Distance from A to D = 7 m
Distance from A to D (via B) = 4 + 5 = 9 m
Distance from A to D (via C) = 2 + 6 = 8 m
Therefore, the route directly from A to D requires minimum cable of length 7 m.

We know that \(\angle\)POR = \(\angle\)POQ + \(\angle\)QOR
62° = 23° + \(\angle\)QOR
Subtracting 23° on both sides
62° – 23° = 23° + \(\angle\)QOR – 23°
\(\angle\)QOR = 39°

i) 4x + 7y = 4 (3) + 7 (2) = 12 + 14 = 26
ii) 3x + 2y – 5 = 3 (3) + 2 (2) – 5 = 9 + 4 – 5 = 8
iii) x – y = 3 – 2 = 1

Given that the adjacent sides of parallelogram are 12 cm and 9 cm
If we choose the shorter side as base, that is b = 9 cm then distance between the shorter sides is height, that is h = 8 cm
Area of parallelogram = b x h sq.units = 9 x 8 = 72 sq. cm.
Again, if we choose longer side as base, that is b = 12 cm then distance between longer sides is height. Let it be 'h' units.

We know that, the area of the parellelogram = 72 sq.cm
b x h = 72
12 x h = 72
\(h=\cfrac { 72 }{ 12 } =6cm\)
Therefore, the distance between the longer sides = 6 cm.

(i) (–70) + (–12) = –70 – 12 = – 82
(ii) 103 + 39 = 142

Price increases every year for item x = Rs.10.
Price decreases every year for item y = Rs. -15
Price of x in 2018 = Rs. 50
Price of x in 2020 = Rs. 50 +2 x Rs. 10 = Rs. 70 
= Rs. 50 + Rs. 20 = Rs. 70
Price of y in 2018 = Rs. 90
Price of y in 2020 = Rs. 90 + 2 x Rs. 15 = Rs. 60
Item x is costlier in the year 2020.

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

(i) (-11) - (-33)
= (-11) + (+33)
= 22
(ii) (-90)-(-50)
= -90 - (-50)
= -90 + 50
= -40

In the given algebraic expression, x,y,z are the variables.
Let us group the like terms.
100x + 99y − 98z + 10x + 10y + 10z − x − y + z
= (100x + 10x − x) + (99y + 10y − y) + ( −98z + 10z + z)
= (100 + 10 − 1)x + (99 + 10 − 1)y + ( −98 + 10 + 1)z
= (110 − 1)x + (109 − 1)y + ( −98 + 11)z
= 109x + 108y + ( −87)z
= 109x + 108y − 87z.

(i) 

∴. 8 + (-12) = -4
(ii) 

∴ (-3) + (-5) = -8
(iii) (-100) + (-10) = -100-10 = -110
(iv) 20 + (-72) = 20 - 72 = -52
(v) 82 + (-75) = 82 - 75 = 7
(vi) -48 + (-15) = -48 - 15 =- 63
(vii) -225 + (-63) = 225 - 63 = -288

\(\frac{\mathrm{P}-\mathrm{Q}}{\mathrm{P}+\mathrm{Q}}=\frac{-15-5}{-15+5}=\frac{-20}{-10}=2\)

Given : Area = 100 sq.cm, d₁ = 8 cm, d₂ = ?
Area of rhombus = 100 sq.cm
\(\frac{1}{2} \times\left(d_{1} \times d_{2}\right)=100 \)
\(\frac{1}{2} \times\left(8 \times d_{2}\right)=100 \)
8 x d₂ = 100 x 2
\(\mathrm{d}_{2}=\frac{100 \times 2}{8} \Rightarrow \mathrm{d}_{2}=25 \mathrm{~cm}\)

Given x = 2; y = 3.
(i) 2x - 3y = 2 (2) - 3 (3) = 4 - 9
= 4 + (Additive inverse of 9)
= 4 + (-9) = - 5
(ii) x + y =2 + 3 = 5 
(iii) 4y - x = 4 (3) - 2 = 12 - 2 = 10
(iv) x + 1- y = 2 + 1 - 3 = 3 - 3 = 0