Given, Q(0, 2) is equidistant from P(5, -3) and R(x, 7), which means PQ = QR.
Find the distance of PQ and QR using distance formula,
\(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}\)
\(\begin{aligned} P Q=\sqrt{(5-0)^2+(-3-2)^2}=5 \sqrt{2} \end{aligned}\)
\(\begin{aligned} Q R=\sqrt{(0-x)^2+(2-7)^2}=\sqrt{x^2+25} \end{aligned}\)
Now, \(\begin{aligned} P Q=Q R \Rightarrow 5 \sqrt{2}=\sqrt{x^2+25} \end{aligned}\)
On squaring both sides of the above equation, we get
50 = x² + 25 \(\Rightarrow\) x² = 25  \(\Rightarrow\) x = \(\pm 5\)
Therefore, the values of x are 5 and -5.

Area of restangle = 3 \(\times\) 2 = 6 m²
and area of circle of radius \(\frac{1}{2} \mathrm{~m}=\pi\left(\frac{1}{2}\right)^2=\frac{\pi}{4} \mathrm{~m}^2\)
\(\left[\because \text { diameter }=1 \mathrm{~m} \Rightarrow \text { radius }=\frac{1}{2} \mathrm{~m}\right]\)
Now, probanility that the die land inside the circle
\(=\frac{\text { Area of circle }}{\text { Area of rectangle }}=\frac{\pi / 4}{6}=\frac{\pi}{24}\)

Let AB is tower and car is at C on the highway.
 
In right \(\Delta\)ABC,
\(\frac { AB }{ BC } =\tan { { 30 }^{ o } } \)
\(\Rightarrow\)  \(\frac { 100 }{ BC } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow\)  \(BC=100\sqrt { 3 }\) m

Given, radius of a circle, AO = 10 cm and \(\angle\)AOC = 90°
Area of \(\triangle A O C=\frac{1}{2} \times O A \times O C=\frac{1}{2} \times 10 \times 10=50 \mathrm{~cm}^2\)
\(\begin{aligned} \text { Area of sector } O A E C O & =\frac{\theta}{360^{\circ}} \times \pi r^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times(10)^2 \\ \end{aligned}\)
\(\begin{aligned} =\frac{314}{4}=78.5 \mathrm{~cm}^2 \end{aligned}\)

(i) Area of minor segment AECDA
= Area of sector OAECO - Area of \(\Delta\)AOC
= 78.5 - 50 = 28.5 cm²
(ii) Area of major sector OAFGCO
= Area of circle - Area of sector OAECO
= 3.14 \(\times\)(10)² - 78.5
= 314 - 78.5 = 235.5 cm²

Let AB be a diameter of a given circle and LM and PQ be the tangent lines drawn to the circle at points A and B, respectively.

To prove LM || PQ
Proof We know that the tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OA \(\perp\) PQ and OB \(\perp\) LM
\(\Rightarrow\) AB \(\perp\) PQ
and AB \(\perp\) LM
\(\Rightarrow\) \(\angle\)PAB = 90°
and  \(\angle\)ABM = 90°
\(\Rightarrow\) \(\angle\)PAB = \(\angle\)ABM
[each = 90°]
But these are alternate angles.
\(\therefore\) PQ || LM
Hence, the tangents drawn at the ends of a diameter of a circle are parallel.
Hence proved.

\(3 \mathrm{~cm} \text { and } \frac{3 \sqrt{3}}{2} \mathrm{~cm}\)

Let p(u) = 4u² + 8u = 4u(u+2)
To find zeroes, put p(u) = 0
\(\Rightarrow\) 4u(u+2) = 0 \(\Rightarrow\) u = 0 or u + 2 = 0 [\(\because\) 4 \(\neq\)0]
\(\Rightarrow\) u = 0 or u = -2
Hence, zeroes of the given polynonial are 0 and -2.
Verification
Here, sum of zeroes = 0 - 2 = -2 = -(8/4)
=-\(\frac{Coefficient \quad of \quad u}{Coefficient \quad of \quad u^{2}}\)
and product of zeroes
=0 \(\times\)-2 = 0 = (0/4) = \(\frac{Constant \quad term}{Coefficient \quad of \quad u^{2}}\)
so, the relationship between the zeroes and its coefficients is verified.

