Try this beautiful Problem on Sum of Sides of Triangle from Triangle from Prmo-2018, Problem 17.
Triangles $\mathrm{ABC}$ and $\mathrm{DEF}$ are such that $\angle \mathrm{A}=\angle \mathrm{D}, \mathrm{AB}=\mathrm{DE}=17, \mathrm{BC}=\mathrm{EF}=10$ and $\mathrm{AC}-\mathrm{DF}=12$
What is $\mathrm{AC}+$ DF ?
,
Geometry
Triangle
Pre College Mathematics
Prmo-2018, Problem-17
\(30\)
In \(\triangle ABC\) & \(\triangle DEF\), given that \(\angle A\)=\(\angle D\) & \(AB=DE\)=$17$ . $BC = EF = 10 $ and $AC – DF = 12.$ we have to find out $AC+DF$.
According to the question Let us assume that the point A coincides with D, B coincides with E. Now if we draw a circle with radius 10 and E(B) as center ....
Can you finish the problem?
Let M be the foot of perpendicular from B(E) to CF. So BM = 8. Hence $\mathrm{AM}=\sqrt{17^{2}-8^{2}}=\sqrt{(25)(9)}=15$
Hence $AF = 15 – 6 = 9$ & $AC = 15 + 6 = 21$
So $AC + DF = 30$
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We can directly apply Cosine law
yes we can..