Rajasthan Board RBSE Class 9 Maths Important Questions Chapter 12 Heron’s Formula Important Questions and Answers.

I. Multiple Choice Questions :

Choose the correct answer from the given four options.

Question 1.

If the area of equilateral triangle is 16√3 cm^{2}, then the perimeter of the triangle is :

(a) 48 cm

(b) 24 cm

(c) 12 cm

(d) 36 cm

Answer:

(b) 24 cm

Question 2.

The edge of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it at the rate of 9 paise per cm^{2} is :

(a) ₹ 2.00

(b) ₹ 2.16

(c) ₹ 2.48

(d) ₹ 3.00

Answer:

(b) ₹ 2.16

Question 3.

If the perimeter of a rhombus is 80 cm and one of its diagonals is 24 cm, then the length of the other diagonal is:

(a) 16 cm

(b) 20 cm

(c) 32 cm

(d) 48 cm

Answer:

(c) 32 cm

Question 4.

If the perimeter of an equilateral triangle is 60 m, then the area is :

(a) 10√3 m^{2}

(b) 15√3 m^{2}

(c) 20√3 m^{2}

(d) 100√3m^{2}

Answer:

(d) 100√3m^{2}

Question 5.

An isosceles right triangle has area 8 cm2. The length of its hypotenuse is :

(a) √32 cm

(b) √l6 cm

(c) √48 cm

(d) √24 cm

Answer:

(a) √32 cm

Question 6.

If the sides of a triangle are 56 cm, 65 cm and 52 cm long, then the area of the triangle is :

(a) 1322 cm^{2}

(b) 1311 cm^{2}

(c) 1344 cm^{2}

(d) 1392 cm^{2}

Answer:

(c) 1344 cm^{2}

Question 7.

If the side of a parallelogram are 9 cm and 4 cm, then the ratio of their corresponding altitudes is :

(a) 2 : 3

(b) 3 : 2

(c) 9 : 4

(d) 4 : 9

Answer:

(d) 4 : 9

Question 8.

The sides of a triangle are 35 cm, 54 cm and 61 cm. The length of its longest altitude is:

(a) 16√5 cm

(b) 10√5 cm

(c) 24√5 cm

(d) 28 cm

Answer:

(c) 24√5 cm

Question 9.

The sides of a triangle are in the ratio of 5 : 12 : 13 and its perimeter is 150 cm. The area of the triangle is :

(a) 375 cm^{2}

(b) 750 cm^{2}

(c) 250 cm^{2}

(d) 500 cm^{2}

Answer:

(b) 750 cm^{2}

Question 10.

Each of the equal sides of an isosceles triangle is 13 cm and its base is 24 cm. The area of the triangle is:

(a) 156 cm^{2}

(b) 78 cm^{2}

(c) 60 cm^{2}

(d) 120 cm^{2}

Answer:

(c) 60 cm^{2}

Question 11.

The base of right triangle is 48 cm and its hypotenuse is 50 cm long. The area of the triangle is :

(a) 168 cm^{2}

(b) 252 cm^{2}

(c) 336 cm^{2}

(d) 504 cm^{2}

Answer:

(c) 336 cm^{2}

Question 12.

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is :

(a) 16√5 cm^{2}

(b) 8√5 cm^{2}

(c) 16√3 cm^{2}

(d) 8√3 cm^{2}

Answer:

(b) 8√5 cm^{2}

Question 13.

Each side of an equilateral triangle is 10 cm long. The height of the triangle is :

(a) 10√3 cm

(b) 5√3 cm

(c) 10√2 cm

(d) 5 cm

Answer:

(b) 5√3 cm

Question 14.

The area of an equilateral triangle is 81 /3 cm2, its height is :

(a) 9√3 cm

(b) 6√3 cm

(c) 18√3 cm

(d) 9 cm

Answer:

(a) 9√3 cm

Question 15.

The lengths of the three sides of a triangular field are 40 m, 24 m and 32 m respectively. The area of the triangle is :

(a) 480 m^{2}

(b) 320 m^{2}

(c) 384 m^{2}

(d) 360 m^{2}

Answer:

(c) 384 m^{2}

II. Fill in the Blanks:

Question 1.

Area of trapezium with parallel sides 'a' and 'b' and the distance between two parallel sides as h = ___________ .

Answer:

\(\frac{1}{2}\)(a + b) × h

Question 2.

