\(\frac{2}{3},\frac{4}{7},\frac{7}{9},\frac{8}{11},\frac{9}{17}\)

Given: A triangle ABC
such that AB = AC.
Bisector of ㄥA is AD.
To prove: ㄥADB = ㄥADC=90° and ​​​​​​​\(\overline { BD } =\overline { CD } \)
Proof: In \(\triangle \)ADB and \(\triangle \)ADC, We have
\(\overline { AD } =\overline { AD } \)
ㄥBAD=ㄥCAD
[\(\because \) AD is the bisector of ㄥBAC]
\(\overline { AB } =\overline { AC } \) [Given]
\(\therefore \)\(\triangle \)ADB ≅\(\triangle \)ADC [SAS congruency]
\(\Rightarrow \) Their corresponding parts are equal.
\(\therefore \quad \overline { BD } =\overline { CD } \)
i.e. the side BC is bisected at D.
Also ㄥADB =ㄥADC
But ADB and ADC form a linear pair.
\(\therefore \)ㄥADB +ㄥADC=180°
ㄥADB+ㄥADB=180°
\(\Rightarrow \) 2ㄥADB=180°
\(\Rightarrow \)\(\angle ADB=\frac { 180^{ \circ } }{ 2 } \Rightarrow \angle ADB={ 90 }^{ \circ }\)
i.e. AD \(\bot \)BC
Hence AD is the perpendicular bisector of BC.
​​​​​​​

Now, for n = - 2
5n² + 5n-2 = 8 and n³ = (- 2)³ = (- 2) x (- 2) x (- 2) = - 8
Combining,
n³ + 5n² + 5n - 2 = - 8 + 8 = o.

The cost of a flower vase = Rs.120
Loss % = 10%
Price at which it is sold =?
\(\frac { 10 }{ 100 } \times 20\)
=Rs.12
\(\therefore \) S.P=C.P-Loss
=Rs.120-Rs.12
=Rs.108

(i) ➝ Side, (ii) ➝ Top, (iii) ➝ Front

In the given figure, a perpendicular is drawn at both Band B'.
\(\angle XBB'=\angle YB'B=90^0\)
∴ The angle between made by rays with BB' is equal to 90°. Hence, both rays BX II B'Y.

we have, \(\frac { { 2 }^{ 8 }\times { a }^{ 5 } }{ { 4 }^{ 3 }\times { a }^{ 3 } } =\frac { { 2 }^{ 8 }\times { a }^{ 5 } }{ { \left( { 2 }^{ 2 } \right) }^{ 3 }\times { a }^{ 3 } } \)    [\(\because\) 4 = 2 x 2 = 2²]
\(=\frac { { 2 }^{ 8 }\times { a }^{ 5 } }{ { 2 }^{ 2\times 3 }\times { a }^{ 3 } } =\frac { { 2 }^{ 8 }\times { a }^{ 5 } }{ { 2 }^{ 6 }\times { a }^{ 3 } } \)                  [\(\because\) (a^m)ⁿ = a^mn]
=2^8-6 x a^5-3 =2² x a²      [\(\because\)a^m \(\div\)aⁿ =a^m-n]
= (2a)²                  [\(\because\)a^m x b^m = (ab)^m]

In an isosceles triangle, two sides are equal in length.
\(\therefore\) Perimeter of an isosceles triangle
                 = Sum of length of all three sides
                 = 2 x 5 + 6 [since, two sides are equal]
                 = 10 + 6 = 16 cm.