Let fixed hostel charge (monthly) = Rs.y
and  cost of food for one day       = Rs.x
In case of student A,
25 x+ y = 4500     ....(i)
In case of student B,
30 x+ y = 5200       ....(ii)
Solve by cross-multiplication method.
x = 140, y = 1000

Using Euclid's algorithm, the HCF (30, 72)
72 = 30 x 2 + 12  ...(i)
30 = 12 x 2 + 6   ...(ii)
12 = 6 x 2 + 0    ...(iii)
HCF (30, 72) = 6
6 = 30-12 x 2   [From (ii)]
6 = 30 - (72 - 30 x 2) x 2
6 = 30 - 2 x 72 + 30 x 4
6 = 30 (1 + 4) - 72 x 2
6 = 30 x 5 + 72 x (- 2)
x = 5,y =-2
Also, 6 = 30 x 5 + 72(-2) + 30 x 72-30 x 72
6 = 30 x (77) + 72 x (-32)
∴ x=77, y=-32
Hence, x and yare not unique

44 cm

Given, side of the cube = Diameter of the hemisphere = l units
\(\therefore\) Radius of the hemisphere, \(r=\frac{l}{2}\) units

Now, required surface area of the remaining solid = TSA of the cube + CSA of hemisphere - Area of circular base of hemisphere
\(\begin{aligned} & =6 \times(\text { Edgc })^2+2 \pi r^2-\pi r^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 \times l^2+2 \pi \times\left(\frac{l}{2}\right)^2-\pi\left(\frac{l}{2}\right)^2 \\ \end{aligned}\)
\(\begin{aligned} & =6 l^2+2 \pi \times \frac{l^2}{4}-\pi \frac{l^2}{4}=6 l^2+\pi \frac{l^2}{4} \\ \end{aligned}\)
\(\begin{aligned} & =\frac{l^2}{4}(\pi+24) \text { sq units } \end{aligned}\)

Let AB be the tangent drawn at a point C on the circle with centre O.
To prove Perpendicular at point C passes through the centre O. If possible, let the perpendicular passing through some other point say O'.
Construction Join OC and O' C.

Proof Since, tangent at any point of a circle is perpendicular to the radius through the point of contact.
\(\therefore\) OC \(\perp\) AB \(\Rightarrow\) \(\angle\)OCB = 90^o
Also, \(\angle\)O'CB = 90^o  [as it is supposed that CO' \(\perp\)AB]
\(\therefore\)  \(\angle\)OCB = \(\angle\)O' CB
which is possible only when points O and O' coincide.
So, our assumption is wrong.
Hence, the perpendicular at the point of contact to the tangent to a circle always passes through the centre.
Hence proved.

Given, LCM (26, 91) = 182
\(\therefore \quad HCF(26,91)=\frac { 26\times 91 }{ LCM(26,91) } =\frac { 26\times 91 }{ 182 } =13\)

Let us draw a perpendicular OV on chord ST. It will bisect the chord ST.
SV = VT
In ΔOVS,
OV/OS = cos 60º
OV/12 = 1/2
OV = 6 cm
\(S \frac{V}{S} O=\sin 60^{\circ}=\frac{\sqrt{3}}{2}\)
\(\frac{S V}{12}=\frac{\sqrt{3}}{2} \)
\(S V=6 \sqrt{3} \mathrm{~cm} \)
\(S T=2 S V=2 \times 6 \sqrt{3}=12 \sqrt{3} \mathrm{~cm}\)
Area of ΔOST = 1/2 x ST x OV
\(\frac{1}{2} \times 12 \sqrt{3} \times 6 \)
\(=36 \sqrt{3}=36 \times 1.73=62.28 \mathrm{~cm}^{2}\)
Area of sector OSUT \(=\frac{120^{\circ}}{360^{\circ}} \times \pi(12)^{2}\)
Area of segment SUT = Area of sector OSUT − Area of ΔOST
= 150.72 − 62.28
= 88.44 cm²

Let ABCD be a parallelogram circumscribing a circle.
To prove ABCD is a rhombus.
i.e. to prove AB = BC = CD= DA
Proof We know that the tangents to circle from an external point are equal in length.