Heron's foumula is stated as :

Area of a triangle ABC = ____________

Where s = ____________

a, b, c, = ____________

Answer:

\(\sqrt{s(s-a)(s-b)(s-c)}\), semiperimeter, sides of the triangle,

Question 3.

The semiperimeter of a triangle having the length of its sides as 20 cm, 15 cm and 9 cm is ____________ .

Answer:

22 cm

Question 4.

The perimeter of an equilateral triangle is 60 m. The area is ____________ .

Answer:

100√3 cm^{2}

Question 5.

The sides of a triangle are 56 cm, 60 cm and 52 cm long, then the area of the triangle is ____________.

Answer:

1344 cm^{2}

III. True/False:

State whether the following statements are True or False.

Question 1.

Heron formula for area of triangle is not valid of all triangles.

Answer:

False

Question 2.

If each side of the triangles is tripled, the area will becomes 9 times.

Answer:

True

Question 3.

Base and corresponding altitude of the parallelogram are 8 and 5 cm respectively. Area of parallelogram is 40 cm^{2}.

Answer:

True

Question 4.

If each side of triangle is doubled, the perimeter will become 4 times.

Answer:

False

Question 5.

If p is the perimeter of the triangle of sides a, b, c, the area of triangle is

A = \(\frac{1}{4}\sqrt{p(p-2 a)(p-2 b)(p-2 c)}\).

Answer:

True

Question 6.

When two triangles are congruent, there areas are same.

Answer:

True

Question 7.

Heron belongs to America.

Answer:

False

Question 8.

If the side of the equilateral triangle is a rational number, the area would always be irrational number.

Answer:

True

IV. Match the Columns:

Match the column I with the column II.

Column I |
Column II |

(1) Perimeter of rectangle of length 24 cm and diagonal 26 cm |
(i) 22 cm |

(2) Perimeter of the square of side 10 cm |
(ii) 17 cm |

(3) Perimeter of triangle of sides 4, 5, 8 cm |
(iii) 40 cm |

(4) Perimeter of parallelogram of two sides 5 and 6 cm respectively |
(iv) 68 cm |

Answer:

Column I |
Column II |

(1) Perimeter of rectangle of length 24 cm and diagonal 26 cm |
(iv) 68 cm |

(2) Perimeter of the square of side 10 cm |
(iii) 40 cm |

(3) Perimeter of triangle of sides 4, 5, 8 cm |
(ii) 17 cm |

(4) Perimeter of parallelogram of two sides 5 and 6 cm respectively |
((i) 22 cm |

V. Very Short Answer Type Questions:

Question 1.

Find the area of ∆ABC in which AB = BC = 4 cm and ∠B = 90°.

Answer:

From the figure, ∆ABC is a right-angled triangle.

∴ Area of AABC = \(\frac{1}{2}\) × base × height

= \(\frac{1}{2}\) × (BC) × (AB) = \(\frac{1}{2}\) × 4 × 4 = 8 cm^{2}

∴ ar(∆ABC) = 8 cm^{2}.

Question 2.

Find the side of an isosceles right triangle of hypotenuse 5√2 cm.

Answer:

According to the question :

PQ - QR = a (say)

PR = 5√2

∴ By Pythagoras theorem: P^{2} + B^{2} = H^{2 }

a^{2} + a^{2} = (5√2)^{2}

⇒ 2a^{2} = 50

⇒ a^{2} = \(\frac{50}{2}\)

⇒ a^{2} = 25

⇒ a = 5 cm

Hence, equal sides of isosceles triangle are 5 cm each.

Question 3.

The perimeter of an equilateral triangle is 60 m. Find its area.

Answer:

Let side of equilateral be 'a'.

Now, perimeter = 3a

⇒ 3a = 60 m

⇒ a = \(\frac{60}{3}\) = 20 m

Then, area of equilateral triangle = \(\frac{\sqrt{3}}{4}\)(side)^{2}

⇒ area of equilateral triangle = \(\frac{\sqrt{3}}{4}\) × (20)^{2} = \(\frac{\sqrt{3}}{4}\) × 20 × 20 = 100√3

∴ Area = 100 √3 m^{2}.

Question 4.

Area of an isosceles right triangle is 8 cm^{2}. Find its hypotenuse.

