Triangle

Geometry Level 3

Given triangle $ABC$ , $AD$ is perpendicular to $BC$ such that $DC = 2$ and $BD = 3$ . If $\angle BAC = {45}^{\circ}$ , then the area of the triangle $ABC$ is:

The answer is 15.

2 solutions

Ajit Athle
Apr 10, 2019

Let AD =x, / BAD = α & / CAD = β. Then tan(45°)= tan(α+ β) = tan(α)+tan(β)/(1-tan(α)tan(β).. In other words, tan(α)+tan(β) = 1-tan(α)tan(β) or 3/x + 2/x = 1 - 6/x² 0r 5/x =1 - 6/x² or x² - 5x - 6 = 0 from which the only admissible value is x= 6. Hence, Area of Tr. ABC = (5x6)/2 = 15 s.u.

Edwin Gray
Apr 2, 2019

Define <ACB = a, <ABC = 135 - a, AD = h. Then AB = sqrt(h^2 + 9), AC = sqrt(h^2 + 4), sin(a) = h/sqrt(h^2 + 4), sin(135 - a) = h/sqrt(h^2 + 9). sin(135 - a) = sin(135)cos(a) - cos(135)sin(a) = h/sqrt(h^2 + 9) [sqrt(2)/2] [2/sqrt(h^2 + 4)] + [sqrt(2)/2] [h/sqrt(h^2 + 4)] = h/sqrt(h^2 + 9) sqrt(2) (h + 2)/[2 sqrt(h^2 + 4) = h/sqrt(h^2 + 9). Squaring both sides, 2(h^2 + 4h + 4)/[4(h^2 + 4) = h^2/(h^2 + 9). Expanding and simplifying, the equation becomes: h^4 - 4h^3 - 5h^2 - 36h - 36 = 0, whicg has a positive root of h = 6. Area of ABC = (1/2)(AD) (BC) = (1/2)(6) (5) = 15

Triangle

The answer is 15.

2 solutions

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