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AMC 10 Geometry Math Olympiad USA Math Olympiad

Right-angled shaped field | AMC 10A, 2018 | Problem No 23

Try this beautiful Problem on triangle from AMC 10A, 2018. Problem-23. You may use sequential hints to solve the problem.

Try this beautiful Problem on Geometry based on Right-angled shaped field from AMC 10 A, 2018. You may use sequential hints to solve the problem.

Right-angled shaped field – AMC-10A, 2018- Problem 23


Farmer Pythagoras has a field in the shape of a right triangle. The right triangle’s legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square $S$ so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from $S$ to the hypotenuse is 2 units. What fraction of the field is planted?

,

  • $\frac{25}{27}$
  • $\frac{26}{27}$
  • $\frac{73}{75}$
  • $\frac{145}{147}$
  • $\frac{74}{75}$

Key Concepts


Geometry

Triangle

Pythagoras

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2018 Problem-23

Check the answer here, but try the problem first

\(\frac{145}{147}\)

Try with Hints


First Hint

Given that ABC is a right-angle Triangle field . Here The corner at \(B\) is shaded region which is unplanted. now we have to find out fraction of the field is planted?

Now if we join the triangle with the dotted lines then it will be divided into three triangles as shown below…

Therefore there are three triangles . Now if we can find out the area of three triangles and area of the smaller square then it will be eassy to say….

Now can you finish the problem?

Second Hint

Let \(x\) be the side length of the sqare then area will be\(x^2\)

Now area of two thin triangle will be $\frac{x(3-x)}{2}$ and $\frac{x(4-x)}{2}$

area of the other triangle will be \(\frac{1}{2}\times 5 \times 2=5\)

area of the \(\triangle ABC =\frac{1}{2}\times 3 \times 4=6\)

Now Can you finish the Problem?

Third Hint

Therefore we can say that $x^{2}+\frac{x(3-x)}{2}+\frac{x(4-x)}{2}+5=6$

\(\Rightarrow x=\frac{2}{7}\)

Therefore area of the small square will be \(\frac{4}{49}\)

Thererfore our required fraction =Area of the \(\triangle ABC\)-area of the smaller square=\(6- \frac{4}{49}\)=\(\frac{145}{147}\)

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