Steps of construction:
I. Draw a line segment QR = 5 cm.
II. At Q, construct \(\angle\)RQX = 60°.
III. At R, construct \(\angle\)QRY = 75°
IV. Let the rays \(\overrightarrow{QX}\) and \(\overrightarrow{RY}\) intersect at P.
Thus \(\triangle\)PQR is the required triangle.

Radius of circular park = \(\frac { 84 }{ 2 } \)=42 m (given)
Width of the road = 3.5m (given)
Radius of outer circle = 42 + 3.5 = 45.5 m

Are of the road = (Area of outer circle) - (Area of inner circle)
= \(\pi\) x (45.5)² - \(\pi\) x (42)²
= \(\pi\) x {(45.5)² - (42)²}
= \(\pi\) x {(45.4 + 42)(45.4 - 42)}
= \(\pi \) x 87.5 x 3.5
= \(\frac { 22 }{ 7 } \)x 87.5 x 3.5 = 11 x 87.5
= 962.5 m²
Cost of the road = 962.5 x Rs. 200
= Rs. 192500

We have \(({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4})\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
= \({x^3\over 5}-{3\over2}x^2+{2\over3}x+{1\over4}\)-\(({8\over 5}x^2-{2\over3}x^3+{3\over2}x-1)\)
 =\(({1\over5}+{2\over3})x^3+(-{3\over2}-{8\over5})x^2+({2\over3}-{3\over2})x+({1\over4}+1)\)
\(={13x^3\over 15}-{31x^2\over10}-{5x\over 6}+{5\over4}\) 

To construct the MBC, we follow the following steps:
(a) Draw a line segment BC of length 4.5 cm 

(b) Draw
(c) With centre Band radius AB = 6.2 cm, draw an arc of the circle to intersect ray CX at A.
(d) Join BA to obtain the desired triangle ABC.

(a)\(\frac{7}{48}-\frac{17}{36}\)=\(\frac{7(3)-17(4)}{144}\)
=\(\frac{21-68}{144}\)
=\(\frac{-47}{144}\)
(b) \(\frac{5}{63}-(\frac{-6}{21})\)=\(\frac{5}{63}+\frac{6}{21}\)
=\(\frac{5+6(3)}{63}\)
=\(\frac{5+18}{63}=\frac{23}{63}\)
(c) \(\frac{-6}{13}-(\frac{-7}{15})\)=\(\frac{-6}{13}+\frac{7}{15}\)
=\(\frac{-6(15)+7(13)}{195}\)
=\(\frac{-90+91}{195}=\frac{1}{195}\)
(d) \(\frac{-3}{8}-\frac{7}{11}\)=\(\frac{-3(11)-7(8)}{88}\)
=\(\frac{-33-56}{88}=\frac{-89}{88}\)
=-1\(\frac{1}{88}\)

\(\frac { 9 }{ 32 } \)

Let ABeD represents the rectangular floor of the room and shaded region represents the verandah of width 2.25 m all along the outside of a room.

Given, length of the room, AB = 5.5 m
and breadth of the room, AD = 4 m
\(\therefore\) Area of the room ABeD = AB x AD = 5 .5 x 4 = 22 m²
Length of the room and verandah, PQ or SR
                 = 5.5 + 2 x 2.25 = 5.5 + 4.5 = 10 m
Breadth of the room and verandah, PS or QR
                 = 4 + 2 x 2.25 = 4 + 4.5 = 8.5 m
\(\therefore\) Area of the room and verandah
                = PQ x PS = 10 x 8.5 = 85 m²
(i) Area of the verandah = Area of the room and verandah - Area of the room
                              = 85 - 22 = 63 m².
(ii) Cost of cementing the floor of the verandah at the rate of Rs 200 per m² = Rs (200 x 63) = Rs 12600.

Given, diameter of a circular table cover (d) = 1.5 m
\(\therefore\) Length of the lace required
                       = Circumference of circular table cover
                       = \(\pi\) x d = 3.14 x 1.5 = 4.71 m
\(\therefore\) Cost of 1 m of lace = Rs 15
\(\therefore\) Cost of 4.71 m of lace = Rs (15 x 4.71) = Rs 70.65
Hence, the length of the lace required is 4.71 m and cost of the lace is Rs 70.65.

Given, length of the door = 2 m and breadth of a door = 1m
Since door is in the shape of a rectangle
\(\therefore\)  Area of the door = Length x Breadth = 2 x 1= 2 m
Also, length of the wall = 4.5 m and breadth of the wall = 3.6 m
\(\therefore\)  Area of the wall including door = Length x Breadth = 4.5 x 3.6 = 16.2 m²
Now, area of the wall for whitewash = Area of the wall including door - Area of the door = 16.2 - 2 = 14.2 m²
\(\therefore\) Cost of whitewashing the wall at the rate of Rs 20 per m²
                                 = Rs (14.2 x 20) = Rs 284
Hence, the cost of whitewashing the wall is Rs 284.

