[5 marks sum]

Question 15 Marks

A square lawn is surrounded by a path 2.5 m wide. If the area of the path is $165 m^2$ find the area of the lawn.

Answer

Area of path $=165 \mathrm{~m}^2$
Width of path $=2.5 \mathrm{~m}$
Let the side of square lawn $=x \mathrm{~m}$
$\therefore$ Outer side $=x+2 \times 2.5$
$ =(x+5) m $
$\therefore$ Area of path $=(x+5)^2-x^2$
$ \Rightarrow x^2+10 x+25-x^2=165 $
$ \Rightarrow 10 x=165-25=140 $
$ \Rightarrow x=\frac{140}{10}=14 \mathrm{~m} $
$\therefore$ Side of lawn $=14 \mathrm{~m}$
and area of lawn $=(14)^2 m^2=196 \mathrm{~m}^2$

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Question 25 Marks

The length and breadth of a rectangular piece of land are in the ratio $5: 3$. If the total cost of fencing it at the rate of ₹24 per meter is ₹9600, find its:
(i) length and breadth
(ii) area
(iii) cost of levelling at the rate of $₹ 60$ per $m ^2$.

Answer

Ratio in length and breadth of a rectangular piece of land $=5: 3$
Cost of fencing = ₹ 9600
And rate = ₹ 24 per $m$
Perimeter $=\frac{\text { Total cost of fencing }}{\text { Rate per } m}$
$=\frac{9600}{24}$
$=400 \mathrm{~m}$
Let length $=5 \mathrm{x}$
Then breadth $=3 \mathrm{x}$
$\therefore$ Perimeter $=2(l+b)$
$400=2 \times(5 x+3 x)$
$\therefore 16 x=400$
$x=\frac{400}{16}=25$
(i) $\therefore$ Length of land $=5 \mathrm{x}=5 \times 25=125 \mathrm{~m}$
and breadth $=3 x=3 \times 25=75 \mathrm{~m}$
(ii) Area $=1 \times b$
$=125 \times 75=9375 \mathrm{~m}^2$
(iii) Cost of levelling at rate ₹ 60 per $\mathrm{m}^2$
$=\text₹ 60 \times 9375 \mathrm{~m}^2=$ ₹ 5,62,500

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Question 35 Marks

The ratio between the areas of two circles is 16 : 9. Find the ratio between their :
(i) radius
(ii) diameters
(iii) circumference

Answer

(i) Let radius of first circle $=r_1$
and radius of second circle $=r_2$
Given that ratio of the areas of circles $=16: 9$
$\Rightarrow \frac{\pi r_1^2}{\pi r_2^2}=\frac{16}{9}$
$\Rightarrow \frac{\pi r_1^2}{\pi r_2^2}=\frac{4^2}{3^2}$
$\Rightarrow \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{4}{3}$
(ii) Let the diameter of first circle $=d_1$
and diameter of second circle $=\mathrm{d}_2$
Since, we know that diameter $=2 \times$ radius
$\therefore d_1=2 \times r_1=2 \times 4 x=8 x$
and $d_2=2 \times r_2=2 \times 3 x=6 x$
Now, the ratio between the diameter of two circles $=d_1: d_2$
$ =8 x: 6 x=4: 3 $
(iii) Now, consider the ratio of circumference of the circles
$ =\frac{2 \pi r_1}{2 \pi r_2}=\frac{r_1}{r_2}=\frac{4}{3} $
$\therefore$ The ratio between the circumference of two circles $=4: 3$

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Question 45 Marks

The sides of a rectangular park are in the ratio $4: 3$. If its area is $1728 m^2$, find
(i) its perimeter
(ii) cost of fencing it at the rate of ₹ $40$ per meter.