On copying the figure with punched holes, the axis of symmetry corresponding to the punched holes are shown by dotted lines in the figures given below

Since, interior opposite angles are 45° and 65°.
We know that, the exterior angle of a triangle is equal to sum of its interior opposite angles.
So, x = 45°+ 65° = 110°

Let the two complement angles be 2x and 7x.
\(\therefore\) 2x + 7x = 90°
[\(\because\) the sum of two complement angles = 90°]
\(\Rightarrow\) 9x = 90° \(\Rightarrow\) x = 10°​​​​​​​ ​​​​​​​
So, the angles are 2\(\times\)10°​​​​​​​ = 20°​​​​​​​ and 7\(\times\)10°​​​​​​​=70°​​​​​​​

We have, (- 41) \(\times\) 102 = - (41 \(\times\)102)
[\(\because\) (- a) \(\times\) b = - (a \(\times\) b)]
= - [41 x (100 + 2)] = -(41 \(\times\)100 + 41 \(\times\) 2)      [ by distributive property of multiplication over addition ]
= - [4100 + 82] = - [4182] = - 4182

0

Mixed fraction = 25\(\frac{1}{3}\)

According to the question,
Number b divided by 5 =\(\frac { b }{ 5 } \) and the quotient = 6  Hence, the required equation is \(\frac { b }{ 5 } \)=6

(i) 60°
(ii) 30°
(iii) 140°
(iv) 50°
(v) 35°
(vi) 60°

(-4)² or 16

Given, =2x + 3,  = \(\frac { 3 }{ 2 } \) + 7 and  = x -3
i) = 2[2x6+3]+\(\frac { 3 }{ 2 } \)x 3+7 - (1-3)
= 2 x 15 +\(\frac { 9 }{ 2 } \) + 7 + 2=30 + \(\frac { 9 }{ 2 } \) + 9
= 39 + \(\frac { 9 }{ 2 } \) = \(\frac { 78+9 }{ 2 } =\frac { 87 }{ 2 } \)
ii) 
\(\frac { 1 }{ 2 } \left[ \frac { 3 }{ 2 } \times 2+7 \right] +[8-3]-3\left[ 2\times 0+3 \right] \)
\(\frac { 1 }{ 2 } \)[10] +5-9 = 5 + 5 -9 = 1

Line of symmetry = 2, Order of symmetry = 4

(i) The given net of a cube shows edge GH meets edge AN.
(ii) The given net of a cube shows, DC meets DE edge.

(a) In the parallelogram ABCD, AD = BC, AB = DC
Base = 300 m and height = 80 m
\(\therefore\)  Area of a parallelogram = Base x Height
so, area of parallelogram = 300 m x 80 m
                                    = 24000 m²
(b) The values are depicted here is that owner of the plot is a very helpful cooperative to the society.

Given, Lalita reads the book for 2\(\frac{4}{8}\)h everyday. If she reads the entire book in 64 days.
∴ The total number of reading hours
= 64 x Reading hours per day
\(=64 \times 2 \frac{4}{8}=64\times \frac{(2\times 8)+4}{8}\)
=\(64 \times \frac{16+4}{8}=64\times \frac{20}{8}= \frac{64}{1}\times \frac{20}{8}\)
=\( \frac{64 \times 20}{8}=8\times 20=160 h\)
Hence, Lalita reads the entire book in 160 h.  

Here, we use the following steps:
Steps of construction
Step I Draw a line I and take any point O on it.

Step II With O as centre and any radius, draw an-arc to intersect line I at P and Q.

Step III Now, with P and Q as centres and radius greater than \(\frac{1}{2}PQ,\) draw two arcs respectively which intersects each other at R.

Step IV Join RO and produce it to B, then OB is perpendicular to I.

Step V With O as centre and radius 4 cm, draw an arc which intersect OB at X, then OX = 4 cm.

Step VI Take O as centre and draw an are, which intersects I and OB at M and N, respectively.

Step VII With X as centre and with same radius as in Step VI, draw an arc JK which cuts OX at j.

Step VIII Place the pointed tip of the compasses at M and adjust the opening so that the pencil tip is at N. With this opening and with J as centre, draw an arc cutting the arc JK at H.

Step IX Now, join XH to draw a line m.

Thus, m || l.

