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Let D be the Intersection of Height of the Triangle and Base of the Triangle BC
From the Figure, We can notice that Triangle ADB is a Right angled Triangle.
We know that, In a Right angled Triangle :
(Hypotenuse)² = (First Leg)² + (Second leg)²
In Triangle, ADB : AB is the Hypotenuse, AD is the First leg and BD is the Second leg
Given : AB = 15 and BD = 9
Substituting the values, We get :
(15)² = (AD)² + (9)²
225 = (AD)² + 81
(AD)² = 225 - 81
(AD)² = 144
(AD)² = (12)²
AD = 12
We know that, In a Right angled Triangle :
Now, Consider Triangle ADC : With respect to 45° Angle, AD is the Opposite Side and DC is the Adjacent Side
DC = 12
Total Length of the Base (BC) = BD + DC
Total Length of the Base (BC) = 9 + 12
Total Length of the Base (BC) = 21
We know that, Area of the Triangle is given by :
In Triangle, ABC : AD is the Height and BC is the Base
