RBSE Solutions for Class 9 Maths Chapter 11 Area of Plane Figures Additional Questions 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 Additional Questions.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Area of Plane Figures |

Exercise |
Additional Questions |

Number of Questions Solved |
11 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 11 Area of Plane Figures Additional Questions

**Multiple Choice Questions**

Question 1.

If the area of an equilateral triangle is 16√3 cm², then the (RBSESolutions.com) perimeter of the triangle is:

(A) 48 cm

(B) 24 cm

(C) 12 cm

(D) 306 cm

Solution

(B) 24 cm

Question 2.

If two sides of a right-angled triangle are 5 cm then the third side is:

(A) 13 cm

(B) 17 cm

(C) 7 cm

(D) 4 cm

Solution

(A) 13 cm

Question 3.

The area of the triangle (RBSESolutions.com) having sides 1 m, 2 m and 2 m respectively is:

Solution

(D) \(\frac { \surd 15 }{ 4 }\) m^{2}

Question 4.

1 Hectare is equal to:

(A) 10 arc

(B) 100 arc

(C) 1000 arc

(D) None of these

Solution

(B) 100 arc

Question 5.

Area of an equilateral triangle (RBSESolutions.com) of side 10 cm is:

(A) 100√3 cm²

(B) 50√3 cm²

(C) 25√3 cm²

(D) 10√3 cm²

Solution

(C) 25√3 cm²

Question 6.

The base and hypotenuse of a right triangle are respectively 12 cm and 13 cm long. Its area is:

(A) 25 cm²

(B) 65 cm ²

(C) 30 cm²

(D) 40 cm²

Solution

(C) 30 cm²

Question 7.

The perimeter of a triangle is 30 m. If its (RBSESolutions.com) sides are in the ratio 1 : 3 : 2, then its smallest side is:

(A) 1 cm

(B) 5 cm

(C) 10 cm

(D) 15 cm

Solution

(B) 5 cm

Question 8.

The area of an isosceles triangle is 12 sq. cm. If one of the equal sides is 5 cm, then the length of the base is:

(A) 4 cm

(B) 5 cm

(C) 6 cm

(D) 12 cm

Solution

(C) 6 cm

Question 9.

The height of an equilateral (RBSESolutions.com) triangle is 10 cm, its area is:

Solution

(D) \(\frac { 100 }{ \surd 3 }\) cm^{2}

Question 10.

The area of a right-angled triangle is 40 times its base, then its height is:

(A) 45 cm

(B) 80 cm

(C) 60 cm

(D) None of these

Solution

(B) 80 cm

**Very Short Answer Type Questions**

Question 1.

The area of an equilateral triangle is 16√3 cm^{2}. Find the (RBSESolutions.com) length of each side of the triangle.

Solution.

Area of an equilateral triangle = \(\frac { \surd 3 }{ 4 }\) (side)^{2}

⇒ 16√3 = \(\frac { \surd 3 }{ 4 }\) (side)^{2}

⇒ (side)^{2} = 64

⇒ side = √64 = 8 cm

Question 2.

Find the altitude of a triangular region whose base is 40 m and area is 560 m^{2}.

Solution.

Area of triangular region = \(\frac { 1 }{ 2 }\) x base x altitude

560 m^{2} = \(\frac { 1 }{ 2 }\) x 40 m x altitude

⇒ 560 m^{2} = 20 m x altitude

⇒ altitude = 28 m

Question 3.

The edges of a triangular board are 6 cm, 8 cm and 10 cm respectively. Find the (RBSESolutions.com) cost of painting at the rate of ₹ 9 per m^{2}.

Solution.

Semi-perimeter of a triangular board = \(\frac { 6+8+10 }{ 2 }\) = 12 cm

Area of triangular board = \(\sqrt { 12(12-6)(12-8)(12-10) } \)

= \(\sqrt { 12\times 6\times 4\times 2 } \)

= 24 cm^{2}

Cost of painting 1 cm^{2} is ₹ 9

Cost of painting 24 cm^{2} is 24 x 9 = ₹ 216.

Question 4.

Find the percentage increase in the area (RBSESolutions.com) of an equilateral triangle if its every side is doubled.

Solution.

Let each side of an equilateral triangle be x, then its area = \(\frac { \surd 3 }{ 4 }\) x^{2}

Now, when the side is doubled i.e. 2x.

Question 5.

Two sides of a triangular field are 85 m and 154 m and (RBSESolutions.com) its perimeter is 324 m. Find the area of the field.

Solution.

Perimeter = 324 m

⇒ Semi-perimeter (s) = 162 m

85 + 154 + third side = 324 m

⇒ third side = 324 m – 239 m = 85 m

Area of triangular field

Question 6.

Find the area of (RBSESolutions.com) an equilateral triangle of side 4√3 cm.

Solution.

