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Explore the Back-StoryTry 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.

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?

,

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

Geometry

Triangle

**Pythagoras**

Pre College Mathematics

AMC-10A, 2018 Problem-23

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

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?

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?

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=\(6- \frac{4}{49}\)=\(\frac{145}{147}\)

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

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.

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?

,

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

Geometry

Triangle

**Pythagoras**

Pre College Mathematics

AMC-10A, 2018 Problem-23

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

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?

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?

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=\(6- \frac{4}{49}\)=\(\frac{145}{147}\)

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=OvduZbqenWU

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