Question

What is the area of ABC?

Answer 1

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 :

$\bigstar$  (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 :

$:\implies$  (15)² = (AD)² + (9)²

$:\implies$  225 = (AD)² + 81

$:\implies$  (AD)² = 225 - 81

$:\implies$  (AD)² = 144

$:\implies$  (AD)² = (12)²

$:\implies$  AD = 12

We know that, In a Right angled Triangle :

$\bigstar\;\;\boxed{\mathsf{Tan\theta = \dfrac{Opposite\;Side}{Adjacent\;Side}}}$

Now, Consider Triangle ADC : With respect to 45° Angle, AD is the Opposite Side and DC is the Adjacent Side

$:\implies \mathsf{In\;Triangle\;ADC,\;Tan45^{\circ} = \dfrac{AD}{DC}}$

$\mathsf{:\implies 1 = \dfrac{12}{DC}}$

$:\implies$  DC = 12

$:\implies$  Total Length of the Base (BC) = BD + DC

$:\implies$  Total Length of the Base (BC) = 9 + 12

$:\implies$ Total Length of the Base (BC) = 21

We know that, Area of the Triangle is given by :

$\bigstar\;\;\boxed{\mathsf{Area = \dfrac{1}{2} \times Base \times Height}}$

In Triangle, ABC : AD is the Height and BC is the Base

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times BC \times AD}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = \dfrac{1}{2} \times 21 \times 12}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = (21 \times 6)}$

$:\implies \mathsf{Area\;of\;the\;Triangle\;ABC = 126}$