The side of an equilateral triangle = 4√3 cm

Area of an equilateral triangle

Question 7.

If the area of a square with side a is equal to the area (RBSESolutions.com) of a triangle with base a, then find the altitude of the triangle.

Solution.

Area of a square with side a = a^{2} sq. units

Area of a triangle with base a = \(\frac { 1 }{ 2 }\) x a x h sq. units

According to question,

Area of square = Area of triangle

a^{2} = \(\frac { 1 }{ 2 }\) x a x h

⇒ 2a^{2} = ah

⇒ h = 2a

Hence, altitude of the triangle is 2a.

Question 8.

The perimeter of a right-angled triangle is 60 cm. Its (RBSESolutions.com) hypotenuse is 26 cm. Find the area of the triangle.

Solution.

Let base and height of a right angled triangle be b and h cm respectively

b + h + 26 = 60

⇒ b + h = 34

⇒ (b + h)^{2} = (34)^{2}

Also b^{2} + h^{2} = (26)^{2} …(ii)

Subtracting (ii) from (i), we get

(b + h)^{2} – (b^{2} + h^{2}) = (34)^{2} – (26)^{2}

⇒ b^{2} + h^{2} + 2bh – b^{2} – h^{2} = (34 – 26)(34 + 26)

⇒ 2bh = 8 x 60

⇒ bh = 240

⇒ Area = \(\frac { 1 }{ 2 }\) x b x h = \(\frac { 1 }{ 2 }\) x 240 = 120

Hence, area =120 cm^{2}.

Question 9.

Using Heron’s formula, find the area of (RBSESolutions.com) an equilateral triangle the length of whose one side is ‘x’.

Solution.

The semi perimeter (s) = \(\frac { x+x+x }{ 2 }\) = \(\frac { 3x }{ 2 }\)

Area of the triangle

Question 10.

An isosceles right triangle has (RBSESolutions.com) an area of 200 cm^{2}. What is the length of its hypotenuse?

Solution.

Suppose ABC be an isosceles right triangle in which AB = BC = a

Area = \(\frac { 1 }{ 2 }\) x base x altitude

200 = \(\frac { 1 }{ 2 }\) x a x a

200 = \(\frac { 1 }{ 2 }\) x a2

⇒ a^{2} = 2 x 200

⇒ a^{2} = 400

⇒ a = 20 cm

i.e. AB = BC = 20 cm

Hypotenuse AC

**Short Answer Type Questions**

Question 1.

Find the area of a triangular field, the length (RBSESolutions.com) of whose sides are 275 m, 660 m and 715 m. What is the cost of cultivating the field at the rate of ₹ 200 per hectare?

Solution.

Suppose sides of a triangular field are

a = 275 m, b = 660 m and c = 715 m

Question 2.

Find the area of the triangle ABC whose (RBSESolutions.com) dimensions are shown in the figure.

Solution.

Question 3.

The area of a triangle is 30 cm^{2}. Find the base (RBSESolutions.com) if the altitude exceeds the base by 7 cm.

Solution.

Let base QR = x cm

altitude (PS) = (x + 7) cm

Now, area of ΔPQR = \(\frac { 1 }{ 2 }\) x base x altitude

30 = \(\frac { 1 }{ 2 }\) x QR x PS

= \(\frac { 1 }{ 2 }\) × x × (x + 7)

⇒ 60 = x^{2} + 7x

⇒ x^{2} + 7x – 60 = 0

⇒ x^{2} + 12x – 5x – 60 = 0

⇒ x (x + 12) – 5 (x + 12) = 0

⇒ (x + 12)(x – 5) = 0

⇒ x = 5 or x = -12

⇒ x = -12 is not admissible

⇒ x = 5 cm

base (x) = 5 cm

and corresponding altitude = x + 7 = 5 + 7 = 12 cm.

Question 4.

The difference between the sides at right angles in (RBSESolutions.com) a right-angled triangle is 14 cm. The area of the triangle is 120 cm^{2}. Calculate the perimeter of the triangle.

Solution.

Suppose the sides containing right angle be x and (x – 14) cm

Area of right angled triangle = \(\frac { 1 }{ 2 }\) x base x altitude

= \(\frac { 1 }{ 2 }\) × x (x – 14) cm^{2}

area = 120 cm^{2}

⇒ 120 = \(\frac { 1 }{ 2 }\) x (x^{2} – 14x)

⇒ x^{2} – 14x – 240 = 0

⇒ x^{2} – 24x + 10x – 240 = 0

⇒ x(x – 24) + 10(x – 24) = 0

⇒ (x – 24)(x + 10) = 0

either x – 24 = 0

⇒ x = 24

or x + 10 = 0

⇒ x = -10 (neglecting)

x = 24 cm (one side)

and another side = (x – 24) = (24 – 14) = 10 cm

Hypotenuse = √(24^{2} + 10^{2}) = √(576 + 100) = √676 = 26 cm

Perimeter of the triangle = 24 + 10 + 26 = 60 cm

Question 5.

