Math Olympiad Test: Heron’s Formula- 2 - Class 9 MCQ

In ∆ABD,

AB² + BD² = AD²
⇒ AD² = AB² – BD²

⇒ AD = √135 = 3√15 cm
∴ area (∆ABC) = 1/2 x 6 x 3√15 
= 9√15 cm²

 Here S = = 120 cm
∴ Area

= 2 × 3 × 2 × 2 × 2 × 5 × 7 
= 240 × 7 cm² = 1680 cm²
Area = 1/2 x base altitude = 1680 cm²


= 30 cm

BC = 



⇒ AC × BD = AB × AC
⇒  13 × BD = 5 × 12 
⇒ 

In ∆DOC,
OD² + OC² = DC²

⇒ OD = 

∴ DB = 2 × OD = 2 × 60 = 120 cm. 
Area of rhombus = 
= 9600 cm²

Area of square = a²
Area of rhombus = a² sin θ
∵ sin θ < 1
∴ a² > a² sin θ
 ar(square) > ar(rhombus)
⇒ S > R

Area of square ABCD = 1/2 × AC × BD
= 1/2 x 32 x 32
= 16 × 32 cm²
= 512 cm²
Here 

Area of ∆ACD = area of ∆ACB
⇒ area of ∆ACD = 



= 4 × 6 × 7 × 2
= 4 × 84 = 336 cm²
∴ area of ∥gm = 2 × ar(∆ACD)
= 2 × 336 cm²
= 672 cm²

In ∆AOB,
AB = 

= 20 cm. 
∴ ar(∆AOB) = 1/2 × 12 × 16 = 96 cm²
For ∆ABC,

∴ ar (∆ABC) = 

= 480 cm²
∴ Area of shaded region
= 480 – 96
= 384 cm²

Length of equilateral triangle = x/3 cm
Length of side of square = x/4 cm
Length of side of square

= 12 cm

Let the length of third side be x
∴ 
Area


⇒ x = 30 cm
∴ Third side = 30 cm

Lengths of sides of triangle

∴ Area

∴ Height corresponding to the longest side

Area of ∥gm = AL × DC = CM × AD
⇒ 8 × DC = 10 × 6
⇒ DC = 60/8 = 7.5 cm = AB.

 Here S = 
∴ Required area

If the sides of ∆ are a, b and c
∴ New sides are 2a, 2b, 2c

∴ New area = 

The two parts of the parallelogram are congruent because they have equal area.