RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.2 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 11.2.
| Board | RBSE |
| Textbook | SIERT, Rajasthan |
| Class | Class 9 |
| Subject | Maths |
| Chapter | Chapter 11 |
| Chapter Name | Area of Plane Figures |
| Exercise | Ex 11.2 |
| Number of Questions Solved | 7 |
| Category | RBSE Solutions |
Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Ex 11.2
Question 1.
Find the area of a quadrilateral when one of the diagonal measure 12 cm and the length (RBSESolutions.com) of the perpendicular drawn from the opposite vertices to the diagonal are 7 cm and 8 cm respectively.
Solution.
Area of quadrilateral
= [latex]\frac { 1 }{ 2 }[/latex] x diagonal x (sum of offsets)
= [latex]\frac { 1 }{ 2 }[/latex] x 12 x (7 + 8)
= [latex]\frac { 1 }{ 2 }[/latex] x 12 x 15 = 90 cm²
Question 2.
The area of a parallelogram shaped field is 2000 sq. m. If its base is 50 m then find its height.
Solution.
Area of parallelogram shaped field
= base x corresponding altitude
2000 m² = 50 m x h
Question 3.
Find the area of (RBSESolutions.com) a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 6 cm and DA = 5 cm and diagonal AC = 5 cm.
Solution.
Area of quadrilateral ABCD
= area of ∆ABC + area of ∆ACD
Since, sides of ∆ABC are Pythagorian triplet
So, area = [latex]\frac { 1 }{ 2 }[/latex] x base x altitude
= [latex]\frac { 1 }{ 2 }[/latex] x 3 x 4 = 6 cm²
Also ∆ACD is an isosceles triangle with equal side (a) = 5 cm and third side (b) = 6 cm.
∴ Area of an isosceles triangle ACD
Question 4.
Find the area of a quadrilateral ABCD whose sides are 9 cm, 40 cm, 28 cm and 15 cm respectively and (RBSESolutions.com) the angle between the first two sides is a right angle.
Solution.
According to question, angle between AB and BC is 90°.
Question 5.
The length of two adjacent sides of a parallelogram are 50 cm and 40 cm respectively and (RBSESolutions.com) its diagonal is 30 cm. Find the area of the parallelogram.
Solution.
Area of parallelogram ABCD = 2 x (ar ∆ABC)
∴ Area of parallelogram ABCD
= 2 x (ar ∆ABC)
= 2 x 600
= 1200 sq. m.
Question 6.
Find the area of a parallelogram, whose one diagonal is 5.2 cm and the perpendicular distance (RBSESolutions.com) on this diagonal from opposite vertices is 3.5 cm.
Solution.
Area of parallelogram
= [latex]\frac { 1 }{ 2 }[/latex] x (sum of offsets) x diagonal
Question 7.
The length of two adjacent sides of a plot in the form (RBSESolutions.com) of parallelogram are 39 meters and 25 meters and its diagonal is 56 metre. Find the cost of levelling the plot at the rate of Rs 100 per sq. metre.
Solution.
Area of parallelogram ABCD
= 2 x (area of ∆ABC) …(i)
∴Using relation (i), we will get area (RBSESolutions.com) of parallelogram ABCD
i.e. area of parallelogram ABCD
= 2 x 420
= 840 sq. m.
∵ Cost of levelling 1 sq. m is Rs 100
∴ Cost of levelling 840 sq. m is 840 x 100
= Rs 84,000
We hope the given RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.2 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Ex 11.2, drop a comment below and we will get back to you at the earliest.
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