A right triangle has the hypotenuse of length q cm and (RBSESolutions.com) one side of length p cm. If q – p = 2 cm, express the length of the third side of the right triangle in terms of q.

Solution.

Let length of the third side of the right triangle be x.

x^{2} + p^{2} = q^{2}

⇒ x^{2} = q2 – p^{2}

⇒ x^{2} = (q + p)(q – p) [∵ q – p = 2]

⇒ x^{2} = (q + q – 2) (2)

⇒ x^{2} = (2q – 2) x 2

⇒ x = √(4q – 4)

⇒ x = 2√(q – 1)

Hence, length of the third side of right triangle is 2√(q – 1)

Question 6.

If each side of an equilateral triangle is increased by 2 cm, then (RBSESolutions.com) its area is increased by 3√3 cm^{2}. Find the length of each side and its area.

Solution.

Suppose side of an equilateral triangle be ‘a’, then

Question 7.

Prove that the area of a triangle having (RBSESolutions.com) sides 8 m, 10 m and 12 m is 15√7 m^{2}.

Solution.

The sides of a triangle are 8 m, 10 m and 12 m respectively.

Let it be a = 8 m, b = 10 m and c = 12 m

**Long Answer Type Questions**

Question 1.

Find the area of (RBSESolutions.com) a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Solution.

Here, we are given sides of a quadrilateral ABCD as

AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Here, we see that AB^{2} + BC^{2} = AC^{2}

i e. (3)^{2} + (4)^{2} = (5)^{2}

Hence, ∆ABC is right angled at B

= √(7 x 2 x 2 x 3)

= 2√21 cm^{2}

Hence, area of quadrilateral ABCD = ar (∆ABC) + ar (∆ACD) = (6 + 2√21) cm^{2}

Question 2.

A floral design on a floor is made up of 16 tiles which (RBSESolutions.com) are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost of polishing the tiles at a rate of 50 paise per cm^{2}.

Solution.

The sides of a triangular tile in floral design are 9 cm, 28 cm and 35 cm respectively.

Semi-perimeter of triangular tile (s) = \(\frac { 9+28+35 }{ 2 }\) = \(\frac { 72 }{ 2 }\) = 36 cm

Area of a triangular tile by using Heron’s formula

Total area of such 16 tiles (RBSESolutions.com) in a floral design

= 16 x 36√6

= 576√6

= 576 x 2.45 cm^{2} [√6 = 2.45 approximate]

= 1411.20 cm^{2} approximate

Cost of polishing the tile = 50 paise/cm^{2}

Cost of polishing the 1411.20 cm^{2} = ₹ 1411.20 x \(\frac { 1 }{ 2 }\) = ₹ 705.60.

Question 3.

A field is in the shape of a trapezium whose parallel (RBSESolutions.com) sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.

Solution.

PQRS is the given trapezium in which PQ || SR

Through R draw RM || SP

Now, PMRS become a parallelogram

PS = RM = 13 m

PM = SR = 10 m

⇒ MQ = PQ – PM = 25 m – 10 m = 15 m

Now for finding the area of the field which is trapezium in shape, we (RBSESolutions.com) first find the area of the ΔRMQ and then the height of the triangular field RMQ.

Semi-perimeter of ΔRMQ

Area of parallelogram PMRS = base x altitude = 10 m x 11.2 m = 112 m^{2}

Area of trapezium PQRS = area (||gm PMRS) + area (ΔRMQ) = 112 m^{2} + 84 m^{2} = 196 m^{2}

Question 4.

A Rangoli design which is triangular in shape is made (RBSESolutions.com) on the floor, with four different colours red, pink, green and yellow (as shown in the figure).

Solution.

Since, the Rangoli design is made of four similar triangles (RBSESolutions.com) with sides 5 cm, 5 cm and 4 cm respectively.

Similarly, other triangular regions: green, yellow and pink are same in area i.e. = 2√21 cm^{2}.

Hence, total triangular areas used in the Rangoli = 4 x 2√21 cm^{2} = 8√21 cm^{2}.

Question 5.

Calculate the area of the shaded portion (RBSESolutions.com) of the triangle as shown in the figure.

Solution.

In right angled triangle ABD

AB^{2} = AD^{2} + BD^{2}

⇒ AB^{2} = (12)^{2} + (16)^{2}

⇒ AB^{2} = 144 + 256

⇒ AB^{2} = 400 cm^{2}

⇒ AB = √400 cm^{2}

⇒ AB = 20 cm

Now in ΔABC, a = 48 cm, b = 52 cm, c = 20 cm.

Thus, area of shaded portion (RBSESolutions.com) of the triangle = (480 – 96) cm^{2} = 384 cm^{2}

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

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