b²

12x² - y² - 8xy

The total number of vertices in a cuboid is equal to 8.

Given, radius of circular card sheet = 14 cm
\(\therefore\) Area of circular card sheet = \(\pi\) x (Radius)²
                                          = \(\frac{22}{7}\times (14)^2=\frac{22}{7}\times 14\times 14\) = 616 cm²
Area of circle of radius 3.5 cm = \(\frac{22}{7}\times\) (3.5)²
                    = \(\frac{22}{7}\times 3.5\times 3.5=\frac{269.5}{7}\) = 38.5 cm²
\(\therefore\)  Area of two circles = 2 x 38.5 = 77 cm²
Now, length of rectangle, L= 3 cm
and breadth of rectangle, b = 1cm
\(\therefore\) Area of the rectangle = Length X Breadth = 3 x 1 = 3 cm²
Now, area of remaining sheet = Area of circular card sheet - (Area of two circles + Area of rectangle)
                                           = 616 - (77 + 3) = 616 - 80 = 536 cm²
Hence, the area of remaining card sheet is 536 cm².

Let the whole quantity be Rs x
\(\therefore\) 12% of \(x=Rs1080 \Rightarrow {12\over100}\times x=1080\)
\( \Rightarrow x ={1080\times100\over12}=90\times100=9000\)
Hence, the whole quantity is Rs. 9000.

Country B

Let the number be x.
Then, one-third of x = \(\frac{x}{3}\)
According to the question,
One-third of number added to 5 gives 8,
i.e,   \(\frac{x}{3}+5=8\)
Now, transposing (+5) from LHS to RHS, we get
        \(\frac{x}{3}=8-5\quad \Rightarrow \frac{x}{3}=3\)
Again, multiplying both sides by 3, we get
       \(\frac{x}{3}\times 3=3\times 3\quad \Rightarrow x=9\)
Hence, the required number is 9.

(i) Two line segments are congruent, if they have same length.
(ii) Among two congruent angles, one has a measure of 70°, i.ie measure of the other angle is 70°.
(iii) When we write ㄥA = ㄥB, we actually mean mㄥA = mㄥB.

 1. m \(\angle \) A + m \(\angle \) B = 85° + 115° = 200° > 180°
This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of the two angles of a triangle cannot exceed 180°. Hence, \(\Delta\) ABC cannot be constructed.
2. Steps of Construction
1. Draw QR of length 4.7 cm.
2. At Q, draw a ray QX making an angle of 30° with QR.
3. At R, draw a ray RY making an angle of 60° with RQ.
 
4. Mark the point of intersection of the two rays as P.
\(\Delta\) PQR is now completed.
3. Steps of Construction
By angle-sum property of a triangle,
m\(\angle\) A + m \(\angle\) B + m \(\angle\) C = 180°
\(\Rightarrow\) 70° + 50° + \(\angle\) C = 180°
\(\Rightarrow\) 120° +  \(\angle\)C = 180°
\(\Rightarrow\) \(\angle\)C = 180° - 120°
\(\Rightarrow\) \(\angle\)C = 60°
1. Draw AC of length 3 cm.
2. At A, draw a ray AX making an angle of 70° with AC.
3. At C, draw a ray CY making an angle of 60° with CA.

4. Mark the point of intersection of two rays at B.
\(\Delta\) ABC is now completed
4. m\(\angle\)L + m \(\angle\) N
= 60° + 120° = 180° = 180°
This is not possible since the sum of the measures of the three angles of a triangle is 180°. As such, the sum of the two angles of a triangle cannot be equal to 180°.
Hence, \(\Delta\) LMN cannot be constructed.
5. We have,
AC + BC = 2 cm + 2 cm = 4 cm = AB
This is not possible since the sum of the lengths of any two sides of a triangle is greater than the length of the third side. As such, the sum of the lengths of two sides of a triangle cannot be equal to the length of the third side. Hence, \(\Delta\) ABC cannot be constructed.
6. Steps of Construction
1. Draw a line segment QR of length 4 cm.
2. With Q as centre, draw an arc of radius 3.5 cm.
3. With R as centre, draw an arc of radius 3.5 cm.

4. Mark the point of intersection of arcs as P.
5. Join PQ and PR.
\(\Delta\) PQR is now ready.
7. Steps of Construction
1. Draw a line segment YZ of length 4 cm.

