Name: History & Derivation of Heron's Formula
Uploaded: 2019-07-19
Duration: 9 min 37 s
Description: Heron from Alexandria derived a formula to calculate the area of a triangle. His formula can be used not only for the triangles but also for all the types of quadrilaterals too.

Question

Kamla has a triangular field with sides

240 m, 200 m, 360 m

, where she grew wheat.
In another triangular field with sides

240 m, 320 m, 400 m

adjacent to the previous field, she wanted to grow potatoes and onions. She divided the field in two parts by joining the mid-point of the longest side to the opposite vertex and grew potatoes in one part and onions in the other part. Find the area in which the wheat is growing.

Accepted Answer

$E$ is the mid - point of $A D$
So, $A E = E D = 200 m$

Area of wheat field $= A r e a △ A B C$
Area of potato field $= A r e a △ A E C$
Area of onion field $= A r e a △ E D C$

Area $△ A B C$
Area of triangle $= \sqrt{s (s - a) (s - b) (s - c)}$
Here , $s$ is the semi - perimeter ,
and $a, b, c$ are the sides of the triangle

$s = \frac{a + b + c}{2} = \frac{240 + 200 + 360}{2} m = \frac{800}{2} m = 400 m$

Area $△ A B C$ $= \sqrt{400 (400 - 200) (400 - 240) (400 - 360)} m^{2}$
$= \sqrt{400 \times 200 \times 160 \times 40} m ² = \sqrt{(4 \times 2 \times 4 \times 16) \times 10^{6}} m ² = \sqrt{(4 \times 4) \times 2 \times (16) \times 10^{6}} m ² = \sqrt{(16) \times 2 \times (16) \times 10^{6}} m ² = \sqrt{(16 \times 16) \times 2 \times 10^{6}} m ² = \sqrt{16 ² \times 2 \times 10^{6}} m ² = \sqrt{16 ²} \times \sqrt{2} \times \sqrt{10^{6}} = 16000 \sqrt{2} m ² (1 h e c t a r e = 10000 m ²) \frac{16000 \sqrt{2}}{10000} h e c t a r e s = 1.6 \times \sqrt{2} h e c t a r e s = 1.6 \times 1.414 h e c t a r e s = 2.26 h e c t a r e s (a p p r o x .)$

Area for growing wheat $= 2.26 h e c t a r e s$ .

Important Questions on Heron's Formula

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