\(\angle QOR+\angle QPR={ 180 }^{ ° }\)
(Supplementary angle)
\(\Rightarrow \quad \angle QOR+{ 46 }^{ ° }={ 180 }^{ ° }\)
\(\Rightarrow \quad \angle QOR={ 180 }^{ ° }-{ 46 }^{ ° }={ 134 }^{ ° }\)

120\(°\)

Given, \({ t }_{ n }=\begin{cases} { n }^{ 2 },where\ n\ in\ even \\ { n }^{ 2 }-1,where\ n\ is\ odd \end{cases}\)
For 19th term, i.e. n = 19 which is odd, we take \({ t }_{ n }={ n }^{ 2 }-1={ \left( 19 \right) }^{ 2 }-1=361-1=360\) 

5.7 cm

As 8 is a root of the equation x²-10x+k=0
∴ (8)²-10(8)+k=0
⇒ 64-80+k=0 ⇒ k=16
Hence, the value of k is 16.

15m

14 cm

40m

32.78m, 20.78m

Since distance between the store and first flag towards right side in 2 m.

\(\therefore\)  Distance covered by Ruchi to place the flag first towards right side of the store = 4 m
Distance covered by Ruchi to place the second flag towards right side of the store = 8 m
Distance covered by Ruchi to place the third flag towards right side of the store = 12 m
Similarly, distance covered by Ruchi to place the thirteenth flag towards right side of the store = a + 12d = 4 + 12 x 4 = 25 m
Now,  total distance covered in completing the job = 2S13
\(=2\{ \frac{13}{2}(2\times4+12\times4) \}=13(8+48)\)
= 13 x 56 = 728 m
Maximum distance covered by Ruchi for carrying a flag is 52 m.
Armed forces flag day is celebrated on 7th December, every year.

Given equation is x² - 4x + 4 = 0.
Here, a = 1, b = -4, c = 4
Discriminant (D) = b² - 4ac = (-4)² - 4(1)(4) = 0

\(k = - {1\over 2}\)

19.8 Km/h ; Keeping save energy in mind, cycle should be preferred as (i) it is good for health (ii) save energy (no fuel) (iii) no pollution.

6

\({15\over2}{\sqrt {3}} \ m \)

a = -1

S_n=2n²+5n
S₂₀=2(20)²+5x20
=2x400+100=900

In right \(\Delta\)ABC
\(\frac { AB }{ BC } =\tan { { 30 }^{ o } } \)
\(\Rightarrow\)  \(\frac { 150 }{ BC } =\frac { 1 }{ \sqrt { 3 } } \)
\(\Rightarrow\)  BC = 150\(\sqrt { 3 } \) feet

AB'C' is the required triangle.

(i) Draw a circle of radius 3 cm with O as its centre
(ii) Draw AB as diameter of thc circle.
(iii) A point P is taken on outer circle and OP is joined.
(iv) Perpendicular bisector of Op is drawn interesting OP at Q.
(v) With Q as centre and OQ as radius a circle is drawn intersecting the smaller circle at A and B
(vi) PA and PB is joined
(vii) PA and PB are the required tangents.
Length of tangent = 4 cm.

CA ⊥ AP
CB ⊥ BP
PA ⊥ PB
Also AP = PB
ஃ BPAC is a square.
⇒ AP = PB = BC = 4 cm

Joined PO and produces it to D.

Here OP ⊥ TP
⇒ \(\angle \)OPT = 90^o
Also TP | | AB
∴ \(\angle \)TPD + \(\angle \)ADP = 180^o
⇒ \(\angle \)ADP = 90^o
⇒ OD bisects AB
[Perpendicular from the centre bisects the chord]
In \(\triangle\)ADP and \(\triangle\)BDP
AD = BD [proved]; \(\angle \)ADP = \(\angle \)BDP
[Each 90o]; PD=PD
ஃ \(\triangle\)ADP\(\cong \) \(\triangle\)BDP
\(\angle \)PAB = \(\angle \)PBA
ஃ  \(\triangle\)PAB is isosceles triangle.

\(\frac { 1 }{ (a+b+x) } =\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ x } \)
\(\Rightarrow \frac { 1 }{ a+b+x } \frac { 1 }{ x } =\frac { 1 }{ a } +\frac { 1 }{ b } \)
\(\Rightarrow \frac { x-(a+b+x) }{ x(a+b+x) } =\frac { a+b }{ ab } \)
\(\Rightarrow \frac { -(a+b) }{ x(a+b+x) } =\frac { a+b }{ ab } \)
\(\Rightarrow x+(a+b+x)=-ab\)
\(\Rightarrow (x+a)(x+b)=0\)
\(\Rightarrow x=-aorx--b\)

Two digit numbers which are divisible by 7 are 14,21,28,..........98.
It forms an A.P.
a = 14, d = 7, a_n = 98
a_n = a + (n - 1)d
98 = 14 + (n -1)7
98 -14 = 7n - 7
84 + 7 = 7n
\(\Rightarrow\) 7n = 91
\(\Rightarrow\) n = 13

Given, radius of circle = 1 cm and distance between point P and centre = 2.2 cm.
(i) Draw a circle of radius 1 cm with centre O.
(ii) Take a point P outside it, such that its distance from centre O is 2.2 cm
(iii) Take O and P as centre and draw arcs of radius more than half of OP on both sides of OP which intersect each other at R and S. Join RS which bisect OP at M. Then, MP = MO.
(iv) Taking M as centre and MO as radius, draw a dotted circle which intersects given circle at Q and Q'.
(v) Join PQ and PQ'. Thus we get the required tangents drawn from point Pro the given circle.