\(\frac { -39 }{ 10 } \)

Perimeter = \(2\times 7+\frac { 22 }{ 7 } \times 7\)
= 14 + 22 = 36 cm

Base =\(\frac { 20 }{ 2 } \)=10 cm

\(4\frac { 4 }{ 5 } \div \frac { 24 }{ 5 } +8\frac { 1 }{ 4 } =\frac { (5\times 4)+4 }{ 5 } \div \frac { 24 }{ 5 } +\frac { (8\times 4)+1 }{ 4 } \)
\(\frac { 24 }{ 5 } \div \frac { 24 }{ 5 } +\frac { 33 }{ 4 } \)
Since , the reciprocal  of  \(\frac { 24 }{ 5 } \) is \(\frac { 5 }{ 24 } \) 
\(\\ \therefore \quad \frac { 24 }{ 5 } \times \frac { 5 }{ 24 } +\frac { 33 }{ 4 } =1+\frac { 33 }{ 4 } =\frac { 37 }{ 4 } \)

(i) The prisms and cylinder, both, have their base and top faces as congruent and parallel to each other. Also, a prism becomes a cylinder as the number of sides of its base becomes larger and larger.
(ii) The pyramids and cones are alike in the sense that their lateral faces meet at a point (called vertex). Also, a pyramid becomes a cone as the number of sides of its base becomes larger and larger.

125

The simplified expression is 11x+7 and its required value is 29.

Rs 720.

2¹⁰x5¹⁰

The mirror image of English letter E will be

\(\frac { 40 }{ -3 } \)

We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2 = 2⁷

Here, base = 2 and exponent = 7, since 2 repeated 7 times.

Perimeter of the rectangle
= 2\(\times\)(l+b)
⇒ 130 = 2\(\times\)(l + 30)
⇒ l+30 = \(\frac { 130 }{ 2 } \)
⇒ l+30= 65
⇒ l = 65 - 30 = 35 cm
Hence, the length of the rectangle is 35 cm.
Area of the rectangle
= l\(\times\)b
= 35\(\times\)30 = 1050 cm²
Hence, the area of the rectangle is 1050 cm²

\(\frac { 8 }{ 3 } \)

a²b + b² - 2a² - 7ab

(i) Net for triangular prism

(ii) Net for tetrahedron,

(iii) Net for cuboid

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.

Steps of construction
(i) Draw a line AB= 5 cm and mark a point C outside it.
(ii) Mark another point D on AB, join CD.
(iii) With point D as centre, draw an arc cutting AB at E and also, CD at F.
(iv) With point C as centre and radius equal to DE, draw another arc cutting CD at M.
(v) With point M as centre and radius equal to EF, mark a point P on the previous arc.
(vi) Now, join PC to draw a line n.
Hence, line n is required line parallel to AB.

Here, Length of the rectangular park = 70 m
Breadth of the rectangular park = 45 m
The cross paths are shown by EFGH and PQRS in the figure.
Now, PQ =3 m and PS = 45 m
EH = 3 m and EF = 70 m
KL = 3 m and KN = 3 m
Now, area of the path = [Area of rectangle PQRS] + [Area of rectangle EFGH]- [Area of square KLMN]
= [PS x PQ] + [EF x EH] - [KL x KN]
= [45 x 3] m² + [70 x 3] m² - [3 x 3] m²
= 135 m² + 210 m² - 9 m² = 336 m²
Now, cost of constructing the path = Rs. 110 x 336 = Rs. 36960.
Note: While finding the area of cross roads, the area of the middle square (here, KLMN) is taken twice
which is to be subtracted once.

(a) (6^-1 - 8^-1)^-1 + (2^-1 - 3^-1)^-1
\(=(\frac{1}{6}-\frac{1}{8})^{-1}+(\frac{1}{2}-\frac{1}{3})^{-1}\)
\(=(\frac{4-3}{24})^{-1}+(\frac{3-2}{6})^{-1}\)
\((\frac{1}{24})^{-1}+(\frac{1}{6})^{-1}\)
= 24 + 6 = 30
(b) \({ \left\{ { 6 }^{ -1 }+{ \left( \frac { 3 }{ 2 } \right) }^{ -1 } \right\} }^{ -1 }\)
\(=(\frac{1}{6}+\frac{2}{3})^{-1}\)
\(=(\frac{1+4}{6})^{-1}=(\frac{5}{6})^{-1}\)
\(=\frac{6}{5}\)