∠x +∠Y + ∠Z = 180°
29° + 900 + (2x + 1) = 180°
120⁰ + 2x = 180°
 2x = 180°-120 °= 60°
\(x=\frac{60^{\circ}}{2}=30^{\circ}\)
x = 30°

8 × 45 × 84 = 28 × 9 × 120

10x² − 3xy + 9y² − (3x² − 6xy − 3y² )
= 10x² − 3xy + 9y² − 3x² + 6xy + 3y² 
= ( 10x² - 3x² ) + (-3xy + 6xy ) + (9y² + 3y²)
= 7x² + 3xy + 12y²
Degree of First term 7x² is 2
Degree of second term 3xy is 2
Degree of Third term 12y² is 2
Thus, the degree of the expression is 2

Unit digit of base 11 is 5 and power is 10 (Even power).
Thus, the unit digit of 11¹⁰ is 1.

Criterion : SSS

Corresponding sides
\(\overline{P Q} \text { matches } \overline{M N}\)
\(\overline{Q R} \text { matches } \overline{L M}\)
\(\overline{P R} \text { matches } \overline{L N}\)
Congruent angles
<P = <N
<R = <L
<Q = <M

In right ΔABC,
∠A + ∠B + ∠C =180°
(x+1) + 90^o + 2x + 5 = 180°
3x + 96 = 180°
3x = 180^o- 96
3x = 84^o
\(x=\frac{84^{\circ}}{3}=28^{\circ}\) 
x = 28^o
∠A = x + 1 = 28°+ 1 = 29°
∠C = 2x + 5 =2 x 28+  5 = 56 + 5 = 61 °
∠LA = 29°
∠C = 61°

Let the three angles are 3x, 5x and 4x
= 3x, 5x, 4x = 180°
12x = 180°
\(x=\frac{180^{\circ}}{12}=15^{\circ}\)
3x = 3 x 15°= 45°
5x = 5 x 15°= 75°
4x = 4x 15°= 60°
The three angles are 45°,75°, 60°

From the figure,
\( \angle O+\angle D+\angle C =180^{\circ} \)
\(\angle O+70^{\circ}+30^{\circ} =180^{\circ} \)
\(\angle O+100^{\circ} =180^{\circ} \)
\(\angle O =180^{\circ}-100^{\circ} \)
\(\angle O =80^{\circ}\)
 \(\Delta AOB,\)
\( \angle O+\angle A+\angle B =180^{\circ} \)
\(80^{\circ}+3 x+2 x =180^{\circ} \Rightarrow 80^{\circ}+5 x=180^{\circ} \)
\(5 x =180^{\circ}-80^{\circ}=100^{\circ} \)
\(x =\frac{100^{\circ}}{5}=20^{\circ} \)
\( \angle A=3 x=3 \times 20^{\circ}=60^{\circ} \)
\(\angle B=2 x=2 \times 20^{\circ}=40^{\circ}\)

The radius ofthe circle = 3 cm
Length of the rectangle= 8 x radius
= 8 x 3=24cm
Breath = 2 x radius
= 2 x 3=6 cm
Radius ofa circle r = 3cm
Area = πr²
The area of each circle = \(\cfrac { 22 }{ 7 } \times 3\times 3=\cfrac { 198 }{ 7 } =28.28^{ 2 }cm\) 

Cost of decorating the circumference circular lawn of a house at the rate of Rs 55 per metre
= Rs 16940
\(55 \times 2 \pi r =16940 \)
\(55 \times 2 \times \frac{22}{7} \times r =16940 \)
\(r =16940 \times \frac{1}{55} \times \frac{1}{2} \times \frac{7}{22}=49 \mathrm{~m} \)
Radius of the lawn r = 49 m.

Radius of school ground (R) = 103 m
Width of a track = 3 m
Width of a tracks = 4 x 3 = 12 m
Inner radius of ground (r) = 103 - 12 = 91 m
Area of tracks \(=\pi\left(R^{2}-r^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}(103 \times 103-91 \times 91) \)
\(=\frac{22}{7}(10609-8281) \)
\(=\frac{22}{7} \times 2328=7316.57 \mathrm{~m}^{2} \)
cost of construction of track of 1 sq.m = Rs. 50
cost of construction of track of 7316.57 sq. units = 50 x 7316.57
= Rs. 365828.50

Given outer radius (R) = 32 cm
Inner radius (r) = 18 cm
Area of circular pathway \(=\pi\left(R^{2}-r^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(32^{2}-18^{2}\right) \)
\(=\frac{22}{7}(32 \times 32-18 \times 18) \)
\(=\frac{22}{7}(1024-324) \)
\(=\frac{22}{7} \times 700 \)
Area of circular pathway = 2200 cm²

Given diameter of flower garden = 49 m
d = 49 m
\(2 r=49 \Rightarrow r=\frac{49}{2}\)
Area of the circular flower garden = \(\pi r^{2}\) sq. units
\(=\frac{22}{7} \times \frac{49}{2} \times \frac{49}{2} \)
\(=\frac{3773}{2} m^{2} \)
The cost of levelling the flower garden per sq.m = Rs. 150
The cost of levelling the flower garden of \(\frac{3773}{2} sq.m \)
\(=Rs. 150 \times \frac{3773}{2}=Rs. 282975 \)

Given length of rope = radius = 4.9 m
Area that the sheep can graze = \(\pi r^{2}\)
\(=\frac{22}{7} \times 4.9 \times 4.9\)
= 75.46 m²

Diameter (d) = 105 cm
\(\mathrm{r}=\frac{105}{2} \mathrm{~cm}\)
Area of the dining table = \(\pi r^{2}\) sq. units
\(=\frac{22}{7} \times \frac{105}{2} \times \frac{105}{2}\)
\(=\frac{11 \times 15 \times 105}{2}\)
Area of the dining table = 8662.5 cm²

Length of wire = 1320 cm
Circumference of a circular frame - 2r units
\(=2 \times \frac{22}{7} \times 7=44 \mathrm{~cm}\)
Number of circular
\(\text { frames }= \frac{\text { Length of wire }}{\text { Circumference of a circular frame }} \)
\(=\frac{1320 \mathrm{~cm}}{44 \mathrm{~cm}} \)
= 30 frames
30 circular frames can be made.

\(\frac{421}{100} =\frac{400+20+1}{100}=\frac{400}{100}+\frac{20}{100}+\frac{1}{100} \)
\(=4+2 \times \frac{1}{10}+1 \times \frac{1}{100}=4.21 \)

\(23+\cfrac { 4\div 2 }{ 10\div 2 } =23+\cfrac { 2 }{ 5 } \) 
= \(23\cfrac { 2 }{ 5 } =\cfrac { 117 }{ 5 } \)

\(\frac{3}{10}=0+3 \times \frac{1}{10}=0.3\)

The place value of 9 in 17.39 is Hundredth = \(\cfrac { 1 }{ 100 } \) 

\(1 \mathrm{~cm}=\frac{1}{100} m=0.01 \mathrm{~m}\)
7 cm = 7 x 1 cm
\(=7 \times \frac{1}{100} m=\frac{7}{100} m\)
7 cm = 0.07 m

\(1 \mathrm{~mm}=\frac{1}{10} \mathrm{~cm}=0.1 \mathrm{~cm}\)
42 mm = 42 x 1 mm
\(=42 \times \frac{1}{10} \mathrm{~cm}=\frac{42}{10} \mathrm{~cm}\)
42 mm = 4.2 cm