Heron’s Formula Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. Write the statement of Heron’s Formula.

Ans. Heron's formula is a method that helps calculate the area of triangles given their three sides. By dividing the quadrilateral into two triangles along its diagonal, this formula can also be used to get the area of the quadrilateral.

Ques. Write the Heron’s Formula.

Ans. If Perimeter = a + b + c
Then, s = Perimeter/2
Thus,

Heron’s Formula:

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

Also Read: MCQ on Heron’s Formula 

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Short Answer Questions [2 Marks Questions]

Ques. A park houses a slide. Recently, a company hired one of the walls for a period of 3 months. It was then painted in a color that read the message, “KEEP THE PARK GREEN AND CLEAN.” Assuming that the sides of the wall are 15m, 11m and 6m, determine the area painted in that color.

Ans. As given in the equation,
a = 15m
b = 11m
c = 6m

Then the Semi-perimeter will be,
Now, if s = (a + b + c)/2
= (15 + 11 + 6)/2
= 32/2
= 16m.
Now, considering the area of the triangular wall surface,

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

= \(\sqrt{16 (16 - 15)(16 - 11)(16 - 6)}\)

=\(\sqrt{16 \times 1 \times 5 \times 10}\)

=\(\sqrt{2 \times 400}\)

Thus, the required area painted in color
= 20√2 m²

Ques. Determine the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.

Ans. Let the sides of the triangle be a =18 cm, b = 10 cm and c = x cm
Since, perimeter of the triangle = 42 cm
∴ 18cm + 10 cm + x cm = 42
x = [42 – (18 + 10)cm = 14cm]
Now, we know that the semi-perimeter, s = √(441)cm = 21 cm

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

= \(\sqrt{21 (21 - 18)(21 - 10)(21 - 14)}\)

=\(\sqrt{21 \times 3 \times 11 \times 7}\)

=\(\sqrt{3 \times 7\times3\times11\times7}\)  = 21√(11) cm²

∴The required area of the triangle = 21√(11) cm²

Ques. Sides of a triangle are represented in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Determine its area.

Ans. Let the sides of the triangle be
a = 12z cm
b = 17z cm
c = 25z cm
Perimeter of the triangle = 540 cm
Now, 12z + 17z + 25z = 540
⇒ 54x = 54 ⇒ z = 10
∴a = (12 x10)cm = 120 cm,
b = (17 x 10) cm = 170 cm
and c = (25 x 10)cm = 250 cm
Now, semi-perimeter, s = 540/2 cm = 270 cm

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

= \(\sqrt{270 (270 - 120)(270 - 170)(270 - 250)}\)

=\(\sqrt{270 \times 100 \times 150 \times 20}\)

=\(\sqrt{10^2 \times 10^2\times3^2\times3^2\times5^2\times2^2}\)  
= (10 * 10 * 3 * 3 * 5 * 2) cm²
= 9,000 cm²

Read More: Area of a Triangle: Formula, Types and Solved Examples

Ques. A triangle and a 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 28 cm, show the height of the parallelogram.

Ans. For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm
Now, as the area of the triangle can be given by = \(\sqrt{s (s - a)(s - b)(s - c)}\)
After replacing the values = \(\sqrt{42 (42 - 28)(42 - 30)(42 - 26)}\) cm²
Therefore, \(\sqrt{42 \times14 \times12\times16}\) cm²
= \( \sqrt{112896}\)cm²
= 336 cm²

Area of the given parallelogram = Area of the given triangle
∴Area of the parallelogram = 336 cm²
⇒ base x height = 336
⇒ 28 x h = 336, where ‘h’ is the height of the parallelogram.
⇒ h = 33628
= 12
Thus, the required height of the parallelogram = 12 cm

Ques. An umbrella is made by stitching 10 triangular pieces of cloth of two different colors (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each color is required for the umbrella?
Ans. Let the sides of each triangular piece be
a = 20 cm,
b = 50 cm,
c = 50 cm
Therefore, the semi-perimeter can be considered as, s = (a + b + c)/2
= (20 + 50 + 50)/2 cm = 120/2 cm
= 60 cm

Semi-perimeter, s = (a+b+c)/2

= (20+50+50)/2 cm

= 60 cm

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

= \(\sqrt{60 (60 - 20)(60 - 50)(60 - 50)} cm^2\)

= \(\sqrt{60 \times 40 \times 10 \times 10}\ cm^2 = 200\sqrt6 cm^2\)

Area of 5 triangular pieces of one color

= 500 x 200\(\sqrt6 cm^2 = 1000\sqrt6 cm^2\)

Therefore, the area of 5 triangular pieces of other colors = \(1000\sqrt6cm^2\)

Read More: Scalene Triangle

Long Answer Questions [3 Marks Questions]

Ques. The triangular side walls of a flyover are recently utilized for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements gain about 5000/m² every year. A company hired one of its walls for 3 months, then determine what rent it paid?

Ans. Let the sides of the triangular will be
a = 122m
b = 22m
c = 120m

Semi-perimeter, s = (a + b + c)/2
s = (122 + 22 + 120)/2 m
= 132 m
The area of the triangular sidewall can be given by = \(\sqrt{s (s - a)(s - b)(s - c)}\)

= \(\sqrt{132 (132 - 122)(132 - 120)(132 - 22)}\)

=\(\sqrt{132 \times 10 \times 12 \times 110}\)

=\(\sqrt{12 \times 11\times10\times12\times11\times10}\)  

Rent for 1 year (i.e. 12 months) per m² = Rs. 5000
∴ Rent for 3 months per m² = Rs. 5000 x 312
= Rent for 3 months for 1320 m² = Rs. 5000 x 312 x 1320 = Rs. 16,50,000.