2. With Y as centre, draw an arc of radius 3 cm
3. With Z as centre, draw an arc of radius 5 cm.
4. Mark the point of intersection of arcs as X.
5. Join XY and XZ.
\(\Delta\) XYZ is now ready.
8. Steps of Construction
1. Draw a line segment EF of length 5.5 cm.
2. With E as centre, draw an arc of radius 4.5 cm.
3. With F as centre, draw an art of radius 4 cm

4. Mark the point of intersection of two arcs as D.
5. Join DE and DF.
\(\Delta\) DEF is now ready.

x³ - 5x² + 7x - 7

A wall of a room is of dimensions 5 m x 4 m.
Length of the room = 5 m
Breadth of the room = 4 m
\(\therefore\) Area of the room = 5 x 4= 20 m²
Length of the window = 1.5 m
Breadth of the window = 1 m
\(\therefore\) Area of the window = 1.5 x 1 = 1.5 m²
Length of the door = 2.25 m
Breadth of the door = 1 m
\(\therefore\) Area of the door = 2.25 x 1 = 2.25 m²
The area of the wall to be painted = Area of the room - Area of the window - Area of the door
                                                 = 20 - 1.5 - 2.25
                                                 = 20 - 3.75 = 16.25 m².

32.4

(a) Clearly, Imperial I, Imperial II and Ashok Towers are three tallest buildings.

(b) Average height of each storey of the buildings given in (a)
= \(\frac { sum\quad of\quad heights\quad of\quad each\quad storey\quad of\quad three\quad tallest\quad buildings }{ 3 } \)
= \(\frac { 4.15+4.15+3.94 }{ 3 } =\frac { 12.24 }{ 3 } =4.08\)
(c) We can clearly see from the data, Mumbai has maximum number of skyscrapers from the list given. It has 9 skyscrapers out of the list of 10 buildings given.
Required percentage = \(\frac { 9 }{ 10 } \times 100=90%\)%
(d) Range of data = Maximum height - Minimum height
= 249 - 156 = 93
(e) Arranging the data in ascending order, we get 156,170,170,170,180,181,184,193,249,249 Since, there are ten observations, median will be the mean of 5th and 6th observations.
\(\therefore\)  Median = \(\frac { 180+181 }{ 2 } =180.5\)
(f) A bar graph is shown below

Let his total journey length be x km.
\(\therefore\) Travelled by train = \(\frac{2}{5}x,\) travelled by bus = \(\frac{1}{3}x\) and travelled by car = \(\frac{1}{4}x\)
\(\therefore\) Total journey travelled by train, bus and car
               = \(\frac{2}{5}x+\frac{1}{3}x+\frac{1}{4}x\)
               = \(\frac{12\times 2x+20\times x+15\times x }{60}\)
               = \(\frac{24x+20x+15x}{60}=\frac{59x}{60}\)
\(\therefore\) Remaining = \(\frac{x}{1}-\frac{59x}{60}=\frac{60x-59x}{60}=\frac{x}{60}\)
According to the question, remaining journey is 3 km.
\(\therefore \quad \frac{x}{60}=3\Rightarrow x=3\times60=180\)
Hence, the length of his total journey is 180 km.

We have, \(\frac { 3p }{ 4 } =6\)
\(\Rightarrow \quad \frac { 3p }{ 4 } \times \frac { 4 }{ 3 } =6\times \frac { 4 }{ 3 } \)
\(\Rightarrow \quad p=2\times 7=8\)
So, p = 8 is the solution of the given equation.

Given, radius = 21 cm
\(\therefore\) Area of a circle = \(\pi\)r² = \(\pi\)(21)² = \(\frac{22}{7}\) x 21 x 21
                          = 22 x 3 x 21 = 1386 cm².

15 cm

\(-\frac { 11 }{ 7 } \)

Given, Principal, P = Rs 3500; Rate of interest, R = 7 %;
Time, T= 2yr
\(\therefore Interest,I={P\times R\times T\over100}={3500\times7\times2\over100}=35\times7\times2\)
=70x7=Rs490 
Hence, interest received at the end of two years will be Rs 490

In ΔABC and ΔPQR, we have
BC = 6 cm and QR = 6 cm, so BC = QR
B = 30° and ㄥQ = 30°, so ㄥB = ㄥQ
ㄥC = 60° and ㄥR = 60°, so ㄥC = ㄥR
Hence, two angles and one side are equal in ΔABC and ΔPOR. So, ΔABC ≅ ΔPOR by ASA congruent rule.