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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of Triangle and Square | AMC 8, 2012 | Problem 25

Try this beautiful problem from AMC-8-2012 (Geometry) based on area of a Triangle and square and congruency.

Area of a Triangle- AMC 8, 2012 – Problem 25


A square with area 4 is inscribed in a square with area 5,  with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length a, and the other of length b. What is the value of ab?

Area of triangle and square

  • \(\frac{1}{5}\)
  • \(\frac{2}{5}\)
  • \(\frac{1}{2}\)

Key Concepts


Geometry

Square

Triangle

Check the Answer


Answer:\(\frac{1}{2}\)

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find the area of four triangles

Can you now finish the problem ……….

Four triangles are congruent

can you finish the problem……..

Area of Triangle and Square 2

Total area of the big square i.e ABCD is 5 sq.unit

and total area of the small square i.e EFGH is 4 sq.unit

So Toal area of the \((\triangle AEH + \triangle BEF + \triangle GCF + \triangle DGH)=(5-4)=1\) sq.unit

Now clearly Four triangles\(( i.e \triangle AEH , \triangle BEF , \triangle GCF , \triangle DGH )\) are congruent.

Total area of the four congruent triangles formed by the squares is 5-4=1 sq.unit

So area of the one triangle is \(\frac{1}{4}\) sq.unit

Now “a” be the height and “b” be the base of one triangle

The area of one triangle be \((\frac{1}{2} \times base \times height )\)=\(\frac{1}{4}\)

i.e \((\frac{1}{2} \times b \times a)\)= \(\frac{1}{4}\)

i.e \(ab\)= \(\frac{1}{2}\)

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AMC 8 Beautiful Mathematics videos Geometry Math Olympiad

Circumference of a Semicircle | AMC 8, 2014 | Problem 25

Try this beautiful problem from AMC-8-2014 (Geometry) based on Circumference of a Semicircle

Circumference of a Semicircle- AMC 8, 2014 – Problem 25


On A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5miles per hour, how many hours will it take to cover the one-mile stretch?

circumference of a semicircle

  • \(\frac{\pi}{11}\)
  • \(\frac{\pi}{10}\)
  • \(\frac{\pi}{5}\)

Key Concepts


Geometry

Semicircle

Distance

Check the Answer


Answer:\(\frac{\pi}{10}\)

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find the circumference of a semi-circle

Can you now finish the problem ……….

If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours

can you finish the problem……..

circumference of a semicircle

If Robert rides in a straight line, it will take him \(\frac {1}{5}\) hours. When riding in semicircles, let the radius of the semicircle r, then the circumference of a semicircle is \({\pi r}\). The ratio of the circumference of the semicircle to its diameter is \(\frac {\pi}{2}\). so the time Robert takes is  \(\frac{1}{5} \times \frac{\pi}{2}\). which is equal to \(\frac{\pi}{10}\)

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AMC 8 Geometry Math Olympiad

Area of a square | AMC 8- 2015| Problem 25

Try this beautiful problem from AMC-8-2015 (Geometry) based on area of square.

Area of a square – AMC 8, 2015 – Problem 25


One-inch Squares are cut from the corners of this 5 inch square.what is the area in square inches of the largest square that can be fitted into the remaining space?

area of a square- AMC 8
  • 9
  • 15
  • 17

Key Concepts


Geometry

Area

Square

Check the Answer


Answer:15

AMC-8, 2015 problem 25

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find the Length of HG……

Can you now finish the problem ……….

square that fits into the area

Draw the big square in the remaining space of the big sqare and find it’s area …….

can you finish the problem……..

square that fits into the area

We want to find the area of the square. The area of the larger square is composed of the smaller square and the four red triangles. The red triangles have base  3 and height 1 .  so the combined area of the four triangles is  $ 4 \times \frac {3}{2} $=6.

The area of the smaller square is  9+6=15.

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