Answer

Ratio in the sides of a rectangle $=4: 3$
Area $=1728 \mathrm{~m}^2$
Let length $=4 \mathrm{x}$, and breadth $=3 \mathrm{x}$
$\therefore$ Area $=1 \times b$
$1728=4 \mathrm{x} \times 3 \mathrm{x}$
$\Rightarrow 12 x^2=1728$
$\Rightarrow \mathrm{x}^2=\frac{1728}{12}$
$\Rightarrow \mathrm{x}^2=144=(12)^2$
$\therefore x=12$
$\therefore$ Length $=4 \mathrm{x}=4 \times 12=48 \mathrm{~m}$
Breadth $=3 \mathrm{x}=3 \times 12=36 \mathrm{~m}$
(i) Now perimeter $=2(l+b)$
$=2(48+36) m$
$=2 \times 84=168 \mathrm{~m}$
(ii) Rate of fencing = ₹ 40 per metre
Total cost =168 $\times 40=$ ₹ 6720

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Question 55 Marks

The ratio between the radius of two circles is $5 : 7.$ Find the ratio between their:
(i) circumference
(ii) areas

Answer

(i) The ratio of the radius of the circles $=5: 7$
Let radius of first circle $=5 \mathrm{x}$
and radius of second circle $=7 \mathrm{x}$
$\therefore$ Circumference of first circle $=2 \pi r$
$ =2 \pi \times 5 x=10 \pi x $
and circumference of second circle
$ =2 \pi \times 7 x=14 \pi x $
$\therefore$ Ratio between their circumference
$ \begin{aligned} & =10 \pi x: 14 \pi x \\ & =10: 14=5: 7 \end{aligned} $
(ii) Area of first circle $=\pi r^2$
$ =\frac{22}{7} \times 5 \mathrm{x} \times 5 \mathrm{x}=\frac{550}{7} \mathrm{x}^2 $
and area of second circle $=\pi r^2$
$ =\frac{22}{7} \times 7 \mathrm{x} \times 7 \mathrm{x}=\frac{1078}{7} \mathrm{x}^2 $
Ratio between their areas
$=\frac{550}{7} x^2: \frac{1078}{7} x^2$
$=550: 1078 \ldots \ldots \ldots . . .(\text { Dividing by } 22)$
$=25: 49$

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Question 65 Marks

The sides of a triangle are in the ratio 15 : 13 : 14 and its perimeter is 168 cm. Find the area of the triangle.

Answer

Perimeter of the triangle $=168 \mathrm{~cm}$
Sum of ratio of sides $=15+13+14=42$
Let the first side $=\frac{168 \times 15}{42}=60 \mathrm{~cm}$
Second side $=\frac{168 \times 13}{42}=52 \mathrm{~cm}$
Third side $=\frac{168 \times 14}{42}=56 \mathrm{~cm}$
Now, $s=\frac{a+b+c}{2}$
$ =\frac{60+52+56}{2}=\frac{168}{2}=84 $
$\therefore$ Area $=\sqrt{s(s-a)(s-b)(s-c)}$
$ =\sqrt{84(84-60)(84-52)(84-56)} $
$ =\sqrt{84 \times 24 \times 32 \times 28} $
$ =\sqrt{2 \times 2 \times 3 \times 7 \times 2 \times 2 \times 2 \times 3 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 7} $
$ =2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 7 $
$ =1344 \mathrm{~cm}^2 $

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Question 75 Marks

The legs of a right-angled triangle are in the ratio $4 : 3$ and its area is $4056 cm^2$. Find the length of its legs.

