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AMC 8 Geometry Math Olympiad USA Math Olympiad

Area of a Triangle | AMC-8, 2000 | Problem 25

Try this beautiful problem from Geometry: Area of the triangle from AMC-8, 2000, Problem-25. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry: Area of a Triangle

Area of the Triangle- AMC-8, 2000- Problem 25


The area of rectangle \(ABCD\) is \(72\) units squared. If point \(A\) and the midpoints of \(BC\) and \(CD\) are joined to form a triangle, the area of that triangle is

Area of a Triangle
  • \(25\)
  • \(27\)
  • \(29\)

Key Concepts


Geometry

Triangle

square

Check the Answer


Answer: \(27\)

AMC-8 (2000) Problem 25

Pre College Mathematics

Try with Hints


Area of the triangle =\(\frac{1}{2} \times base \times height \)

Can you now finish the problem ……….

Area of Shaded Triangle

Therefore area of the shaded region i.e area of the \(\triangle AEF\)=area of the square- area of \((\triangle ADE +\triangle EFC +\triangle ABF)\)

can you finish the problem……..

Area of Shaded Triangle

Given that area of the rectangle ABCD=72

Let length AB=\(x\) and length of CD=\(y\)

Therefore DE=EC=\(\frac{y}{2}\) and BF=FC=\(\frac{x}{2}\)

Area of ABCD=\(xy\)=72

Area of the \(\triangle ADE=\frac{1}{2}\times DE \times AD= \frac{1}{2}\times \frac{x}{2} \times y =\frac{xy}{4}=\frac{72}{4}=18\)

Area of the \(\triangle EFC=\frac{1}{2}\times EC \times FC= \frac{1}{2}\times \frac{x}{2} \times \frac{y}{2} =\frac{xy}{8}=\frac{72}{8}=9\)

Area of the \(\triangle ABF=\frac{1}{2}\times AB \times BF= \frac{1}{2}\times y\times \frac{x}{2}=\frac{xy}{4}=\frac{72}{4}=18\)

Therefore area of the shaded region i.e area of the \(\triangle AEF\)=area of the square- area of \((\triangle ADE +\triangle EFC +\triangle ABF)=72-(18+8+18)=27\)

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