2=\mathrm{AC}^2-\mathrm{AB}^2=\left(25^2-24^2\right)=(625-576)=49$
$\mathrm{BC}=(\sqrt{49}) \mathrm{cm}=7 \mathrm{~cm}$
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times \mathrm{BC} \times \mathrm{AB}$ squints
$=\frac{1}{2} \times 7 \times 24 \mathrm{~cm}^2=84 \mathrm{~cm}^2$
Hence, area of the right angled triangle is $84 \mathrm{~cm}^2$.
View full question & answer→Question 673 Marks
The area of a rhombus is $441cm^2$ and its height is $17.5\ cm$. Find the length of each side of the rhombus.
AnswerGiven:
Height of the rhombus $=17.5 \mathrm{~cm}$
Area of the rhombus $=441 \mathrm{~cm}^2$
We know:
Area of a rhombus $=$ Base $\times$ Height
$\therefore$ Base of the rhombus $=\frac{\text { Area }}{\text { Height }}$
$=\left(\frac{441}{17.5}\right) \mathrm{cm}=25.2 \mathrm{~cm}$
Hence, each side of a rhombus is $25.2 cm .$
View full question & answer→Question 683 Marks
The area of a square is $16200m^2$. Find the length of its diagonal.
AnswerWe know:
Area of a square $=\Big\{\frac{1}{2}\times(\text{Diagonal})^2\Big\}\text{sq. units}$
Diagonal of the square $=\sqrt{2\times\text{Area of square}}\text{ units}$
$=\big(\sqrt{2\times16200}\big)\text{m}=180\text{m}$
$\therefore$ Length of the diagonal of the square $= 180m$
View full question & answer→Question 693 Marks
Find the distance covered by the wheel of a bus in $2000$ rotations if the diameter of the wheel is $98\ cm.$
AnswerIt may be noted that in one rotation, the bus covers a distance equal to the circumference of the wheel.
Now, diameter of the wheel $= 98cm$
$\therefore$ Circumference of the wheel $=\pi\text{d}=\Big(\frac{22}{7}\times98\Big)\text{cm}=308\text{cm}$
Thus, the bus travels $308cm$ in one rotation.
$\therefore$ Distance covered by the bus in $2000$ rotations $= (308 \times 2000)cm = 616000cm = 6160m$ [since $1m = 100cm]$
View full question & answer→Question 703 Marks
The circumference of a circle is $35.2m.$ Find its area.
AnswerLet the radius of the circle be $r m$.
Then, its circumference will be $(2\pi\text{r})\text{m}$
$\therefore(2\pi\text{r})=35.2$
$\Rightarrow\Big(2\times\frac{22}{7}\times\text{r}\Big)=35.2$
$\Rightarrow\text{r}=\Big(\frac{35.2\times7}{2\times22}\Big)=5.6$
$\therefore$ Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times5.6\times5.6\Big)\text{m}^2=98.56\text{m}^2$
View full question & answer→Question 713 Marks
The cost of carpeting a room $15m$ long with a carpet of width $75cm$ at $Rs 80$ per metre is $Rs 19200$. Find the width of the room.
AnswerCost of carpeting a room $= Rs. 19200$
Rate $= Rs. 80$ per m
Length of carpet $=\frac{19200}{80}\text{m}=240\text{m}$
Width of carpet $=75\text{cm}=\frac{75}{100}=\frac{3}{4}\text{m}$
Area of carpet $=240\times\frac{3}{4}=180\text{m}^2$
Length of a room $=15\text{m}$
Width $=\frac{\text{Area}}{\text{Length}}$
$=\frac{180}{15}=12\text{m}$
View full question & answer→Question 723 Marks
A wire in a circular shape of radius $28\ cm$. If it is bent in the form of a square, what will be the area of the square formed?
AnswerRadius of circular wire $=28 \mathrm{~m}$
Circumference $=2 \pi \mathrm{r}=2 \times \frac{22}{7} \times 28 \mathrm{~cm}=176 \mathrm{~cm}$
Perimeter of the square formed by this wire $=176 \mathrm{~cm}$
Side $(a)=\frac{176}{4}=44 \mathrm{~cm}$
Area of square so formed $=a^2=(44)^2 \mathrm{~cm}^2=1936 \mathrm{~cm}^2$
View full question & answer→Question 733 Marks
Find the area of the triangle in which $a = 52m, b = 56cm, c = 60cm.$
Answer$a = 52m, b = 56cm, c = 60cm$
$\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{52+56+60}{2}=\frac{168}{2}=84$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{84(84-52)(84-56)(84-60)}$
$=\sqrt{84\times32\times28\times24}$
$=\sqrt{2\times2\times3\times7\times2\times2\times2\times2\times2\times2\times2\times7\times2\times2\times2\times3}$
$=2\times2\times2\times2\times2\times2\times3\times7=1344\text{cm}^2$
View full question & answer→Question 743 Marks
The area of right triangular region is $129.5 \mathrm{~cm}^2$. If one of the sides containing the right angle is $14.8cm$, find the other one.
