Question

In △ABC, AB = 4 cm, BC = 6 cm, and AC = 8 cm. Let D be a point on AC so that BD = AB. Find AD.
The correct answer was 11/2 cm and I tried to find out the reason why the answer was such.
I first tried looking for similar triangles but it didn't work. Then I also assumed that the height of the isosceles triangle is equal to the height of △ABC, but it was totally different from the correct answer. I would appreciate it if someone could help me find the solution to the problem.

Answer 1

Answer:

$\sf\\\text{See the image for reference. Here, the height of the isosceles triangle and }\\\triangle\text{ABC are actually same. So, you don't need to assume that.}$

Step-by-step explanation:

$\text{Solution:}\\\text{Construct BE}\perp\text{AC.}\\\text{Now,}\\\text{Perimeter of }\triangle \text{ABC}(P)=\text{AB + BC + AC}=4+6+8=18\text{cm}\\\therefore\ \text{Semiperimeter}(s)=P/2=9\text{cm}$

$\text{Using Heron's formula,}\\\text{Area of }\triangle \text{ABC}=\sqrt{s(s-\text{AB})(s-\text{BC})(s-\text{AC})}\\=\sqrt{9(9-4)(9-6)(9-8)}\\=3\sqrt{15}\text{cm}^2$

$\text{Now, we have another formula for area of triangle:}\\\text{Area of }\triangle \text{ABC}=\frac{1}{2}\times \text{sin A}\times \text{AB}\times \text{AC}\\\text{or,\ }3\sqrt{15}=\frac{1}{2}\times \text{sin A}\times4\times8\\\text{or,\ }\text{sin A}=3\sqrt{15}/16$

$\therefore\ \text{cos A}=\sqrt{1-\text{sin }^2 \text{A}}=\sqrt{1-135/256}=11/16$

$\text{In triangle ABE,}\\\text{cos A}=\frac{\text{AE}}{\text{AB}}=\frac{\text{AE}}{4}\\\\\text{or,\ }\frac{11}{16}=\frac{\text{AE}}{4}\\\\\text{or,\ }\text{AE = }11/4\text{cm}$

$\text{BE is the height of isosceles triangle ABE, so BE bisects AD.}\\\text{Therefore, DE = AE=11/4cm.}\\\therefore \text{AD = AE + DE = 11/4+11/4 = 11/2cm}$

Do not hesitate to ask if you have any doubts...Hope this helps!