CLASS 9TH

SUBJECT: MATHEMATICS

CHAPTER 12: HERON FORMULA (MCQ)

1. Which of the following denotes semi perimeter(s) in Heron’s formula?

a. s = sum of sides of triangle.

b. s = product of sides of triangle.

c. s = sum of opposite sides of triangle.

d. s = half of the sum of sides of triangle.

2. Space occupied by a three-dimensional solid are named as

a. Perimeter.

b. Surface area.

c. Density.

d. Volume.

3. Which defines uniform cross-section

a. Opposites sides which are of same lize but of different shape.

b. Opposite faces which are of same shape but of different size.

c. Adjacent sides are of similar size and shape.

d. Opposite faces are of same size and same shape.

4. 2 RH is used to determine which of the following

a. Area of curved surface.

b. Volume of curved surface.

c. Perimeter of curved surface.

5. Major use of Heron’s formula is

a. Determine the perimeter of triangle.

b. Determine the volume of triangle.

c. Determine the area of triangle.

d. Determine the dimensions of the triangle.

6. The semi perimeter of a triangle when the sides of the triangle are of 25 cm, 15 cm

and 8 cm is

a. 44 cm

b. 21 cm

c. 24 cm

d. 48 cm

7. An isosceles triangle with equal sides of 12 cm. Perimeter of the triangle is 42 cm

then determine the area of the triangle.

a. 108 sq. cm

b. 220 sq. cm

c. 390 sq. cm

8. Circumference of a lawn in the shape of rectangle can be determined using

a. Sum of all diagonals.

b. Sum of all sides.

c. Product of length and width.

d. Product of diagonals.

9. Surface enclosed by its perimeter is known as

a. Area.

b. Volume.

c. Surface volume.

d. Density.

10. Which of the following is used to find area of an equilateral triangle

a. Product of length of base side and height.

b. Square of sides.

c. 0.5 times of the sum of sides.

d. 0.5 times of product of base length and height.

11. Identify Heron’s formula from the following

a. Area =

b. Area =

c. Area =

12. If the side of an equilateral triangle is 2 cm then area will be

a. cm2

b. cm2

c. cm2

d. cm2

13. If area of an equilateral triangle is a2 cm

2 then what could be the length of side

a. a2

2 cm

b. a 4 cm

c. a 2 cm

d. a2 cm

ANSWERS:

1. d

2. d

3. d

4. a

5. c

6. c

7. a

8. b

9. a

10. d

11. a

12. b

13. c

CLASS: 9th

SUBJECT: MATHEMATICS

CHAPTER 12: HERON’S FORMULA

VERY SMALL ANSWERS QUESTIONS

1. What is perimeter of triangle whose sides are a, b and c?

Ans. Perimeter of triangle = a+b+c.

2. What is the area of triangle?

Ans. Area of triangle = ½ X base X height

3. What is the area of equilateral triangle?

Ans. √3/4.

4. What is the area of an isosceles triangle with length of two equal sides a and base

b?

Ans. Area of triangle = ½ X b X √a2 - b

2 /4.

5. What is Heron’s formula?

Ans. If a, b, c denotes the lengths of the sides of a triangle, then area of triangle.

CLASS 9th

SUBJECT: MATHEMATICS

CHAPTER 12: HERON'S FORMULA

SHORT ANSWER TYPE QUESTIONS

1. The perimeter of Equilateral triangle is 60 cm. What will be it’s area?

ANS: Perimeter of Equilateral triangle = 3a = 60 cm

a = 60/3 = 20 m

Area of Equilateral triangle = √3/4 × (Side)2 = √3/4 × (20)

2 =100 √3 m

2

2. Find the length of each side of an Equilateral triangle having an area of 9√3 cm2

ANS: Area of equilateral triangle = √3/ 4 × (side)2

9√3 = √3/4 × (Side)2

(Side) 2

= 9 × 4 = 36

Side = 6 cm

3. The sides of a triangle are 35 cm , 54 cm and 61 cm . What is the length of the

longest altitude?

ANS: s =( 35 + 54 + 61 ) / 2 = 75 cm.

Area of a triangle = √s(s-a)(s-b)(s-c)

√75(75-35)(75-54)(75-61)

420 √5 cm2Ok

Length of longest altitude = ( 2 × Area of triangle) / 35

(2 × 420 √5 ) / 35 = 24 √5 cm.

4. What is the area of an isosceles triangle having base = 2 cm and length of equal sides

is 4 cm.