Answer:

Let ABC be an isosceles right triangle, where

∠B = 90°

Let AB = BC = a

Now, area of ∆ABC = \(\frac{1}{2}\) × base × height

⇒ 8 = \(\frac{1}{2}\) × a × a

⇒ a^{2} = 16

⇒ a = 4

Now, by Pythagoras theorem

AB^{2} + BC^{2} = AC^{2}

⇒ a^{2} = a^{2} = AC^{2}

⇒ (4) + (4) = AC^{2}

⇒ 16 + 16 = AC^{2}

⇒ AC^{2} = 32

⇒ AC = √32

⇒ AC = 4√2 cm

∴ Hypotenuse = 4√2 cm.

VI. Short Answer Type Questions:

Question 1.

If the length of median of an equilateral triangle be x cm, then find its area.

Answer:

Let ABC be an equilateral triangle of side ‘a’ cm. A

Also, we know that in an equilateral triangle, the median and perpendicular coincide with each other.

∴ AD ⊥ BC and D is mid-point of BC.

Question 2.

The sides of a triangle are 7 cm, 24 cm and 25 cm. What is its area?

Answer:

Since, sides are given, we use Heron’s formula to find area of triangle.

Let a = 7 cm, b = 24 cm and d = 25 cm

s = \(\frac{a+b+c}{2}\) = \(\frac{7+24+25}{2}=\frac{56}{2}\) = 28

∴ Area = \(\sqrt{s(s-a)(s-b)(s-c)}\)

∴ Area = A = \(\sqrt{28(28-7)(28-24)(28-25)}\)

⇒ A = \(\sqrt{28 \times 21 \times 4 \times 3}\)

⇒ A = \(\sqrt{2 \times 2 \times 7 \times 7 \times 3 \times 2 \times 2 \times 3}\)

⇒ A = 2 × 7 × 3 × 2 = 84

Hence, area of triangle = 84 cm^{2}

Question 3.

The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the cost of painting it at the rate of 9 paise per cm^{2}.

Answer:

Let a = 10 cm, b = 8 cm and c = 6 cm

Then, s = \(\frac{a+b+c}{2}\) = \(\frac{10+8+6}{2}\) = \(\frac{24}{2}\) = 12

∴ Area = \(\sqrt{12(12-10)(12-8)(12-6)}\)

⇒ Area = \(\sqrt{12 \times 2 \times 4 \times 6}\) = 24

∴ Area = 24 cm^{2}

Now, cost of painting the board at the rate of 9 paise per cm^{2}

= 24 × 0.09 = ₹ 2.16 (∵ 9 paise = ₹ 0.09)

Hence, cost of painting = ₹ 2.16.

Question 4.

In figure, P and Q are two lamp posts. If the area of the ∆PDC is same as that of the rectangle ABCD, find the distance between the two lamp posts.

Answer:

In ∆PCD,

a = 12 cm, 6 = 6 cm, c = 10 cm.

s = \(\frac{12+6+10}{2}\) cm = 14 cm.

Question 5.

A municipal corporation wall on road side has dimensions as shown in figure. The wall is to be used for advertisements and it yields an earning of ₹ 400 per m^{2} in a year. Find the total amount of revenue earned in a year.

Answer:

We can find the total area of the wall in two parts. One parts triangular, i.e. ∆ABC at the top and the second part is rectangular, i.e. rectangle PQCB at the bottom.

In triangular part, a = 14 m, b = 13 m, c = 15 m.

The area of rectangle PQCB = 14 × 3 m^{2} = 42 m^{2}.

Total area of the wall = 84 m^{2} + 42 m^{2} = 126 m^{2}

Total amount of revenue earned in a year at the rate of ₹ 400/m^{2}

= ₹ 400 × 126 = ₹ 50,400

Question 6.

Find the area of the pentagon ABCDE shown in figure.

Answer:

In triangle BCD, ∠C = 90°, therefore, by Pythagoras theorem,

BD = \(\sqrt{B C^{2}+C D^{2}}\) = \(\sqrt{4^{2}+3^{2}}\) cm

= \(\sqrt{16+9}\) cm = √25 cm = 5 cm

Again in ∆ABD, ∠ABD = 90°, therefore, by Pythagoras theorem,

Next, we find area ∆ADE using Heron’s formula

Hence, the area of the pentagon ABCDE

= area ∆ABD + area ∆BCD + area ∆ADE

= (30 + 6 + 84) cm^{2} = 120 cm^{2}

Question 7.