Justification
Join OQ. Then \(\angle PQO\) is an angle in the semicircle and therefore.\(\angle PQO=90°\)This shows that \(OQ\bot PQ\) 
Since OQ is the radius of given circle, so PQ has to be a tangent of given circle.
Similarly, PQ' is also a tangent to the given circle.

1605

PR = 3.2 cm, QR = 4.6 cm

10.41 m

\(\frac { 8 }{ 3 } and\frac { 5 }{ 2 } \)

2

AD is the lighthouse, Find AD=?
In right ΔABD,
h=x
\(\frac { h }{ x } \)=tan45^o
\(\frac { h }{ 100-x } \)=tan30^o ......(ii)
Solve for h and x.
⇒ \(\frac { h }{ 100-x } =\frac { 1 }{ \sqrt { 3 } } \Rightarrow \sqrt { 3 } h\)=100-x
⇒ \(\sqrt { 3 } \)x=100-x   [Using eq (i)]
⇒ (\(\sqrt { 3 } \)+1)x=100 ⇒ x=\(\frac { 100 }{ \sqrt { 3 } +1 } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ (\sqrt { 3 } +1)(\sqrt { 3 } -1) } \)
⇒ x=\(\frac { 100(\sqrt { 3 } -1) }{ 2 } =50(\sqrt { 3 } -1)\)m
∴ h=height of lighthouse=\(50(\sqrt { 3 } -1)\)m

(ii) Clearly \(OB\bot AB\)
[\(\because\) tangents is perpendicular and encouraging to the radius through the point of contact greenery]
\(\therefore\angle AOB=90°\)
Now, in \(\Delta\)AOB, we have
OA² = OB² + AB²
\(\Rightarrow\) OA² = 5² + 12²    [\(\because\) OB=radius = 5 cm]
\(\Rightarrow\) OA² = 25 + 144 \(\Rightarrow\) OA = \(\sqrt { 169 } \) = 13 cm
(iii) Creativity, saving water, team work

(a + 2) + (b + 1),(a + 2) + (b + 2),(a + 3) + (b + 2)

(i) \((i)\frac { 1 }{ 2 } ,\frac { 3 }{ 4 } \)
(ii)\((i)-\frac { 1 }{ 2 } ,2\)

\(6\frac { 2 }{ 3 } \)cm

7200km/h

Let length of the shortest side = x m.
Then, hypotenuse = (2x + 6)m and third side
= (2x + 6 - 2)m = (2x + 4)m
By Pythagoras theorem,
(2x + 6)² = x² + (2x + 4)² [∵ H² = p2 + B²]
⇒ 4x² + 24x + 36 = x² + 4x² + 16x + 16 [∵ (a + b)² = a² + 2ab + b²)
⇒ x² + 4x² + 16x + 16 - 4x² - 24x - 36 = 0
⇒ x² - 8x - 20 = 0
By quadratic formula,
\(x=\frac{-(-8) \pm \sqrt{(-8)^{2}-4 \times 1 \times(-20)}}{2 \times 1}\)
\(\left[\because x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} ; \text { here } a=1, b=-8 \text { and } c=-20\right]\)
\(\Rightarrow \quad x=\frac{8 \pm \sqrt{64+80}}{2}\)
\(\Rightarrow \quad x=\frac{8 \pm \sqrt{144}}{2} \)
\(\Rightarrow \quad x=\frac{8 \pm 12}{2}\)
\(\Rightarrow \quad x=\frac{8+12}{2} \text { or } x=\frac{8-12}{2}\)
\(\Rightarrow \quad x=\frac{20}{2} \text { or } x=\frac{-4}{2}\)
⇒ x = 10 or x = - 2
But length of side cannot be negative.
∴ x = 10
Hence, shortest side is 10 m, hypotenuse is 2 x 10 + 6 = 26 m
and third side = 2 x 10 + 4 = 24 m.

Steps of construction:
Draw a line segment BC = 7 cm.

1. Draw a line segment BC of length 5 cm.
2. At B, draw LMBC = 60° and produced line BM.
3. From point C draw a line making an angle of 30°.
4. Both the lines intersect at A.
5.MBC is the given triangle.
6. Draw a ray BXmaking an acute angle.
7. Locate three points B1, Bz, and B3on line segment BX.
8. Join BC
9. Draw a parallel line through B3to B3C intersecting extended line BCat C.
10. Through C' draw a line parallel to AC intersecting extended line segment BA at A'. A'BC is the required triangle.

Let C₁ and C₂ be the two concentric circles, with centre at O and respective radii being r₁ = 4cm and r₂ = 5cm.
Draw a chord AC of circle C₂, to touch circle C₁ at B.
Join OB.

Here OB 丄 AC [As, tangent at any point of circle is perpendicular to radius through the point of contact]
Thus, in right angled OAB ,we have:
OA² = AB² + OB²     [By using phythagoras theorem]
⇒  52 =  AB² + 4²
⇒  AB² = 25 - 16  = 9
∴ Length of chord AC = 2 AB = 2 x 3 = 6 cm