RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.3 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.3.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Area of Plane Figures |

Exercise |
Ex 11.3 |

Number of Questions Solved |
6 |

Category |
RBSE Solutions |

## 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 (RBSESolutions.com) 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 (RBSESolutions.com) perimeter and area.

Solution.

Area of a rhombus

= [latex]\frac { 1 }{ 2 }[/latex] x product of its diagonals

= [latex]\frac { 1 }{ 2 }[/latex] 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 (RBSESolutions.com) 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 = [latex]\frac { 40 }{ 4 }[/latex] = 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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 (RBSESolutions.com) 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

We hope the given RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.3 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.3, drop a comment below and we will get back to you at the earliest.

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