If the perimeter of an equilateral triangle is 342 units. Then find its area.

It is given that side of equilateral triangle = 8 units
Area of equilateral triangle =  a²
=  8²  = 16 square units
Since there are 4 triangles, so area of four equilateral triangles
= 4  16 = 64 sq. units
= 110.85 sq. units
In the figure, length of the side of square = 8 units
Now, Area of square = (side)² = 8² = 64 sq. units
Surface Area of the star
= Area of 4 equilateral triangles + Area of the square
=  110.85 + 64 = 174.85 sq. units

It is given that Base, b = AC = 18 units and Height, h = 10 units
Area of the triangle = 
=  = 90 sq. units

It is given that base of the triangle = 12 m
area of the triangle = 90 m²
 = 90
 = 90

It is given that height to base ratio is 3 : 4
⇒ Height, h = 3x and Base, b = 4x
Area of triangle = 
864 = 
864  = 6x²
144 = x²
12 = x
Height = 3x = 3(12) = 36 units and Base = 4x = 4(12) = 48 units

Area of an equilateral triangle =  a²
=  20² = 100 cm²
Half of the area

Let Base = b and Height = h
It is given that base = 3  height
⇒ b = 3h
Area of the triangle = 
= 
=h²

It is given that side of the triangle = 16 units
Area of an equilateral triangle =  a² =  16²
= 64 cm²
Double of the area = 2  64
= 128 = 221.7 cm²

Base, AB = 8 cm and Height, H = CD = 4 cm
Area of the triangle with altitude corresponding to AB
= 
=  = 16 cm²
With Base = BC , Height, h = AE = 5 cm
Area of the triangle with altitude corresponding to BC is
 = 16 cm²
 = 16
BC = 6.4 cm

It is given that Base, b = 18 cm and Height, h = 6 cm
Area of the parallelogram = b  h
= 18  6 = 108 cm²

It is given that ratio of bases of two triangle is a : b
⇒ Bases of the triangles = ax , bx
Similarly, it is given that ratio of altitudes is c : d
⇒ Altitudes of the triangle = cy , dy
Area of first triangle = 
= 
Area of second triangle = 
Ratio of the areas of two triangle
= 
= 
Hence, ratio of area of two triangles is ac : bd

 It is given that sides are in ratio 3 : 4 : 5
⇒ Length of the side a = 3x , b = 4x and c = 5 x
Perimeter of the triangle = 144 m

3x + 4x + 5x = 144

12x = 144 ⇒  x = 12
Now Using Pythagoras theorem,

(5x)² = (3x)² + (4x)²

25x² = 9x² + 16x²

25x² = 25x²  i.e. Pythagoras holds true
⇒ Given triangle is a right angled triangle
Base, b = 3x =3(12) = 36 m and Height, h = 4x =4(12) = 48 m

⇒ Area of the triangle = 

It is given that a = 11 cm , b = 15 cm and c = 16 cm.
Using Heron’s formula
Area of triangle =  where s = 
s =  = 21

                                                              Area of triangle= 

Since we have to find altitude to the largest side, base of the triangle = 16 cm

Also, area of triangle = 

⇒ 

⇒          ( = 2.64)

Hint:
We use principle of Archimedes to find the length of the cube.
Explanations:
Step 1 of 1:
Let the length of the edge of the metal cube be a.
Volume of cube = a³
Given,  r = 30/2 cm and h = 
By the principle of Archimedes,
Volume of risen milk = volume of cube


, discarding the negative value since distance cannot be negative.
Final Answer:
The length of the edge of the cube is 10cm.

Hint:
We plug in the values in formulae and solve the problem.
Explanations:
Step 1 of 2:
Let the radius of vessel base be
We have,  cm
Height  h = 25cm
Step 2 of 2:
Volume of the vessel =
Final Answer:
The radius is 21cm and volume of the cylinder is 34650cm³.

Hint:
We find the internal radius and subtract it from external radius to get the thickness.
Explanations:
Step 1 of 1:
Let the internal radius of the pipe be r .
Given, external radius R  = 9cm
Length = height of the pipe h = 14cm
We have volume of pipe = 748





Final Answer:
The thickness of the pipe is 8 cm.