AnswerArea of the right angled triangle = $129.5 \mathrm{~cm}^2$
Base (one side) $= 14.8cm$
$\therefore$ Altitude (second side)
$=\frac{\text{Area}\times2}{\text{Base}}$
$=\frac{129.5\times2}{14.8}$
$=\frac{1295\times2\times100}{100\times148}=17.5\text{cm}$ View full question & answer→Question 753 Marks
Find the area of the triangle in which $a = 13m, b = 14m, c = 15m.$
Answer$a = 13m, b = 14m, c = 15m$
$\therefore\text{s}=\frac{\text{a + b + c}}{2}$
$=\frac{13+14+15}{2}=\frac{42}{2}=21$
$\therefore\text{Area}=\sqrt{\text{s(s}-\text{a})\text{(s}-\text{b})\text{(s}-\text{c})}$
$=\sqrt{21(21-13)(21-14)(21-15)}$
$=\sqrt{21\times8\times7\times6}=\sqrt{7056}=84\text{cm}^2$ View full question & answer→Question 763 Marks
The area of the $4$ walls of a room is $168m^2$. The breadth and height of the room are $10m$ and $4m$ respectively. Find the length of the room.
AnswerArea of 4 walls of a room $=168 \mathrm{~m}^2$
Breadth of the room (b) $=10 \mathrm{~m}$
and height $(h)=4 \mathrm{~m}$.
Let I be the length of room
$2(l+b) h=168$
$\Rightarrow 2(l+10) \times 4=168$
$\Rightarrow 1+10=\frac{168}{2 \times 4}=21$
$\Rightarrow l=21-10=11 \mathrm{~m}$
Length of the room $=11 \mathrm{~m}$
View full question & answer→Question 773 Marks
The ratio of the radii of two circle is $5 : 3$. Find the ratio of their circumferences.
AnswerLet the radii of the given circles be $5 x$ and $3 x$, respectively.
Let their circumferences be $\mathrm{C}_1$ and $\mathrm{C}_2$, respectively.
$\mathrm{C}_1=2 \times \pi \times 5 \mathrm{x}=10 \pi \mathrm{x}$
$\mathrm{C}_2=2 \times \pi \times 3 \mathrm{x}=6 \pi \mathrm{x}$
$\therefore \frac{\mathrm{C}_1}{\mathrm{C}_2}=\frac{10 \pi \mathrm{x}}{6 \pi \mathrm{x}}=\frac{5}{3}$
$\Rightarrow \mathrm{C}_1: \mathrm{C}_2=5: 3$
Hence, the ratio of the circumference of the given circle is $5: 3$.
View full question & answer→Question 783 Marks
A rectangular grassy plot is $75m$ long and $60m$ broad. If has path of width $2m$ all around it on the inside. Find the area of the path and cost and of constructing it at $Rs 125$ per $m^2$.
AnswerOuter length of plot $(L) = 75m$ and breadth $(B) = 60m$ Width of path inside $= 2m$
Inner length $(l)=75-2 \times 2=75-4=71 \mathrm{~m}$
and width (b) $=60-2 \times 2=60-4=56 \mathrm{~m}$
Area of the path $=\mathrm{L} \times \mathrm{B}-\mathrm{l} \times \mathrm{b}=(75 \times 60-71 \times 56) \mathrm{m}^2=4500-3976=524 \mathrm{~m}^2$
Rate of constructing it $= Rs. 125$ per $\mathrm{m}^2$
Total cost $=$ Rs. $524 \times 125=$ Rs. 65500 View full question & answer→Question 793 Marks
One circle has radius of $98\ cm$ and a second concentric circle has a radius of $1m \ 26\ cm$. How much longer is the circumference of the second circle than that of the first?
AnswerWe know that the concentric circles are circles that form within each other, around a common centre point.
