If the height to base ratio of a triangle ABC is 3:4 and the area is 864 square units. Determine the height and base of this triangle.

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.

Hint:
Forming the equations based on the given information, we will get two equations. Then we will find the radii by solving those two equations.
 Explanations:
Step 1 of 3:
Given, outer CSA – inner CSA = 88
, where  = length of cylinder,  R = outer radius, r = inner radius

…(i)
Step 2 of 3:
Also, given volume with  R – volume with  r = 176



…(ii)
Step 3 of 3:
Adding (i) and (ii), we get



Putting R = 5/2 in equation (ii), we get


Final Answer:
The outer and inner radii of the cylinder are 2.5 cm and 1.5 cm

It is given that a = 16 cm , b = 30 cm and c = 34 cm
Using Heron’s formula
Area of triangle =  where s = 
s =  = 40
Area of triangle
= 

It is given that Base of the triangle = 30 cm
We know that in a right isosceles triangle, base and height are equal and hypotenuse is the third side.
So, Area of the triangle = 
= 450 cm²

Hint:-
1. Extra charges = number of miles driven × rate per mile
2. Total cost = Constant fees + Extra charges.
Step-by-step solution:-
Let x be the number of miles driven and y be the total fees charged.
We know that-
Extra charges = Number of miles driven × Rate per mile
∴ Extra charges = x * rate per mile ..................................................... (Equation i)
Now, for Driver A-
Total fees charged = y = Constant fees + Extra charges
∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
∴ Total fees charged = y = 25 + x * 0.10 .................................................... (From given information)
∴ Total fees charged = y = 25 + 0.10 x ....................................................... (Equation ii)
For Driver B-
Total fees charged = y = Constant fees + Extra charges
∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)
∴ Total fees charged = y = 0 + x × 0.60 .................................................... (From given information)
∴ Total fees charged = y = 0.60 x ....................................................... (Equation iii)
Since we need to find the distance for which cost of both drivers is the same-
Equating Equations ii & iii, we get-
25 + 0.10 x = 0.60 x
∴ 25 = 0.60 x - 0.10 x
∴ 25 = 0.50 x
∴ 25 / 0.50 = x ...................................................................................... (Dividing both sides by 0.50)
∴ 50 = x
For a distance of 50 miles, the cost charged by both the drivers will be the sam.
Now, at 50 miles i.e. x = 50,
y = 0.60 x ….............................................. (Using Equation iii)
∴ y = 0.60 × 50
∴ y = 30
Note:-
Alternatively, we can find the total cost by using Equation ii,
y = 25 + 0.10 x
∴ y = 25 + 0.10 × 50
∴ y = 25 + 5
∴ y = 30.
Hence, the answer remains the same.