(i) Given, ray AZ i.e. AC is the bisector of ㄥDAB as well as ㄥDCB.
∴ ㄥDAC = ㄥBAC and ㄥDCA = ㄥBCA
Now, three pairs of equal parts in ΔBAC and ΔDAC are
ㄥDAC = ㄥBAC
[since, AC is the bisector of ㄥDAB]
AC = AC [common]
and ㄥDCA = ㄥBCA [since,AC is the bisector of ㄥDAB]
(ii) Yes, in ΔBAC and ΔDAC, we have
ㄥDAC = ㄥBAC, AC = AC, ㄥDCA = ㄥBCA
So, by ASA congruence rule, two triangles are congruent.
The correspondence is A ↔️ A, C ↔️ C and D ↔️ B.
In symbolic form, ΔBAC  ≅ ΔWAC
iii) Yes, here ΔBAC ≅ ΔDAC
We know that, the corresponding parts of two congruent triangles are equal.
So, AB = AD [corresponding sides]
(iv) Yes, Here, ΔBAC ≅ ΔDAC
We know that, the corresponding parts of two triangles are equal.
So, CD = CB [corresponding sides]

Rs 12246.15

We have, x-(y+z)
 \(=\frac { -4 }{ 9 } -\left( \frac { 5 }{ 12 } +\frac { 7 }{ 18 } \right) \\ =\frac { -4 }{ 9 } -\left( \frac { 5\times 3+7\times 2 }{ 36 } \right) =\frac { -4 }{ 9 } -\left( \frac { 15+14 }{ 36 } \right) \\ =\frac { -4 }{ 9 } -\frac { 29 }{ 36 } =\frac { -4\times 4-29 }{ 36 } \\ =\frac { -16-29 }{ 36 } \\ =\frac { -45 }{ 36 } =\frac { -5 }{ 4 } \)

For a pair of integers whose product is -36 and whose difference = 15.
So, first integer = -3 and second integer = 12
Their product = (-3) \(\times\) 12 = -(3 \(\times\) 12)= -36
and the difference between these two integer is 15.

Given, the greater angle, the better chance of the player scoring a goal.
(a) From the figure, seven football players are practicing their kicks. They are lined up in a straight line in front of the goalpost. So, fourth player have the best (the greatest) kicking angle.

(b) From the figure, fourth player has the best kicking angle.

(c) We know that, when the sum of the measures of two angles is 90°, then the angles are called complementary angles.
So, two players are complement to each other (i.e. 45°).

As per the given information in the question, the perimeter of the rectangle is 180 cm.
Let x be the width of the rectangle.
So, length of the rectangle will be 2x.
Perimeter of a rectangle = 2 length + 2 width
\(\therefore\)   2x + 2(2x) = 180  \(\Rightarrow\) 6x = 180
              \(x=\frac{180}{6}=\) 30cm
Hence, width of the rectangle is 30 cm and length of the rectangle is 2 x 30 = 60 cm.

(a) For company A Number of tested watches=40
Number of leaked watches = 20
Now, required fraction = \(\frac { Leaked\quad watches }{ Tested\quad watches } =\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \)  [dividing numerator and denominator by 10]
For company B Number of tested watches = 40
Number of leaked watches = 10
Required fraction  = \(\frac { 10 }{ 40 } =\frac { 1 }{ 4 } \) [dividing numerator and denominator by 10]                     
For company C Number of tested watches = 40
Number of leaked watches = 15
Required fraction  = \(\frac { 15 }{ 40 } =\frac { 3 }{ 8 } \) [dividing numerator and denominator by 5]For company D Number of tested watches = 40
Number of leaked watches = 25
Required fraction = \(\quad \frac { 25 }{ 40 } =\frac { 5 }{ 8 } \) [dividing numerator and denominator by 5]
(b) From above discussion, it is clear that company B has better watches, because company B has least fraction of the number of watches that leaked to the number of watches that were tested.

