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 :
[tex]\bigstar[/tex] (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 :
[tex]:\implies[/tex] (15)² = (AD)² + (9)²
[tex]:\implies[/tex] 225 = (AD)² + 81
[tex]:\implies[/tex] (AD)² = 225 - 81
[tex]:\implies[/tex] (AD)² = 144
[tex]:\implies[/tex] (AD)² = (12)²
[tex]:\implies[/tex] AD = 12
We know that, In a Right angled Triangle :
[tex]\bigstar\;\;\boxed{\mathsf{Tan\theta = \dfrac{Opposite\;Side}{Adjacent\;Side}}}[/tex]
Now, Consider Triangle ADC : With respect to 45° Angle, AD is the Opposite Side and DC is the Adjacent Side
[tex]:\implies \mathsf{In\;Triangle\;ADC,\;Tan45^{\circ} = \dfrac{AD}{DC}}[/tex]
[tex]\mathsf{:\implies 1 = \dfrac{12}{DC}}[/tex]
[tex]:\implies[/tex] DC = 12
[tex]:\implies[/tex] Total Length of the Base (BC) = BD + DC
[tex]:\implies[/tex] Total Length of the Base (BC) = 9 + 12
[tex]:\implies[/tex] Total Length of the Base (BC) = 21
We know that, Area of the Triangle is given by :
[tex]\bigstar\;\;\boxed{\mathsf{Area = \dfrac{1}{2} \times Base \times Height}}[/tex]
In Triangle, ABC : AD is the Height and BC is the Base
[tex]:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times BC \times AD}[/tex]
[tex]:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times 21 \times 12}[/tex]
[tex]:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = (21 \times 6)}[/tex]
[tex]:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = 126}[/tex]
