Cheenta
How Cheenta works to ensure student success?
Explore the Back-Story

# 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 ?

,

• $$28$$
• $$30$$
• $$21$$
• $$26$$
• $$26$$

Geometry

Triangle

## Suggested Book | Source | Answer

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 $AC + DF = 30$

## Subscribe to Cheenta at Youtube

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 ?

,

• $$28$$
• $$30$$
• $$21$$
• $$26$$
• $$26$$

Geometry

Triangle

## Suggested Book | Source | Answer

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 $AC + DF = 30$

## Subscribe to Cheenta at Youtube

This site uses Akismet to reduce spam. Learn how your comment data is processed.

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

1. Rajvardhan singh Sisodiya says:

We can directly apply Cosine law

1. KOUSHIK SOM says:

yes we can..