Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.3
Question 1.
The sides of a ground in the form of a cyclic quadrilateral are 72 m, 154 m, 80 m and 150 m respectively. Find the cost of paving the tile on the ground at Rs 5 per sq. metre.
Solution.
Here a = 72 m, b = 154 m, c = 80 m and d = 150 m
= 78 x 74 x 4
= 11,544 sq. m.
∵ Cost of paving the tiles on the ground 1 sq. m is Rs 5.
∴ Cost of paving the tiles on the ground 11,544 sq. m is 11,544 x 5 = Rs 57,720.
The sides of a ground in the form of a cyclic quadrilateral are 72 m, 154 m, 80 m and 150 m respectively. Find the cost of paving the tile on the ground at Rs 5 per sq. metre.
Solution.
Here a = 72 m, b = 154 m, c = 80 m and d = 150 m
= 78 x 74 x 4
= 11,544 sq. m.
∵ Cost of paving the tiles on the ground 1 sq. m is Rs 5.
∴ Cost of paving the tiles on the ground 11,544 sq. m is 11,544 x 5 = Rs 57,720.
Question 2.
The diagonals of a rhombus are 25 cm and 42 cm. Find its perimeter and area.
Solution.
Area of a rhombus
= x product of its diagonals
= x 25 x 42
= 25 x 21
= 525 sq. cm.
∴ Perimeter = 4 x one side
= 4 x 24.43
= 97.72 cm
Hence, perimeter = 97.72 cm and area = 525 cm².
The diagonals of a rhombus are 25 cm and 42 cm. Find its perimeter and area.
Solution.
Area of a rhombus
= x product of its diagonals
= x 25 x 42
= 25 x 21
= 525 sq. cm.
∴ Perimeter = 4 x one side
= 4 x 24.43
= 97.72 cm
Hence, perimeter = 97.72 cm and area = 525 cm².
Question 3.
The perimeter of a rhombus is 40 m and one of its diagonal is 12 m. Find the area of the rhombus.
Solution.
Perimeter of a rhombus
= 4 x one side
=> 40 m = 4 x one side
∴ One side = = 10 m
Let one of its diagonal be BD.
i. e. BD = 12 m
∴ OB = OD = 6 m
Now in ∆OCD
OC² + OD² = CD² => OC² = CD² – OD²
=> OC² = (10)² – (6)²
=> OC² = 100 – 36
=> OC = √64 = 8 m
∴ Other diagonal AC = 2 x OC
= 2 x 8 = 16 m
And area of rhombus ABCD
The perimeter of a rhombus is 40 m and one of its diagonal is 12 m. Find the area of the rhombus.
Solution.
Perimeter of a rhombus
= 4 x one side
=> 40 m = 4 x one side
∴ One side = = 10 m
Let one of its diagonal be BD.
i. e. BD = 12 m
∴ OB = OD = 6 m
Now in ∆OCD
OC² + OD² = CD² => OC² = CD² – OD²
=> OC² = (10)² – (6)²
=> OC² = 100 – 36
=> OC = √64 = 8 m
∴ Other diagonal AC = 2 x OC
= 2 x 8 = 16 m
And area of rhombus ABCD
Question 4.
Find the area of a trapezium shaped field, the lengths of whose parallel sides are 42 metre and 30 metre and other sides are 18 metre and 18 metre.
Solution.
In the figure, ABCD is a trapezium shaped field in which parallel sides AB = 42 m and CD = 30 m and non-parallel sides AD and BC equal to 18 m.
Draw EC || AD and CF ⊥ AB
∴ Area of trapezium ABCD = area of parallelogram AECD + area of ABCE = 507 + 101.82 = 610.92 m².
Find the area of a trapezium shaped field, the lengths of whose parallel sides are 42 metre and 30 metre and other sides are 18 metre and 18 metre.
Solution.
In the figure, ABCD is a trapezium shaped field in which parallel sides AB = 42 m and CD = 30 m and non-parallel sides AD and BC equal to 18 m.
Draw EC || AD and CF ⊥ AB
∴ Area of trapezium ABCD = area of parallelogram AECD + area of ABCE = 507 + 101.82 = 610.92 m².
Question 5.
If area of trapezium is 350 sq. cm and its parallel sides are 26 cm and 44 cm then find the distance between the parallel sides.
Solution.
Let distance between the parallel sides be x Area of trapezium
Hence, distance between the parallel sides be 10 cm.
If area of trapezium is 350 sq. cm and its parallel sides are 26 cm and 44 cm then find the distance between the parallel sides.
Solution.
Let distance between the parallel sides be x Area of trapezium
Hence, distance between the parallel sides be 10 cm.
Question 6.
A table is in the shape of a trapezium. Its parallel sides are 8 m and 16 m respectively. Area of table is 108 sq. m. Find the width of the table i.e. Distance between the parallel sides.
Solution.
Area of trapezium
A table is in the shape of a trapezium. Its parallel sides are 8 m and 16 m respectively. Area of table is 108 sq. m. Find the width of the table i.e. Distance between the parallel sides.
Solution.
Area of trapezium
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