Let the other number be x
Given, \(\frac { 2 }{ 15 } +x=\frac { 4 }{ 5 } \)
\(\Rightarrow x=\frac { 4 }{ 5 } -\frac { 2 }{ 15 } =\frac { 12-2 }{ 15 } =\frac { 10 }{ 15 } \)
\(\Rightarrow x=\frac { 2 }{ 3 } \)

\(\therefore\) (3a + 4c)² = (3a)² + 2(3a)(4c) + (4c)² (replacing a and b values)
= 3² a² + (2 x 3 x 4)(a x c) + 4²c²
(3a + 4c)² = 9a² + 24ac + 16c²

Guna fixed a single door of 3 feet wide.
Radius of this single door = 3 feet.
Nathan fixed a double door each of 1 1/2 feet wide.
Radius of each of this double door
\(=\frac{3}{2} \text { feet. } \)
The area required for the single door
\(=\frac{\theta}{360} \times \pi r^{2} \)
\(=\frac{120}{360} \times 3.14 \times 3 \times 3 \)
= 9.42 m²    ............(i)
The area required for the double door
\(=2 \times \frac{\theta}{360} \times \pi r^{2} \)
\(=2 \times \frac{120}{360} \times 3.14 \times \frac{3}{2} \times \frac{3}{2} \)
= 4.71 m²       ........................(ii)
From (i) and (ii), the double door requires minimum area.

Assuming that the question Q₁ is answered True, questions Q₂ and Q₃ can be answered as TT, TF, FT, and FF in 4 ways.
Similarly, assuming that the question Q₂ is answered False, Q₂ and Q₃ can also be answered as TT, TF, FT, and FF in 4 ways.
Thus, as each question has only two options (True or False), the number of ways of answering these 3 questions in a slip test is 2 x 2 x 2 = 8 possible ways

Now, from Fig \(\angle\)140^o+ \(\angle\)z = \(\angle\)180^o (linear pair)
⇒ \(\angle\)z = 180^o −140^o = 40^o
Also \(\angle\)x+ \(\angle\)z= \(\angle\)70 + \(\angle\)z (exterior angle property)
⇒\(\angle\)x=70^o
Also \(\angle\)z+ \(\angle\)y + 70^o = 180^o (angle sum property in ΔABC)
⇒ 40^o + \(\angle\)y + 70^o = 180^o
⇒ \(\angle\)y = 180^o − 110^o = 70^o

For the recipe rice required = 1\(\frac{1}{2}\)cup
water required = 2\(\frac{3}{4}\) cups
\(\frac { rice }{ water } =\frac { 1\frac { 1 }{ 2 } }{ 2\frac { 3 }{ 4 } } \)
fraction \(=\frac { \frac { 3 }{ 2 } }{ \frac { 11 }{ 4 } } =\frac { 3 }{ 2 } \times \frac { 4 }{ 11 } \)
\(=\frac { 6 }{ 11 } \)
∴ rice : water = 6 : 11

\(\frac{9}{132} \times \frac{-11}{3}=\frac{3}{12} \times \frac{-1}{1}=\frac{-1}{4} \)

The rational number for the point marked on the number line is -\(\frac{2}{5}\)

\(\frac { 8 }{ 9 } =\frac { 8\times 2 }{ 9\times 2 } =\frac { 16 }{ 18 } \)
\(\frac { 8 }{ 9 } =\frac { 8\times 3 }{ 9\times 3 } =\frac { 24 }{ 27 } \)
\(\frac { 8 }{ 9 } =\frac { 8\times 4 }{ 9\times 4 } =\frac { 32 }{ 36 } \)
\(\frac { 8 }{ 9 } =\frac { 8\times 5 }{ 9\times 5 } =\frac { 40 }{ 45 } \)
Four equivalent rational numbers of \(\frac { 8 }{ 9 } \) are \( \frac { 16 }{ 18 } ,\frac { 24 }{ 27 } ,\frac { 32 }{ 36 } ,\frac { 40 }{ 45 } \)

By Euler's formula F + V - E = 2 for a polyhedron
Here F = 12, V = 17, E = 22
F + V - E = 12 + 17 - 22
= 29 - 22
= 7≠ 2
∴ The polyhedron cannot have 12 faces 22 edges and 17 vertices.

From the figure
diameter = 10 cm
radius = 5 cm
Area of the shaded part
= Area of the square - Area of the four circular quadrant
\( =a^{2}-\left(4 \times \frac{\theta}{360} \times \pi r^{2}\right) \)
\(=(10 \times 10)-\left(4 \times \frac{90}{360} \times 3.14 \times 5 \times 5\right) \)
= 100 - 78.5 = 21.5 cm² (approximately)

(2p²q³)\(\times\)(-9pq)² = (+)(-) \(\times\) (2\(\times\)9) (p²\(\times\) p (q³\(\times\)q²)) = -18p³q⁵

From the figure the tile is in a circular quadrant, Then the central angle is 90^o
Area of the sector = \(\frac{\theta}{360} \times \pi r^{2}\)
\(=\frac{90}{360}\) x 3.14 x 30 x 30
  = 706.5 cm²

Length of the arc of the sector l = 50 mm
Radius r = 14mm
Area of the sector = \(\frac { lr }{ 2 } \) sq. units
= \(\frac { 50\times 14 }{ 2 } \) mm² = 50 x 7 mm² = 350 mm²
Area of the sector = 350mm²

The given field has four triangles (I, III, IV & V) and a trapezium (II).
Area of the triangle (I) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 5\times 6=15{ m }^{ 2 }\)
Area of the trapezium (II) \(=\frac { 1 }{ 2 } h(a+b)=\frac { 1 }{ 2 } \times 13\times (6+4)=65{ m }^{ 2 }\)
Area of the triangle (III) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 8\times 4=\frac { 32 }{ 2 } =16{ m }^{ 2 }\)
Area of the triangle (IV) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
Area of the triangle (V) \(=\frac { 1 }{ 2 } \times b\times h=\frac { 1 }{ 2 } \times 13\times 10=65{ m }^{ 2 }\)
∴ The total area of the field = 15 + 65 + 16 + 65 + 65 = 226 m²

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