We have
\(\frac{4}{5}=\frac{4\times7}{5\times7}=\frac{28}{35}\)
\(\frac{3}{7}=\frac{3\times5}{7\times 5}=\frac{15}{35}\)
\(\frac{28}{35}>\frac{15}{35}\Rightarrow\frac{4}{5}>\frac{3}{7}\)
\(\Rightarrow \frac{-4}{5}<\frac{-3}{5}\Rightarrow\frac{-3}{7}>\frac{-4}{5}\)

(i) Take a = 3, b = 4, c = 5
a \(\times\) (b - c) = 3 \(\times\) (4 - 5)
= 3 \(\times\) (- 1) = - 3 
a\(\times\)b-a\(\times\)c=3\(\times\)4-3\(\times\)5
= 12 -15 =-3
\(\therefore\) a \(\times\) (b - c) = a \(\times\) b - a \(\times\) c
(ii) Take a = - 3, b = - 4, c = - 5
(iii) Take a = 3, b = 4, c = - 5
(iv) Take a = 3, b = - 4, c = 5
(v) Take a = - 3, b = 4, c = 5
(vi) Take a = 3, b = - 4, c = - 5
Verify the remaining parts yourself exactly as in (i).

- Multiply 27 and 135. We get 3645.
- In 2.7, there is 1 digit to the right of the decimal point and in 1.35, there are 2 digits to the right of the decimal point.
- So, count 1 + 2 = 3 digits starting from the right most digit (i.e., 5) in 3645 and move towards left.
- We get 3.645.

Multiplying both sides by 3, we have
\(\frac{-p}{3}\)\(\times\)3=5\(\times\)3
or -p=15
or -p\(\times\)(-1)=15\(\times\)(-1)
[multiplying both sides by(-1)]
Thus, p=-15 is the required solution.

In the figure, two sides of the triangle are equal.
\(\therefore\)The base angles opposite to the equal sides are equal.
Since, one of the base angles is y,
\(\therefore\) Other base angle = y
Now, y and 120° form a linear pair,
\(\therefore\)y + 120° = 180°
or y = 180° - 120° = 60°
Now, sum of the three angles = 180°
\(\therefore\) x + Y + Y = 180°
or x + 60° + 60° = 180°
or x + 120° = 180°
or x = 180° - 120° = 60°
Thus, x = 60° and y = 60°.

In ratio, we would write this as
Rice : Urad dal = 2 : 1
= 2 + 1 = 3 total parts
Now, thi s means \(\frac { 2 }{ 3 } \)X100%
=\(\frac { 200 }{ 3 } =66\frac { 2 }{ 3 } \)%
and,  urad dal in % =\(\frac { 1 }{ 3 } \times 100\)% = 33 \(\frac { 1 }{ 3 } \)%

6234269

Given, circumference of the circular sheet = 154 m
\(\therefore\) Circumference of a circular sheet = 2\(\pi\)r
\(\Rightarrow \)         154 = 2\(\pi\)r
\(\Rightarrow \)             r = \(\frac{154}{2\pi}=\frac{154\times 7}{2\times 22}=\frac{7\times 7}{2}=\frac{49}{2}\)m
\(\therefore\)  Diameter = 2 x Radius
                   = 2 x \(\frac{49}{2}\)
                   = 49 cm.

The order of rotation is 4 and angle of rotation of given figure is \(\frac { 360° }{ 4 } =90°\)

Given, 5 abc + 7p³q³
Total terms = 2, i.e. 5abc and 7p³q³
Tree diagram

(i) Oblique sketch of the cube of edge 4 cm is as follows:

(ii) An isometric sketch of this cube is as follows:

Since, \(\angle \)b and \(\angle \)132° are pair of interior angles on the same side of transversal line.
\(\therefore\) \(\angle \)b + 132° =180° [\(\because\) cointerior angles]
\(\Rightarrow\) \(\angle \)b=180°-132°\(\Rightarrow\)  \(\angle \)b=48°
Also, \(\angle \)a + \(\angle \)b and \(\angle \)65° are pair of interior angles
on the same side of transversal line.
\(\therefore\) \(\angle \)a + \(\angle \)b + 65° =180° [by linear pair]
\(\Rightarrow\) \(\angle \)a + 48° + 65° =180°
\(\Rightarrow\) \(\angle \)a=180°-113°  \(\Rightarrow\)\(\angle \)a=67°
Hence, \(\angle \)a=67° and \(\angle \)b=48°