\(\therefore\)   AM = AP, BM = BN, CO = CN and DO = DP
On adding all above equations, we get
(AM + BM) + (CO + DO) = AP + BN + CN + DP
\(\Rightarrow\) AB + CD = (AP + PD) + (BN + NC)
= AD + BC                          ....(i)
Given, ABCD is a parallelogram.
\(\therefore\) AB = CD and BC = AD   ....(ii)
[\(\because\) opposite sides of a parallelogram are equal]
Then, from Eq. (i), we get
2 AB = 2BC
\(\Rightarrow\)   AB = BC  ...(iii)
From Eqs. (ii) and (iii), we get
AB = BC = CD = DA
\(\Rightarrow\) ABCD is a rhombus.
Hence, the parallelogram circumscribing a circle is a rhombus.
Hence proved.

Total surface area (TSA) of toy = CSA of hemisphere + CSA of cone
This is because the toy is obtained by joining the plane surfaces of hemisphere and cone.

If Assertion is incorrect but Reason is correct.

Given matchstick pattern

(i) The AP corresponding to number of squares in each figure is 1, 5, 9 .........
First term of AP = 1
Common difference of AP = 5 - 1 = 4
(ii) The AP corresponding to the number of sticks used in each figure is 4, 16, 28 ..........
First term of AP = 4
Common difference of AP = 16 – 4 = 12
(iii) (a) From part (i), the AP corresponding to number of squares is 1, 5, 9, ......
Here, a = 1 and 5 - 1 = 4
We have to find a₁₀ 
We know that a_n = a + (n - 1)d
= 1 + (10 - 1)4
=1 + 9  \(\times\) 4 = 37
\(\therefore\) Fig. (10) will have 37 squares.
From part (ii), the AP we get corresponding to number of matchstick used is 4, 16, 28, ...
Here, a = 4 and d = 16 - 4 = 12
We have to find a₁₀.
We know that
a_n = a + (n - 1)d
a₁₀ = 4 + (10 - 1) 12
= 4 + 9 \(\times\) 12
= 4 + 108 = 112
\(\therefore\) Fig. (10) will have 112 matchsticks.
Or
(b) We have given, a_m = 88
a + (m - 1)d = 88
\(\Rightarrow\) 4+(m - 1)12 = 88      [\(\because\) a = 4, d = 12]
\(\Rightarrow\)    (m - 1)12 = 84
\(\Rightarrow\)      m - 1 = 7
m = 8
\(\therefore\)   Fig. (8) will have 88 matchsticks.
To find number of squares in 8th figure we have to find a₈ for the AP 1, 5, 9, .......
a₈ = a + (8 - 1)d
a₈ = 1 + 7 \(\times\) 4 = 29
Therefore, 8th figure will have 29 squares.

(i) (b): Volume of the earth taken out
\(=\pi\left(\frac{7}{2}\right)^{2} \times 12=\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 12=462 \mathrm{~m}^{3}\)
(ii) (d): Area of the rectangular field = 30 x 15 = 450 m²
(iii) (a): Area of top of the pit = \(=\pi\left(\frac{7}{2}\right)^{2} =\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \)
\(=\frac{77}{2}=38.5 \mathrm{~m}^{2}\)
(iv) (d): Area of the remaining field = Area of rectangular field - area of top of pit
= 450 - 38.5 = 411.5 m²
(v) (c): The rise in the level of field = \(=\frac{462}{411.5}=1.12 \mathrm{~m}\)

(i) (b): Here 80 = 2⁴ x 5, 85 = 17 x 5
and 90 = 2 x 3² x 5
L.C.M of 80, 85 and 90 = 2⁴ x 3 x 3 x 5 x 17 = 12240
Hence, the minimum distance each should walk when they at first time is 12240 cm.
(ii) (c): Here 594 = 2 x 3³ x 11 and 189 = 3³ x 7
HCF of 594 and 189 = 3³= 27
Hence, the maximum number of columns in which they can march is 27.
(iii) (c) : Here 768 = 2⁸ x 3 and 420 = 2² x 3 x 5 x 7
HCF of 768 and 420 = 2² x 3 = 12
So, the container which can measure fuel of either tanker exactly must be of 12litres.
(iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm
Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5
HCF = 5 x 5 x 3 = 75
Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm.
(v) (a): LCM of 8 and 12 is 24.
\(\therefore \)The least number of pack of pens = 24/8 = 3
\(\therefore \)The least number of pack of note pads = 24/12 = 2