Ques. A rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?

Ans. Here, each side of the rhombus = 30 m.
Let ABCD be the given rhombus and the diagonal, BD = 48 m

Sides ABC
a = AB = 30m,
b = AD = 30m,
c = BD = 48m
Now, considering Semi-perimeter, s = (a + b + c)/2
Then,

Semi perimeter, s = (a+b+c)/2

= 54 m

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

= \(\sqrt{54 (54 - 30)(54 - 30)(54 - 48)} m^2\)

= 432 m²
Since, a diagonal divides the rhombus into two congruent triangles.
∴Area of triangle II = 432 m²
Now, total area of the rhombus = Area of triangle I + Area of triangle II
= 432 m² + 432 m²
= 864 m²

Area of grass for 18 cows to graze = 864 m²
⇒ Area of grass for 1 cow to graze = 86418 m²
= 48 m²

Read Also:  Pythagoras Theorem

Ques. A ground is shaped as a trapezium that has parallel sides, 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Determine the area of the field.

Ans. Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m
Non-parallel sides are AD = 13 m and BC = 14 m.
Now, we need to draw BE || AD, in a way BE = 13 m.

The given field is divided into two shapes (i) †BCE
(ii) parallelogram ABED for BCE:
Sides of the triangle are
a = 13 m
b = 14 m
c = 15 m

Semi perimeter, s = (a+b+c)/2

= (13+14+15)/2

= 21 m

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

= \(\sqrt{21 (21 - 13)(21 - 14)(21 - 15)} m^2\)

= 84 m²

(ii) Now, for parallelogram ABED,
Let the height of the △BCE corresponding to the side EC be ‘h’ m.
Area of a triangle = 12 x base x height
∴ 12 x 15 x h = 84
⇒ 10 + 82×215 = 565
Now, area of a parallelogram = base x height
= (10 x 565) = (2 x 56) m² = 112 m²
So, area of the field
= area of BCE + area of parallelogram ABED
= 84 + 112  = 196 m²

Read More: Trapezoid

Ques. Sides of a Triangle are in the ratio of 20: 25: 14 and its perimeter is 590cm. Find its area.

Ans. The ratio of the sides of the triangle is given as 20: 25: 14
Let us consider the common ratio between the sides of the triangle be “a”
∴The sides are 20a, 25a and 14a
It is also given that the perimeter of the triangle = 590 cm
20a + 25a + 14a = 590
∴59a = 590
So, a = 10

Now, the sides of the triangle can be classified into, 200 cm, 250 cm, 1400 cm.
So, the semi perimeter of the triangle (s) = 590/2 = 295 cm
Using Heron’s formula for Area of the triangle
= \(\sqrt{s (s - a)(s - b)(s - c)}\) cm²
= \(\sqrt{295(295−200)(295−250)(295−140)}\)} cm²
= \(\sqrt{(295×95×45×155)}\) cm²
= \(\sqrt{195,474,375}\) cm²
= 13981.21cm²

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Very Long Answer Questions [5 Marks Questions]

Ques. Evaluate the area of a quadrilateral ABCD using the following measurements: AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.

Ans. As per the given equation quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

For △ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm
Now, for the semi-perimeter, s = (a + b + c )/2
Thus, after substituting the values, we get, 3 + 4 + 52 cm
= 6 cm
Hence, the Area of ΔABC = \(\sqrt{s (s - a)(s - b)(s - c)}\)

Semi perimeter, s = (a+b+c)/2

= (3+4+5)/2

= 6 cm

Area of triangle ABC = \(\sqrt{6 (6 - 3)(6 - 4)(6 - 5)}\)

= \(\sqrt{6 \times 3 \times 2 \times 1}\)

= 6 cm²

For triangle ACD, a = AD = 5 cm

b = CD = 4 cm

c = AC = 5cm

Semi perimeter, s = (5+4+5)/2 = 7 cm

Area of triangle ACD = \(\sqrt{7 (7 - 5)(7 - 4)(7 - 5)}\)

= \(2\sqrt{21}\)

= 9.2 cm² (approx)

Therefore, Area of ABCD = Area of ABC + Area of AC
ACD = 6 cm² + 9.2 c\(\sqrt{s (s - a)(s - b)(s - c)}\)m²
= 15.2 cm² (in approx.)

Ques. As portrayed below, a kite which is shaped as a square with a diagonal of 32 cm and an isosceles triangle with a base of 8 cm and sides of 6 cm each is to be created of three different hues. Determine the amount of color each side can use.

Ans. Each shade of paper is divided into 3 triangles i.e., I, II, III

For triangle I:
ABCD is a square [Given]
Diagonals of a square are equal and bisect each other.
∴ AC = BD = 32 cm
Height of ABD = OA = (12 x 32 )cm
= 16 cm
Area of triangle I = (12 x 32 x 16 ) cm²
= 256cm²

For triangle II:
Since, the diagonal of a square divides it into two congruent triangles.
So, area of triangle II = area of triangle I
∴ Area of triangle II = 256 cm²

For triangle III:
The sides are given as a = 8 cm, b = 6 cm and c = 6 cm

Semi perimeter, s = (a+b+c)/2

= (8+6+6)/2

= 10 cm

Area of triangle III = \(\sqrt{10 (10 - 8)(10 - 6)(10 - 6)}\)

= \(8\sqrt{5}\)

= 17.92 cm²

Thus, the area of different shades are:
Area of shade I = 256 cm²
Area of shade II = 256 cm²
And, area of shade III = 17.92 cm²

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