Triangles are those closed convex polygons having three sides. We have already familiar with the different types of triangles. Right angled triangle is the one in which one of the interior angle of a triangle is a right angle. The longest side of the right triangle is called its Hypotenuse and the other two sides including the right angle are called as its legs or the base and the perpendicular. It is very much necessary to learn more about the right angled triangle, as it is very much helpful to find the height of a building, distance between the two buildings, vertical height of an aeroplane at a given instant etc. In this section let us discuss about the different formulae we use for a given right angled triangle.

Area of Right Angle Triangle:

The following diagram shows the right triangle.
RT

The area of a right angled triangle is given by the formula,
Area = $\frac{1}{2}$ x Base x Height
= $\frac{1}{2}$ x leg 1 x leg 2



Right Angle Triangle Equations

If a, b, c are the sides of a right angled triangle such that c > b > a, then the sides of the triangle will satisfy the equation,

c2 = a2 + b2


Solved Examples

Question 1: Find the area of a right angled triangle whose sides including the right angles measure 10 cm and 9 cm respectively.
Solution:
 
From the given data, the legs measure, 10 cm and 9 cm respectively.
Therefore, Area of the right triangle = $\frac{1}{2}$ x 10 x 9

                                                   = $\frac{90}{2}$ = 45 cm2
 

Question 2: Find the length of the hypotenuse of a right triangle whose measures of the other two sides are 8 cm and 6 cm respectively.
Solution:
 
We are given that a = 6 cm and b =  8 cm.
These measures, will satisfy the equation, c2 = a2 + b2
                                                      =>    c2 = 62 + 82 
                                                                  = 36 + 64
                                                                  = 100
                                                                  = 102
                                                     =>      c = 10 cm
 

Question 3: A ladder of 17 m long, reaches the window of a building 15 m above the ground. Find the distance of the foot of the ladder from the building.
Right Angled Triangle Example

Solution:
 
From the above figure, the sides of the triangle will satisfy the equation,
                                a2 + b2  = c2
                      =>    152 + b2   = 172 [ substituting, a = 15 and c = 17 ]
                      =>    225 + b2   = 289
                      =>             b2   = 289 - 225
                                             = 64
                                             = 82
                     =>                 b = 8 cm
Therefore, the foot of the ladder from the building is 8 cm.
 

Let us discuss with the following questions in solving the right angled triangle.

Solved Examples

Question 1: Solve the right angled triangle for b, when a = 5 cm and c = 13 cm.
Solution:
 
We have a = 5 cm and c = 13 cm
The given data will satisfy the equation,
                                a2 + b2  = c2
                        =>    52 + b2   = 132
                        =>   25  + b2   = 169
                        =>             b2 = 169 - 25
                                             =  144
                                             = 122
                         =>             b = 12 cm
 

Question 2: In the given figure, $\angle PSR$ = 90o ,  PQ = 10 cm, QS = 6 cm and RQ= 9 cm, calculate the length of PR.Solving Right Triangles
Solution:
 
In Right triangle PQS, $\angle PSQ$ = 90o.
Therefore applying Pythagoras theorem, we get
                                            PS2 + QS2 = PQ2
                                 =>       PS2 + 62    = 102
                                 =>      PS2 + 36     = 100
                                 =>                 PS2  = 100 - 36
                                                              = 64
                                                              = 82
                                 =>                   PS = 8 cm
In triangle PSR, $\angle PSR$ = 90o
 Therefore we have                              RS = RQ + QS
                                                              = 9 + 6
                                                              = 15 cm
                                                        PS = 8 cm (proved)
Applying Pythagoras Theorem, we get
                                                      PR2 = PS2 + RS2
                                =>                 PR2  = 82 + 152
                                                             = 64 + 225
                                                             = 289
                                                             = 172
                                 =>                  PR = 17 cm
Therefore, the length of PR = 17 cm.
 

Question 3: In an isosceles triangle the equal sides measure 15 cm. If the unequal side measure 18 cm, find the area of the triangle.
Solution:
 

Right Angled Triangle Examples
In the above triangle we have AB = AC = 15 cm [ since the triangle is isosceles ]
It is given that the unequal side = 18 cm
                               =>     BC = 18 cm
Since the perpendicular from the vertex bisects the third side, we have BD = DC = 9 cm
In $\Delta ABC$, $\angle ADC$ = 90o
Therefore, applying Pythagoras Theorem, we get
                             AD2 + BD2 = AB2
                        => AD2 + 92    = 152
                        => AD2 + 81    = 225
                        =>           AD2 = 225 - 81
                                              = 144
                        =>             AD = $\sqrt 144$
                                               = 12 cm
Therefore, Height of the triangle = 12 cm
                     Area of the triangle = $\frac{1}{2}$ x Base x Height
               
                                                 = $\frac{1}{2}$ x 18 x 12

                                                 = 108 cm2
 

Question 4: Find the height of an equilateral triangle whose equal sides measure 12 cm.
Solution:
 

Right Angled Triangle Problem
In the above triangle KL = LM = MN = 12 cm
The perpendicular from the vertex bisects the base.
                    ( i. e ) KD $\perp$ LM.
                         Therefore, LD = DM = 6 cm
In $\Delta KLD$, $\angle KDL$ = 90o
Therefore, applying Pythagoras Theorem, we get,
                             KD2 + LD2 = KL2
                    =>     KD2 + 62   = 122
                    =>     KD2 + 36   = 144
                    =>              KD2 = 144 - 36
                                             = 108
                    =>               KD = $\sqrt108$
                                            = 10.392 cm
Therefore, Height of the Equilateral triangle of side 12 cm is 10.392 cm
 

Question 5: Find the length of the sides of a filed of rhombus in shape, whose diagonals measure 30 m and 16 m.
Solution:
 

Right Angle Triangle Problem
The above figure shows the Rhombus PQRS, whose diagonals measure 30 cm and 16 cm respectively.
Since the diagonals of a rhombus bisect each other at right angles we have,
                                 PO = OR = 8 cm and
                                 QO = OS = 15 cm
In $\Delta POQ$, $\angle POQ$ = 90o
Applying Pythagoras Theorem, we have
                                        PQ2 = PO2 + QO2
                                               = 82 + 152
                                               = 64 + 225
                                               = 289
                              =>       PQ = $\sqrt 289$
                                               = 17
Since the sides of a rhombus measure same, we have PQ = QR = RS = SP = 17 cm