Radius of the inner circle, $r = 98cm$
$\therefore$ Circumference of the inner circle $=2\pi\text{r}$
$=\Big(2\times\frac{22}{7}\times98\Big)\text{cm}=616\text{cm}$
Radius of the outer circle, $R = 1m 26cm = 126cm$ [since $1m = 100cm]$
$\therefore$ Circumference of the outer circle $=2\pi\text{R}$
$=\Big(2\times\frac{22}{7}\times126\Big)\text{cm}=792\text{cm}$
$\therefore$ Difference in the lengths of the circumference of the circles $= (792 - 616)\ cm = 176\ cm$
Hence, the circumference of the second circle is $176\ cm$ larger than that of the first circle.
View full question & answer→Question 803 Marks
The base of a triangular field is three times its height. If the cost of cultivating the field at $Rs \ 1080$ per hectare is $Rs \ 14580$, find its base and height.
AnswerTotal cost of cultivating the field $= Rs. 14580$
Rate of cultivating the field $= Rs. 1080$ per hectare
Area of the field $=\Big(\frac{\text{Total cost}}{\text{Rate per hectare}}\Big)\text{hectare}$
$=\Big(\frac{14580}{1080}\Big)\text{hectare}$
$=(13.5 \times 10000) \mathrm{m}^2=135000 \mathrm{~m}^2\left[\right.$ Since $1$ hectare $\left.=10000 \mathrm{~m}^2\right]$
Let the height of the filde be $x m.$
Then, its base will be $3x m.$
Area of the filde $=\Big(\frac{1}{2}\times3\text{x}\times\text{x}\Big)\text{m}^2=\Big(\frac{3\text{x}^2}{2}\Big)\text{m}^2$
$\therefore\Big(\frac{3\text{x}^2}{2}\Big)=135000$
$\Rightarrow\text{x}^2=\Big(135000\times\frac{2}{3}\Big)=90000$
$\Rightarrow\text{x}=\sqrt{90000}=300$
$\therefore\text{Base}=(3\times300)=900\text{m}$
$\text{Height}=300\text{m}$
View full question & answer→Question 813 Marks
Find the length of the largest pole that can be placed in a hall $10m$ long, $10m$ wide and $5m$ high.
AnswerLength of hall $(l) = 10m$
Breadth $(b) = 10m$
and height $(h) = 5m$
Longest pole which can be placed in it
$=\sqrt{\text{l}^2+\text{b}^2+\text{h}^2}$
$=\sqrt{(10)^2+(10)^2+(5)^2}=\sqrt{100+100+25}\text{m}$
$=\sqrt{225}=15\text{m}$
View full question & answer→Question 823 Marks
One side of a right-angled triangular scarf is $80\ cm$ and its longest side is $1m$. Find its cost at the rate of $Rs. 250$ per $m^2$.
AnswerOne side $BC$ of a right triangular scarf $= 80\ cm$
and longest side $\mathrm{AC}=1 \mathrm{~m}=100 \mathrm{~cm}$
By Pythagoras Theorem,
$A C^2=A B^2+B C^2$
$\Rightarrow(100)^2=A B^2+(80)^2$
$\Rightarrow 10000=A B^2+6400$
$\Rightarrow A B^2=10000-6400$
$\Rightarrow A B^2=3600=(60)^2$
$\Rightarrow A B=60$
Second side $=60 \mathrm{~cm}$
Area of the scarf $=\frac{1}{2} \times \mathbf{b} \times \mathbf{h}$
$=\frac{1}{2} \times 80 \times 60 \mathrm{~cm}^2=2400 \mathrm{~cm}^2$
Rate of cost $= Rs. 250$ per $\mathrm{m}^2$
Total cost $=\frac{2400}{100 \times 100} \times 250= Rs. 60$ View full question & answer→Question 833 Marks
The area of a square plot is $6084m^2$. Find the length of the wire which can go four times along the boundary of the plot.
AnswerArea of the square plot $=6084\text{m}^2$
Side of the square plot $=(\sqrt{\text{Area}})$
$=(\sqrt{6084})\text{m}$
$=(\sqrt{78\times78})\text{m}=78\text{m}$
$\therefore$ Perimeter of the square plot $= 4 \times $ side $= (4 \times 78)m = 312m$
312m wire is needed to go along the boundary of the square plot once.
