Sum of Sides of Triangle | PRMO-2018 | Problem No-17

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

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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 comments on “Sum of Sides of Triangle | PRMO-2018 | Problem No-17”

Rajvardhan singh Sisodiya says:
September 19, 2020 at 10:46 am
We can directly apply Cosine law
Reply
1. KOUSHIK SOM says:
  September 19, 2020 at 11:45 am
  yes we can..
  Reply

Sum of Sides of Triangle | PRMO-2018 | Problem No-17