The sides of a triangular park are 8 m, 10 m and 6 m respectively. A small circular area of diameter 2 m is to be left out and the remaining area is to.be used for growing roses. How much area is used for growing roses? (take π = 3.14)

Answer:

Let sides of a triangle be a = 8 m, b = 10 m and c = 6 m

Now, semi-perimeter of a triangle, s = \(\frac{a+b+c}{2}\)

= \(\frac{8+10+6}{2}\) = \(\frac{24}{2}\)= 12 m

Area of a triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{12(12-8)(12-10)(12-6)}\)

= \(\sqrt{12 \times 4 \times 2 \times 6}\) = 24 m^{2}

and area of a circle = πr^{2} = 3.14 × 1^{2}

= 3.14 m^{2}

∴ Ares for growing roses = Area of a triangle - area of a circle

= 24 - 3.14 = 20.86 m^{2}

VII. Long Answer Type Questions:

Question 1.

The length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 144 cm. Find :

(i) the area of the triangle, and

(ii) the height corresponding to the longest side.

Answer:

Given, perimeter = 144 cm

Ratio of sides = 3:4:5

Let the sides be 3x, 4x and 5x

∴ 3x + 4x + 5x = 144

12x = 144

x = 12

1st side, 3x = 3 × 12 = 36 cm

Ilnd side, 4x = 4 × 12 = 48 cm

Illrd side, 5x = 5 × 12 = 60 cm

∴ a = 3x, b = 4x, c = 5x

Now, semi-perimeter of the triangle,

s = \(\frac{a+b+c}{2}\) = \(\frac{36+48+60}{2}\) = \(\frac{144}{2}\) = 72cm

(i) Area of triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\) (By Heron’s formula)

= \(\sqrt{72 \times(72-36)(72-48)(72-60)}\)

= \(\sqrt{72 \times 36 \times 24 \times 12}\)

= \(\sqrt{(36)^{2} \times(24)^{2}}\)

= 36 × 24 = 864 cm^{2} .

Hence, the area of the given triangle is 864 cm^{2}.

(ii) Let height of a triangle be h cm.

Then, area of triangle = \(\frac{1}{2}\) × base × height

⇒ 864 = \(\frac{1}{2}\) × 60 × h

(Since, longest side of a triangle is 60 cm, so we consider it as base of the triangle.)

⇒ 864 = 30h ⇒ h = 28.8 cm

Hence, the height corresponding to the longest side is 28.8 cm.

Question 2.

A rectangular plot is given for constructing a house having a measurement of 40 m long and 15 m in the front.

According to the laws, a minimum of 3 m wide space should be left in the front and back each and 2 m wide space on each of other sides. Find the largest area, where house can be constructed.

Answer:

Let ABCD be a rectangular plot having a measurement of 40 m long and 15 m front.

⇒ Length of inner rectangle

= EF = 40 - 3 - 3 = 34 m

and breadth of inner rectangle

= FG = 15 - 2 - 2 = 11 m

Now, when space are left according to the law, then

∴ Area of inner rectangle, EFGH

= EF × FG = 34 × 11 = 374 m^{2 }(∵ Area of a rectangle = length × breadth)

Hence, the largest area, where the house can be constructed, is 374 m^{2}.

Question 3.

A design is made on a rectangular tile of dimensions 50 cm x 70 cm as shown in figure. Find the total area of the design and the remaining area of the tile.

Answer:

We have

Area of the rectangular tile

= (length × breadth) sq. units

= (50 × 70) cm^{2} = 3500 cm^{2}

For area of each triangle

Let a = 26 cm, b = 17 cm, c = 25 cm

s = \(\frac{a+b+c}{2}\) units

s = \(\frac{26+17+25}{2}\) = \(\frac{68}{2}\) = 34 cm

s - a = 34 - 26 = 8cm

s - b = 34 - 17 = 17 cm

s - c = 34- 25 = 9 cm

Area of each triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)

= \(\sqrt{34 \times 8 \times 17 \times 9}\)9 = \(\sqrt{2 \times 17 \times 2 \times 2 \times 2 \times 17 \times 9}\)

= 12 × 17 = 204 cm^{2}

∴ Total area of design = 8 × area of one triangle

= (8 × 204) cm^{2} = 1632 cm^{2}

So,remaining area of the tile = Area of rectangular tile - total area of the design

= 3500 - 1632 = 1868 cm^{2}

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