Given height (h) = 4 cm
Parallel sides are (a) = 17 cm and (b) = 14 cm
Area of the trapezium = \(\cfrac { 1 }{ 2 } \times h\left( a+b \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 4(17+14)\)
= \(\cfrac { 1 }{ 2 } \times 4(31)\)
= 62 sq. cm
Therefore, the area of canvas used is 62 sq. cm.

Let, the length of the required side be ‘x’ cm.
Then, area of the trapezium = \(\cfrac { 1 }{ 2 } \times h\left( a+b \right) sq.units\)
= \(\cfrac { 1 }{ 2 } \times 16(25+x)\)
= 200 + 8x
But, the area of the trapezium = 352 sq. cm (given)
Therefore, 200 + 8x = 352
\(\Rightarrow\) 8x = 352 – 200
\(\Rightarrow\) 8x = 152
 \(\Rightarrow\) \(x=\cfrac { 152 }{ 8 } \)
x = 19
Therefore, the length of the other side is 19 cm.

Given, Area of the Trapezium = 828 cm²
\(\cfrac { 1 }{ 2 } \times h(a+b)=828\)
\(\cfrac { 1 }{ 2 } \times h(19.6+16.4)=828\)
\(\cfrac { 1 }{ 2 } \times h(36)=828\)
h (18) = 828
\(h=\cfrac { 828 }{ 18 } \)
h - = 46 m
Therefore, distance between the parallel sides = 46 cm

Given, height (h) = 9 cm
Parallel sides are (a) = 23 cm and (b) = 12 cm
Area of the trapezium = \(\cfrac { 1 }{ 2 } \times h(a+b)\)
= \(\cfrac { 1 }{ 2 } \times 9\left( 23+12 \right) \)
= \(\cfrac { 1 }{ 2 } \times 9(35)\)
= 157.5 sq. cm
Therefore, Area of the trapezium is 157.5 sq. cm.

Given, height (h) = 14 cm
parallel sides are (a) = 18 cm and (b) = 9 cm
Area of the trapezium = \(\cfrac { 1 }{ 2 } \times h(a+b)\)
= \(\cfrac { 1 }{ 2 } \times 14\left( 18+9 \right) \)
= 7(27)
= 189 sq. cm
Therefore, area of the trapezium is 189 sq. cm.