Math Olympiad Test: Heron’s Formula- 1

Let s be the semi-perimeter and a, b and c are sides of a triangle
Using Heron's formula,
Area of triangle = √s(s – a)(s – b)(s – c)
a = 8 cm, b = 11 cm c = 32 - (8 + 11) = 13 cm
s = (8 + 11 + 13)/2 = 16
Area of triangle = √16(16 – 8)(16 – 11)(16 – 13)
⇒ Area of triangle = √16 × 8 × 5 × 3
⇒ √2 × 2 × 2 × 2 × 2 × 2 × 2 × 5 × 3
⇒ 2 × 2 × 2 × √30 =  8√30 cm²

Let the sides be a, b, c
∴ New sides = 3a, 3b, 3c
∴ s_new = 3s
New area = 

= 9∆
∴ Increase in area = 9∆ – ∆ = 8∆
∴ % Increase in area = 

= 5 × 2 × 3√7
=  √cm²
1/2 × Altitude to the largest side × largest side = 30√7 cm²
⇒ Altitude to the largest side

Length of median of equilateral triangle

⇒ a = 2 cm
∴ Area of triangle

⇒     ar (∆PQT) = 24 cm²
∴ area (∆TQR) = ar (∆PRT) – ar (∆PQT) = 40 – 24 = 16cm²

Given, the side of an equilateral triangle = 2√3 cm
We have to find the area of the equilateral triangle.
Area of an equilateral triangle = √3/4 (side)²
= √3/4 (2√3)²
= √3/4 (4 × 3)
= 3√3
Consider √3 = 1.732
= 3 × 1.732
= 5.196 cm²
Therefore, the area of an equilateral triangle is 5.196 cm²

Perimeter of square = Perimeter of equilateral ∆ = x
∴ length of side of square = x/4
Length of side of equilateral ∆ = x/3
Clearly, A₁ > A₂

AE = x, FB = (17 – x) cm
From ∆AED, and ∆CFB
CF² = DE²
⇒ (26)² – (17 – x)² = (25)² – (x)²
⇒ (26)² – (25)² = (17 – x)² – (x)²
⇒ (26 – 25) (26 + 25) = (17 – x + x) (17 – x – x)
⇒ 51 = 17 (17 – 2x)
⇒ 17 – 2x = 3
⇒ 2x = 14 ⇒ x = 7cm

∴ area = 1/2 x (60 + 77) x 24 cm² 
= 137 × 12 cm² = 1644 cm²

Side of rhombus = 80/4 = 20 cm 
∴ Length of other diagonal

∴ 
= 16 × 24 cm² = 384 cm²

∴ For ∆DBC,



= 14 × 3 × 3
= 126 cm²
Ar(∆ABD) = 1/2 x 9 x 40
= 180 cm²
∴ Total area = (126 + 180) cm²
= 306 cm²