Five years

In Δ ABC and Δ PQR, there are side possible matchings or correspondences. Out of them, four correspondences are as follow:
(i) ABC ↔️ PRQ
(ii) ABC ↔️ RPQ
(iii) ABC ↔️ RQP
(iv) ABC ↔️ QPR
Yes, all these correspondences may lead to congruence.

(-3)⁴ x (3)⁶ = (-3)⁴⁺⁶ = (3)¹⁰

200 : 400 = 1 : 2

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°

ABCD is the rhombus such that its diagonals AC and BD intersect at O. Here BD = 12 cm.

Suppose AC = x cm.
Since, the area of a rhombus =\(\frac{1}{2}\)\(\times\)[Product of 2 its diagonals]
\(\therefore\)Area of the rhombus ABCD =\(\frac{1}{2}\)\(\times\) 12 \(\times\) x cm²
2 But the area of the rhombus ABCD = 96 cm²
\(\Rightarrow\)\(\frac{1}{2}\)\(\times\)12 \(\times\) x = 96
\(\Rightarrow\)x=\(\frac{96\times2}{12}\)cm \(\Rightarrow\)x=8\(\times\)2cm
\(\Rightarrow\) x=16 cm
Since, the diagonals of a rhombus, bisect each other at right angles.
\(\therefore\)\(\angle \)COD = 90°, OC =\(\frac{1}{2}\)\(\times\) 16 cm=8 cm
and Od=\(\frac{1}{2}\)\(\times\) 12 cm =6 cm
Now, in right \(\Delta\)COD, we have
\(\Rightarrow\)OD² + OC² = CD² \(\Rightarrow\) 6² + 8² = CD²
\(\Rightarrow\)36 + 64 = CD² \(\Rightarrow\)100 = CD²
\(\Rightarrow\) 10² = CD² \(\Rightarrow\) CD = 10 cm.
Since, all the sides of rhombus are equal and perimeter of a rhombus = 4 x Side.
\(\therefore\) Perimeter ofthe rhombus = 4 x 10 em = 40 cm

Substituting a = 3 and b = 2 in
(a) a + b, we get
a + b = 3 + 2 = 5
(b) 7a - 4b,  we get
7a - 4b = 7 x 3 - 4 x 2
= 21 - 8 = 13
(c) a² + 2ab + b², we get
a² + 2ab + b² = 3² + 2 x 3 x 2 + 2²
= 9 + 2 x 6 + 4
= 9 + 12 + 4 = 25.
(d) a3 - b3, we get
a³ - b³ = 3³ - 2³ = 3 x 3 x 3 - 2 x 2 x 2
= 27 -8 = 19.

(25)^n-1 + 100 = 5^(2n-1)
⇒ (5²)^n-1 + 100 = 5^(2n-1)
⇒ 5^2n-2 + 100 = 5^2n-1
⇒ 5^2n-2 - 5^2n-1 = - 100
⇒ 5^{2n - 1} - 5^{2n- 2} = 100
⇒ 5^2n-2 x (5 -1) = 100
⇒ 5^2n-2 x 4 = 100
\(⇒\ 5^{2n-2}={100\over 4}=25\)
Thus, 5^{2n- 2} = 5²
As base is same on both the sides
∴ 2n-2 = 2
⇒ 2n=2+2
⇒ 2n = 4
\(⇒\ n={4\over 2}=2\)

