Try this beautiful Problem on Sum of Sides of Triangle from Triangle from Prmo-2018, Problem 17.
Sum of Sides of Triangle – 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 ?
,
- \(28\)
- \(30\)
- \(21\)
- \(26\)
- \(26\)
Key Concepts
Geometry
Triangle
Suggested Book | Source | Answer
Suggested Reading
Pre College Mathematics
Source of the problem
Prmo-2018, Problem-17
Check the answer here, but try the problem first
\(30\)
Try with Hints
First Hint
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?
Second Hint
Let M be the foot of perpendicular from B(E) to CF. So BM = 8. Hence $\mathrm{AM}=\sqrt{17^{2}-8^{2}}=\sqrt{(25)(9)}=15$
Third Hint
Hence $AF = 15 – 6 = 9$ & $AC = 15 + 6 = 21$
So $AC + DF = 30$
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2 replies on “Sum of Sides of Triangle | PRMO-2018 | Problem No-17”
We can directly apply Cosine law
yes we can..