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AMC-8 Geometry Math Olympiad

Area of a Triangle -AMC 8, 2018 – Problem 20

Try this beautiful problem from AMC 8, 2018 based on the area of triangle. You may use sequential hints to solve the problem.

Try this beautiful problem from Geometry based on Area of a Triangle Using similarity

Area of Triangle – AMC-8, 2018 – Problem 20


In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.

What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?

  • $\frac{2}{3}$
  • $\frac{4}{9}$
  • $\frac{3}{5}$

Key Concepts


Geometry

Area

similarity

Check the Answer


Answer:$\frac{4}{9}$

AMC-8, 2018 problem 20

Pre College Mathematics

Try with Hints


$\triangle ADE$ $\sim$ $\triangle ABC$

Can you now finish the problem ……….

$\triangle BEF$ $\sim$ $\triangle ABC$

can you finish the problem……..

Area of triangle- figure

Since $\triangle ADE$$\sim$ $\triangle ABC$

$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$

i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$

Again $\triangle BEF$ $\sim$ $\triangle ABC$

Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$

i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$

Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$

i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$

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