Given, \(\frac { 2x-3 }{ 5 } +\frac { x+3 }{ 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 4(2x-3) }{ 5\times 4 } +\frac { 5(x+3) }{ 5\times 4 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12 }{ 20 } +\frac { 5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 8x-12+5x+15 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow \quad \frac { 13x+3 }{ 20 } =\frac { 4x+1 }{ 7 } \)
\(\Rightarrow\) 7(13x + 3) = 20(4x + 1)
\(\Rightarrow\) 91x + 21 = 80x + 20
\(\Rightarrow\) 91x - 80x = 20 - 21
\(\Rightarrow\) 11x = -1
Thus, \(x=-\frac { 1 }{ 11 } \)

Let the total marks = x
Then, marks for accent and presentation
= 20% of x
=\(\frac { 20 }{ 100 } \times x=\frac { x }{ 5 } \)
Remaining marks = x - \(\frac { x }{ 5 } ={5x-x\over5}={4x\over 5}\)
Now, marks for subject matter =60% of \(\frac { 4x }{ 5 } \)
\(\frac { 60 }{ 100 } \times \frac { 4x }{ 5 } =\frac { 12x }{ 25 } \)
Remaining marks \(={4x\over 5}-{12x\over 25}={5\times4x-12x\times1\over25}\)
\(={20x-12x\over25}={8x\over 25}\)
According to the question, there are 8 marks for rebuttal.
so,  \(\frac { 8x }{ 25 } =8\Rightarrow 8x=25\times 8\)
\(\Rightarrow \quad x=\frac { 25\times 8 }{ 8 } 25\)

Arranging the observations (marks) in ascending order:
2, 2, 5, 5, 5, 8, 8, 8, 8, 13, 13, 13, 13, 16, 16, 16, 17, 17, 18,18,18,18,18,19,20
Sum of observations
\(2\times2=4\\5\times3=15\\8\times4=32\\13\times4=52\\16\times3=48\\17\times2=34\\18\times5=90\\19\times1=19\\20\times1=20\\\quad\quad\quad\quad\_\_\_\_\\\ Total=314\\ \quad\quad\quad\quad\_\_\_\_\)Number of students = 25
\(\therefore\) Mean marks \(=\frac{Sum\ of\ the\ observations}{Total\ number\ of\ students}\)
\(=\frac{314}{25}=12.5\)
So, the number of students who scored marks more than mean marks is
4 + 3 + 2 + 5 + 1 + 1 = 16.

Given pattern is
\(-\frac { 1 }{ 3 } ,\frac { 2 }{ 6 } ,\frac { 3 }{ 9 },-\frac { 4 }{ 12 } ...\)
Here, \(-\frac { 1 }{ 3 } =\frac { (-1)\times 1 }{ 3\times 1 } \)
\(-\frac { 2 }{ 6 } =\frac { (-1)\times 2 }{ 3\times 2 } \)
\(-\frac { 3 }{ 9 } =\frac { (-1)\times 3 }{ 3\times 3 } \)
and \(-\frac { 4 }{ 12 } =\frac { (-1)\times 4 }{ 3\times 4 } \)
Hence, next four numbers are
\(\frac { (-1)\times 5 }{ 3\times 5 } =-\frac { 5 }{ 15 } \)
\(\frac { (-1)\times 6 }{ 3\times 6 } =-\frac { 6 }{ 18 } \)
\(\frac { (-1)\times 7 }{ 3\times 7 } =-\frac { 7 }{ 21 } \)
\(\frac { (-1)\times 8 }{ 3\times 8 } =-\frac { 8 }{ 24 } \).

In the above figure, AB and CD are two poles whose heights are 9 m and 14m respectively.
\(\Rightarrow\) AB = EC = 9m
and BD = 13m
DE = 14 - 9
= 5m
Now in right ΔBDE, by Pythagoras
BD² = BE² + DE²
13² = BE² + 5²
\(\Rightarrow\) BE² = (13)² - (5)²
= 169-25
BE² = 144
\(\Rightarrow\) BE = \(\sqrt { 144 } \)
\(\Rightarrow\) BE = 12m.
Hence, distance between their feet = 12 m.