ANS: s = ( 4 + 4 + 2) / 2 = 5 cm.

Area = √s(s-a)(s-b)(s-c)

= √5(5-4)(5-4)(5-2)

= √5 × 1 ×1 ×3

= √15 cm2

5. The edge of a triangular board are 6 cm, 8 cm and 10 cm. The cost of painting it is 9

paise / cm2. Find the cost of painting the whole board.

ANS: s = ( 6 + 8 + 10) / 2 = 12 cm

Area of board = √12(12-6)(12-8)(12-10)

= √12 × 6 × 4 × 2

= √576

= 24 cm

Cost = 24 × 0.09 = Rs 2.16

6. Find the area of an isosceles triangle having perimeter 11 cm and base 5 cm.

ANS: Perimeter of isosceles triangle = a + a + b =11 cm

2a + 5 = 11 cm

2a = 11 – 5

2a = 6 cm

a = 6/2 = 3 cm

Area of an isosceles triangle = b/4 √4a2-b

2

= 5/4 √4(3)2- (5)

2

5/4√11 cm2

7. If the area of an equilateral triangle is 20√3 cm2 . Find the length of its side.

ANS: Area of an equilateral triangle = √3 / 4 (side)2

20√3 = √3 / 4 (side)2

(Side)2 = 20 ×4 = 80 cm

Side = √80 cm = 4√5 cm.

CLASS 9TH

SUBJECT: MATHEMATICS

CHAPTER 12 – HERON'S FORMULA (LAQ)

1. The triangular side walls of a flyover have been used for advertisements. The sides

of a walls are 122 m, 22 m, 120 m. The advertisement yields an earning of Rs 5000

per m2 per year. A company hired one of its walls for 3 months. How much rent did

it pay?

ANS: - s = (122 + 22 +120) / 2 = 264/2 = 132 m

Area = √132(132-122) (132-22) (132-120)

=√132 × 10 × 110 × 12

= √12 × 11 × 10 × 11 × 10 × 12

= 12 × 11 × 10

= 1320 m2.

Rent paid for 3 months = 1320 × 5000 × 3/12

= Rs 1650000.

2. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its

area.

ANS : - Let the sides be 12x , 17x and 25x.

12x + 17x + 25x = 540

54x = 540

x = 10 cm

Sides are 12x = 12 × 10 = 120 cm

17x =17 × 10 = 170 cm

25x = 25 × 10 = 250 cm

Semi perimeter = s = 540/2 = 270 cm

Area = √270(270-120)(270-170)(270-250)

= √270 × 150 × 100 × 20

= √9 × 30 × 30 × 5 × 100 × 2 × 1 = 9000 cm2

3. A park is in the shape of a quadrilateral ABCD , has DC = 90°, AB = 9 m , BC = 12

m, CD = 5 m and AD = 8 m. How much area does it occupy?

ANS : - Triangle BCD is right angled at C.

BD = √(5)2 + (12)2 = √25 + 144 = √169 = 13 m

ar(ABCD) = ar(ABD) + ar(BCD)

ar(BCD) = ½ × 12 × 5 = 30 m2

For triangle ABD ,

s =( 8 + 9 +13)/2 = 15 m

ar(ABD) = √15(15-8)(15-9)(15-13)

= √15 × 7 × 6 × 2

= 6 √35 = 35.5 m2

ar(ABCD) = 35.5 + 30 = 65.5 m2

4. A triangle and the parallelogram have the same base and the same area. If the sides

of the triangle are 26 cm , 28 cm and 30 cm and the parallelogram stands on the

base of 28 cm. Find the height of the parallelogram.

ANS : - Area of parallelogram = area of triangle

Let h be the height of the parallelogram

s = (26 + 28 + 30) / 2 = 42 cm

Area of the triangle = √42(42-26)(42-28)(42-30)

= √42 × 16 × 14 × 12

= 336 cm2

Area of parallelogram = Base × height

28 × h = 336

h = 336/28 = 12 cm

5. An umbrella is made by stitching 10 triangular pieces of clothes each piece is

measuring 20 cm , 50 cm, 50 cm. How much clothe is required for the umbrella?

ANS : - s = (20 + 50 + 50 ) / 2 = 60 cm

Area of one piece = √60(60-20)(60-50)(60-50)

=√60 × 40 × 10 × 10

= 200 √6 cm2

Cloth required for umbrella = 10 × 200√6 = 2000√6 cm2