Required length of the wire that can go four times along the boundary $= 4 \times $ Perimeter of the square plot
$= (4 \times 312)m = 1248m$
View full question & answer→Question 843 Marks
The radius of the wheel of a car is $35\ cm$. How many revolutions will it make to travel $33\ km?$
AnswerRadius of the wheel $= 35\ cm$
Circumference of the wheel $=2\pi$
$=\Big(2\times\frac{22}{7}\times35\Big)\text{cm}=(44\times5)\text{cm}$
$=220\text{cm}$
Distance covered by the wheel in $1$ revolution $=\Big(\frac{11}{5}\Big)\text{m}$
Now, $\Big(\frac{11}{5}\Big)\text{m}$ is covered by the car in $1$ revolution.
Thus, $(33 \times 1000)m$ will be covered by the car in $\Big(1\times\frac{5}{11}\times33\times1000\Big)$ revolutions,
i.e. $15000$ revolutions.
$\therefore$ Required number of revolutions $= 15000$
View full question & answer→Question 853 Marks
The length and breadth of a rectangular park are in the ratio $5 : 2. A 2.5-m$-wide path running all around the outside of the aprk has an area of $305m^2$. Find the dimensions of the park.
Answer$\text { Ratio in length and breadth of a park }=5: 2$
$\text { Width of path outside it }=2.5 \mathrm{~m}$
$\text { Area of path }=305 \mathrm{~m}^2$
$\text { Let Inner length }(\mathrm{l})=5 \mathrm{x}$
$\text { and breadth }(\mathrm{b})=2 \mathrm{x}$
$\text { Outer length }(\mathrm{L})=5 \mathrm{x}+2 \times 2.5=(5 \mathrm{x}+5) \mathrm{m}$
$\text { Width }(B)=2 \mathrm{x}+2 \times 2.5=(2 \mathrm{x}+5) \mathrm{m}$
$\text { Area of path }=0 \mathrm{uter} \text { area - Inner area }$
$\Rightarrow(5 \mathrm{x}+5)(2 \mathrm{x}+5)-5 \mathrm{x} \times 2 \mathrm{x}=305$
$\Rightarrow 10 \mathrm{x}^2+10 \mathrm{x}+25 \mathrm{x}+25-10 \mathrm{x}^2=305$
$\Rightarrow 35 \mathrm{x}=305-25=280$
$\Rightarrow \mathrm{x}=8$
$\text { Length of park }=5 \mathrm{x}=5 \times 8=40 \mathrm{~m}$
$\text { and breadth }=2 \mathrm{x}=2 \times 8=16 \mathrm{~m}$
$\text { Dimensions of park }=40 \mathrm{~m} \text { by } 16 \mathrm{~m}$
View full question & answer→Question 863 Marks
The circumference of a circle is $264\ cm$. Find its area.
AnswerLet the radius of the circle be $r cm.$
Circumference $=(2\pi\text{r})\text{cm}$
$\therefore(2\pi\text{r})=264$
$\Rightarrow\Big(2\times\frac{22}{7}\times\text{r}\Big)=264$
$\Rightarrow\text{r}=\Big(\frac{264\times7}{2\times22}\Big)=42$
$\therefore$ Area of the circle $=\pi\text{r}^2$
$=\Big(\frac{22}{7}\times42\times42\Big)\text{cm}^2$
$=5544\text{cm}^2$
View full question & answer→Question 873 Marks
The height of a parallelogram is one-third of its base. If the area of the parallelogram is $108cm^2$, find its base and height.
AnswerLet the base of the parallelogram be $x cm .$
Then, the height of the parallelogram will be $\frac{1}{3} \mathrm{x} \mathrm{cm}$.
It is given that the area of the parallelogram is $108 \mathrm{~cm}^2$.
Area of a parallelogram $=$ Base $\times$ Height
$\therefore 108 \mathrm{~cm}^2=\mathrm{x} \times \frac{1}{3} \mathrm{x}$
$108 \mathrm{~cm}^2=\frac{1}{3} \mathrm{x}^2$
$\Rightarrow \mathrm{x}^2=(108 \times 3) \mathrm{cm}^2=324 \mathrm{~cm}^2$
$\Rightarrow \mathrm{x}^2=(18 \mathrm{~cm})^2$
$\Rightarrow \mathrm{x}=18 \mathrm{~cm}$
$\therefore \text { Base }=\mathrm{x}=18 \mathrm{~cm}$
$\text { Height }=\frac{1}{3} \mathrm{x}=\left(\frac{1}{3} \times 18\right) \mathrm{cm}$
$=6 \mathrm{~cm}$
View full question & answer→