Answer

Area of right-angled triangle $=4056 \mathrm{~cm}^2$
Legs of a right-angled triangled are in the ratio i.e. $4: 3$
Let one leg (Base) $=3 x$
Then second leg (altitude) $=4 x$

Area $=\frac{1}{2} \times$ Base $\times$ Altitude
$ \begin{aligned} & =\frac{1}{2} \times 3 \mathrm{x} \times 4 \mathrm{x}=6 \mathrm{x}^2 \\ & \therefore 6 \mathrm{x}^2=4056 \end{aligned} $
$ x^2=\frac{4056}{6}=676 $
$ x=\sqrt{676}=\sqrt{26 \times 26} $
$\therefore \mathrm{x}=26 \mathrm{~cm}$
$\therefore$ One leg $($ base $)=3 x=3 \times 26=78 \mathrm{~cm}$
and second leg (altitude) $4 x=4 \times 26=104 \mathrm{~cm}$

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Question 85 Marks

Find the area of an isosceles triangle whose base is 16 cm and the length of each of the equal sides is 10 cm.

Answer

In isosceles $\triangle A B C$
Base $B C=16 \mathrm{~cm}$
and $A B=A C=10 \mathrm{~cm}$

$\begin{aligned} & \text { Let } A D \perp B C \text { and } B D=\frac{1}{2} B C=\frac{16}{2} \\ & \therefore B D=8 \mathrm{~cm} \\ & \text { In right } \triangle A B D \\ & A B^2=A D^2+B D^2 \ldots \ldots \ldots \ldots .(\text { Pythagoras Theorem) } \\ & (10)^2=A D^2+(8)^2 \\ & 100=A D^2+64 \\ & 100-64=A D^2 \\ & 36=A D^2 \\ & A D=\sqrt{36}=\sqrt{6 \times 6} \\ & \therefore A D=6 \mathrm{~cm} \\ & N o w \text { the area of triangle }=\frac{\text { Base } \times \text { Altitude }}{2} \\ & =\frac{16 \times 6}{2}=48 \mathrm{~cm} 2\end{aligned}$

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Question 95 Marks

Find the area of a right-angled triangle whose hypotenuse is 13 cm long and one of its legs is 12 cm long.

Answer

In right-angled $\triangle A B C$,
Base $B C=12 \mathrm{~cm}$
and hypotenuse $\mathrm{AC}=13 \mathrm{~cm}$

Applying Pythagoras Theorem,
$ \begin{aligned} & (A C)^2=(A B)^2+(B C)^2 \\ & (13)^2=(A B)^2+(12)^2 \\ & 169=(A B)^2+144 \\ & (A B)^2=169-144 \\ & (A B)^2=25 \\ & \therefore A B=\sqrt{25} \\ & =\sqrt{5 \times 5}=5 \mathrm{~cm} \end{aligned} $
Now, area of $\triangle A B C=\frac{1}{2}$ base $\times$ altitude
$ =\frac{1}{2} \times 12 \times 5=30 \mathrm{~cm}^2 $

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Question 105 Marks

The adjacent sides of a parallelogram are 15 cm and 10 cm. If the distance between the longer sides is 6 cm, find the distance between the shorter sides.

Answer

In parallelogram $A B C D$
$ A B=D C=15 \mathrm{~cm} $
$ B C=A D=10 \mathrm{~cm} $

Distance between longer sides $A B$ and $D C$ is $6 \mathrm{~cm}$
i.e., perpendicular $\mathrm{DL}=6 \mathrm{~cm}$
$\mathrm{DM} \perp \mathrm{BC}$
Area of parallelogram $=$ Base $\times$ Altitude
$ =A B \times D L=15 \times 6=90 \mathrm{~cm}^2 $
Again let $\mathrm{DM}=\mathrm{xcm}$
$\therefore$ Area of parallelogram $A B C D=B C \times D M$
$ =10 \times x=10 x \mathrm{~cm}^2 $
$\therefore 10 \times \mathrm{cm}^2=90 \mathrm{~cm}^2$
$\Rightarrow x=\frac{90}{10}=9 \mathrm{~cm}$

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Question 115 Marks

Find the area of the rhombus, if its diagonals are 30 cm and 24 cm.

