Chapter 10: Heron’s Formula

First, construct a quadrilateral ABCD and join BD.

We know that

C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

The diagram is:

Now, apply Pythagoras theorem in ΔBCD

BD² = BC² +CD²

⇒ BD² = 12²+5²

⇒ BD² = 169

⇒ BD = 13 m

Now, the area of ΔBCD = (½ ×12×5) = 30 m²

The semi perimeter of ΔABD

(s) = (perimeter/2)

= (8+9+13)/2 m

= 30/2 m = 15 m

Using Heron’s formula,

Area of ΔABD

= 6√35 m² = 35.5 m² (approximately)

∴ The area of quadrilateral ABCD = Area of ΔBCD+Area of ΔABD

= 30 m²+35.5m² = 65.5 m²

First, construct a diagram with the given parameter.

Now, apply Pythagorean theorem in ΔABC,

AC² = AB²+BC²

⇒ 5² = 3²+4²

⇒ 25 = 25

Thus, it can be concluded that ΔABC is a right angled at B.

So, area of ΔBCD = (½ ×3×4) = 6 cm²

The semi perimeter of ΔACD (s) = (perimeter/2) = (5+5+4)/2 cm = 14/2 cm = 7 m

Now, using Heron’s formula,

Area of ΔACD

= 2√21 cm² = 9.17 cm² (approximately)

Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD = 6 cm² +9.17 cm² = 15.17 cm²

For the triangle I section:

It is an isosceles triangle and the sides are 5 cm, 1 cm and 5 cm

Perimeter = 5+5+1 = 11 cm

So, semi perimeter = 11/2 cm = 5.5 cm

Using Heron’s formula,

Area = √[s(s-a)(s-b)(s-c)]

= √[5.5(5.5- 5)(5.5-5)(5.5-1)] cm²

= √[5.5×0.5×0.5×4.5] cm²

= 0.75√11 cm²

= 0.75 × 3.317cm²

= 2.488cm² (approx)

For the quadrilateral II section:

This quadrilateral is a rectangle with length and breadth as 6.5 cm and 1 cm respectively.

∴ Area = 6.5×1 cm²=6.5 cm²

For the quadrilateral III section:

It is a trapezoid with 2 sides as 1 cm each and the third side as 2 cm.

Area of the trapezoid = Area of the parallelogram + Area of the equilateral triangle

The perpendicular height of the parallelogram will be

= 0.86 cm

And, the area of the equilateral triangle will be (√3/4×a²) = 0.43

∴ Area of the trapezoid = 0.86+0.43 = 1.3 cm² (approximately).

For triangle IV and V:

These triangles are 2 congruent right angled triangles having the base as 6 cm and height 1.5 cm

Area triangles IV and V = 2×(½×6×1.5) cm² = 9 cm²

So, the total area of the paper used = (2.488+6.5+1.3+9) cm² = 19.3 cm²

Given,

It is given that the parallelogram and triangle have equal areas.

The sides of the triangle are given as 26 cm, 28 cm and 30 cm.

So, the perimeter = 26+28+30 = 84 cm

And its semi perimeter = 84/2 cm = 42 cm

Now, by using Heron’s formula, area of the triangle =

= √[42(42-26)(42-28)(42-30)] cm²

= √[42×16×14×12] cm²

= 336 cm²

Now, let the height of parallelogram be h.

As the area of parallelogram = area of the triangle,

28 cm× h = 336 cm²

∴ h = 336/28 cm

So, the height of the parallelogram is 12 cm.

Draw a rhombus-shaped field first with the vertices as ABCD. The diagonal AC divides the rhombus into two congruent triangles which are having equal areas. The diagram is as follows.

Consider the triangle BCD,

Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m

Using Heron’s formula,

Area of the ΔBCD =

= 432 m²

∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m² = 864 m²

Thus, the area of the grass field that each cow will be getting = (864/18) m² = 48 m²

For each triangular piece, The semi perimeter will be

s = (50+50+20)/2 cm = 120/2 cm = 60cm

Using Heron’s formula,

Area of the triangular piece

=

= √[60(60-50)(60-50)(60-20)] cm²

= √[60×10×10×40] cm²

= 200√6 cm²

∴ The area of all the triangular pieces = 5 × 200√6 cm² = 1000√6 cm²

As the kite is in the shape of a square, its area will be

A = (½)×(diagonal)²

Area of the kite = (½)×32×32 = 512 cm^2.

The area of shade I = Area of shade II

512/2 cm² = 256 cm²

So, the total area of the paper that is required in each shade = 256 cm²

For the triangle section (III),

The sides are given as 6 cm, 6 cm and 8 cm

Now, the semi perimeter of this isosceles triangle = (6+6+8)/2 cm = 10 cm

By using Heron’s formula, the area of the III triangular piece will be

=

= √[10(10-6)(10-6)(10-8)] cm²

= √(10×4 ×4×2) cm²

= 8√5 cm² = 17.92 cm² (approx.)

The semi perimeter of the each triangular shape = (28+9+35)/2 cm = 36 cm

By using Heron’s formula,

The area of each triangular shape will be

= 36√6 cm² = 88.2 cm²

Now, the total area of 16 tiles = 16×88.2 cm² = 1411.2 cm²

It is given that the polishing cost of tiles = 50 paise/cm²

∴ The total polishing cost of the tiles = Rs. (1411.2×0.5) = Rs. 705.6

First, draw a line segment BE parallel to the line AD. Then, from B, draw a perpendicular on the line segment CD.

Now, it can be seen that the quadrilateral ABED is a parallelogram. So,

AB = ED = 10 m

AD = BE = 13 m

EC = 25-ED = 25-10 = 15 m

Now, consider the triangle BEC,

Its semi perimeter (s) = (13+14+15)/2 = 21 m

By using Heron’s formula,

Area of ΔBEC =

= 84 m²

We also know that the area of ΔBEC = (½)×CE×BF

84 cm² = (½)×15×BF

BF = (168/15) cm = 11.2 cm

So, the total area of ABED will be BF×DE i.e. 11.2×10 = 112 m²

∴ Area of the field = 84+112 = 196 m²