56 × 210 × 126 = 70 × 84 × 252

The relationship between x and y is y = 2x - 1

Degree 1

In ΔABC. Exterior angle ΔC = 115°
∠BCD + 115° = 180° (Linear angles)
∠BCD = 180°- 115° = 650
BD = BC (Given)
So, ∠BCD  = ∠BDC
∠BCD = 65°
So, ∠BDC = 65°
The exterior angle of ∠BAD is ,ΔBDC.
The exterior angle ∠BDC = The sum of two interior opposite angles.
65° = 35°+ x
x = 65° - 35°= 30°

In \(\triangle B A C \text { and } \triangle D A C\)
Sides AC is common (equal)
<BAC = <CAD
<BCA = <DCA
By ASA property
\(\triangle B A C \cong \triangle D A C\)

(i)  
Step 1: Draw a line. Mark X and Y on the line such that XY = 6.4 cm

Step 2: With X as centre, draw an arc of radius 5 cm above the line XY.

Step 3: With Y as centre, draw an arc of radius 7.7 cm to intersect arc drawn in step 2. Mark the point of intersection as Z.

Step 4: Join XZ and YZ.
Now XYZ is the required triangle

(ii)
Step 1: Draw a line. Mark X and Y on the line such that XY = 7.5 cm

Step 2: With X as centre, draw an arc of radius 7.5 cm above the line XY.

Step 3: With Y as centre, draw an arc of radius 7.5 cm to intersect arc drawn in step 2. Mark the point of intersection as Z.

Step 4: Join XZ and YZ.
Now XYZ is the required triangle

(iii)
Step 1: Draw a line. Mark X and Y on the line such that XY = 6.5 cm

Step 2: With X as centre, draw an arc of radius 4.6 cm above the line XY.

Step 3: With Y as centre, draw an arc of radius , 4.6 cm to intersect arc drawn in step 2. Mark the point of intersection as Z.

Step 4: Join XZ and YZ.
Now XYZ is the required triangle

SSS criterion

Criterion: ASA

2⁸ = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
8² = 8 x 8 = 64
256 > 64
2⁸ > 9²
So, The greater number is 2⁸.

9 x 9 x 9 = 729

Corresponding sides
\(\overset { \_ \_ }{ PQ } =\overset { \_ \_ }{ LN } ;\)
\(\overset { \_ \_ }{ PR } =\overset { \_ \_ }{ LM } ;\)
\(\overset { \_ \_ }{ RQ } =\overset { \_ \_ }{ MN } \)
Congruent angles
<p = < l
<R = <M
<Q = <N

In△XYZ ,∠Y = 72°
∠X:∠Z = 5:4
Let ∠X = 5x, ∠Z = 4x
∠x + ∠y + ∠z = 180°
5x + 4x + 72° = 180°
9x + 72° = 180°
9x = 180^o - 72^o = 108^o
\(x=\frac{108^{\circ}}{9}=12^{\circ}\)
∠X = 5x = 5 x 12 = 60°
∠X = 60°
∠Z = 4x = 4 x 12 = 48°
∠Z = 48°
∠X = 60° ; ∠Z = 48°

\(\angle A+\angle N+\angle G+\angle L+\angle E+\angle S\)
\(=\underbrace{\angle A+\angle E+\angle G}_{\text {Sum of three angles of } \Delta}+\underbrace{\angle S+\angle N+\angle L}_{\text {Sttm of three angles of } \Delta}\)
= 180^o + 180^o = 360^o

\(\angle E+\angle F+\angle G={ 180 }^{ 0 }\) 
80⁰ + 55^o + x = 180^o
135^o + x = 180^o
x = 180° - 135° = 45° 
x = 45^o

The radius ofthe circle = 3 3m
Length of the rectangle = 8 x radius
= 8 x 3 = 24 cm
Breath = 2 x radius
Area of a circle = 28.28cm²
Area of 4 circles = 4 x 28.28cm²
= 113.12cm²
The shaded area inside = Area of the rectangle - Area of 4 circles.
= (144 - 113.12) cm²
= 30.88cm²

Area of pathway = Area of outer rectangle - Area of inner rectangle
= LB - lb
= 80 x 50 - 70 x 40
= 4000 - 2800 = 1200 m²

Given, Radius of circular carpet (r) = 106 crn
Radius of circular hall (R) = 120 cm
Area of the hall uncovered by the carpet
\(=\pi\left(R^{2}-r^{2}\right) \text { sq. units } \)
\(=\frac{22}{7}\left(120^{2}-106^{2}\right) \)
\(=\frac{22}{7}(120 \times 120-106 \times 106) \)
\(=\frac{22}{7} \times(14400-11236) \)
\(=\frac{22}{7} \times 3164=9944 \mathrm{~cm}^{2} \)

\(1 \mathrm{~cm} =\frac{1}{100} m \)
\(1328 \mathrm{~cm} =1328 \times 1 \mathrm{~cm} \)
\(=1328 \times \frac{1}{100} m=\frac{1328}{100} m=13.28 \mathrm{~m} \)

The whole number parts of the two decimal numbers are equal.
The digits at tenth place are 0 and 3.
3 > 0
7.3 > 7.01
The smaller nurnber is 7.01

Comparing the whole number parts of the two decimal numbers.
We get 0 < 1
0.99 < 1.9
Smaller number is 0.99.

\(3\cfrac { 3 }{ 5 } =3+\cfrac { 3 }{ 5 } =3+\cfrac { 3\times 2 }{ 5\times 2 } \)
\(=3+\cfrac { 6 }{ 10 } =3.6\)

The place value of 2 in 263.271 is Tenths = \(\cfrac { 1 }{ 10 } \) 

\(1 \mathrm{~cm}=\frac{1}{100} m=0.01 \mathrm{~m}\)
16 cm = 16 x 1 cm
\(=16 \times \frac{1}{100} m \)
\(=\frac{16}{100} m=0.16 \mathrm{~m} \)

\(1 \mathrm{~mm}=\frac{1}{10} \mathrm{~cm}=0.1 \mathrm{~cm}\)
9 mm = 9 x 1 mm
\(=9 \times \frac{1}{10} \mathrm{~cm}=\frac{9}{10} \mathrm{~cm}\)
9 mm = 0.9 cm

21.3