Answer

Given, diagonal $\left(d_1\right)=30 \mathrm{~cm}$
Other diagonal $\left(\mathrm{d}_2\right)=24 \mathrm{~cm}$

If $A C$ and $B D$ are the diagonals of a rhombus its
Area $=\frac{1}{2} \times$ Product of it diagonals
$ \begin{aligned} & =\frac{1}{2} \times \mathrm{AC} \times \mathrm{BD} \\ & =\frac{1}{2} \times \mathrm{d}_1 \times \mathrm{d}_2 \\ & =\frac{1}{2} \times 30 \times 24 \mathrm{~cm}^2 \\ & =15 \times 24=360 \mathrm{~cm}^2 \end{aligned} $
$\therefore$ Area of rhombus $=360 \mathrm{~cm}^2$

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Question 125 Marks

Find the perimeter of a rectangle with length = 24 cm and diagonal = 25 cm

Answer

Length of a rectangle $(\mathrm{I})=24 \mathrm{~cm}$, Diagonal $=25 \mathrm{~cm}$

Let breadth of the rectangle $=\mathrm{b} \mathrm{m}$
Applying Pythagoras Theorem in triangle $A B C$,
We get, $(A C)^2=(A B)^2+(B C)^2$
$ \begin{aligned} & (25)^2=(24)^2+(b)^2 \\ & 625=576+(b)^2 \\ & 625-576=b^2 \\ & 49=b^2 \\ & \sqrt{7 \times 7}=b \\ & \therefore b=7 \mathrm{~cm} \end{aligned} $
Now, perimeter of the rectangle
$ \begin{aligned} & =2(1+b) \\ & =2(24+7) \\ & =2(31) \\ & =62 \mathrm{~cm} \end{aligned} $

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Question 135 Marks

The length of a rectangular field is 30 m and its diagonal is 34 m. Find the breadth of the field and its perimeter.

Answer

Length $=30 \mathrm{~m}$
Diagonals $=34 \mathrm{~m}$

Let the breadth of the rectangle $=b \mathrm{~m}$
Applying Pythagoras Theorem in triangle $A B C$,
We get,
$ \begin{aligned} & A C^2=A B^2+B C^2 \\ & (34)^2=(30)^2+b^2 \\ & 1156=900+b^2 \\ & 1156-900=b^2 \\ & 256=b^2 \\ & \Rightarrow b \sqrt{256}=16 \mathrm{~m} \\ & \text { Perimeter }=2(1+b) \\ & =2(30+16) \\ & =2 \times 46 \\ & =92 \mathrm{~m} \end{aligned} $

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Question 145 Marks

Find the radius of the circle whose circumference is equal to the sum of the circumferences of the circles having radius 15 cm and 8 cm.

Answer

For circle with radius $=15 \mathrm{~cm}$
Circumference of circle $=2 \pi r$
$=2 \times \frac{22}{7} \times 15 \mathrm{~cm}$
$ =\frac{660}{7} \mathrm{~cm} $
For circle with radius $=8 \mathrm{~cm}$
Circumference of circle $=2 \pi \mathrm{r}$
$=2 \times \frac{22}{7} \times 8 \mathrm{~cm}$
$=\frac{352}{7} \mathrm{~cm}$
Sum of the circumferences of these two circles
$ =\frac{660}{7} \mathrm{~cm}+\frac{352}{7} \mathrm{~cm}=\frac{1012}{7} \mathrm{~cm} $
If the required radius $=R \mathrm{~cm}$
$ \begin{aligned} & \text { Its circumference }=2 \pi R \\ & =2 \times \frac{22}{7} \times \mathrm{R} \mathrm{cm} \\ & =\frac{44}{7} \mathrm{R} \mathrm{cm} \end{aligned} $
Given, $\frac{44}{7} \mathrm{R}=\frac{1012}{7}$
$ \Rightarrow \mathrm{R}=\frac{7}{44} \times \frac{1012}{7} \mathrm{~cm} $
$ =23 \mathrm{~cm} $
$\therefore$ Required radius $=23 \mathrm{~cm}$

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MATHS [5 marks sum]

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