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……..
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 $\frac{4}{9}$
$\textbf{(A) }3\sqrt{3}-\pi\qquad\textbf{(B) }4\sqrt{3}-\frac{4\pi}{3}\qquad\textbf{(C) }2\sqrt{3}\qquad\textbf{(D) }4\sqrt{3}-\frac{2\pi}{3}\qquad\textbf{(E) }4+\frac{4\pi}{3}$[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version=”4.0″][et_pb_column type=”4_4″ _builder_version=”3.25″ custom_padding=”|||” custom_padding__hover=”|||”][et_pb_accordion open_toggle_text_color=”#0c71c3″ _builder_version=”4.2.2″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px”][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”on”]American Mathematical Contest 2017, AMC 8 Problem 25[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″ inline_fonts=”Abhaya Libre”]
Therefore, The required area = Area of $\triangle UXY$ – $2 \times$ Area of the sector $SXR$. [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]Area of equilateral triangle $\triangle UXY= 4\sqrt{3}$ And the are of sector $SXR= \frac{2\pi}{3}$ ANS : $4\sqrt{3}-\frac{4\pi}{3}$[/et_pb_tab][et_pb_tab title=”Formulas Used ” _builder_version=”4.2.2″]Area of an equilateral triangle =$\frac{a^2\sqrt{3}}{4}$ [where $a$ is a sied of the triangle] Area of a sector of a circle of angle $\theta$ = $\frac{\theta}{360}\pi r^2$ [where $r$ is the radius of the circle][/et_pb_tab][/et_pb_tabs][/et_pb_column][/et_pb_row][/et_pb_section][et_pb_section fb_built=”1″ fullwidth=”on” _builder_version=”4.2.2″ global_module=”50833″][et_pb_fullwidth_header title=”AMC – AIME Program” button_one_text=”Learn More” button_one_url=”https://www.cheenta.com/amc-aime-usamo-math-olympiad-program/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/matholympiad.png” _builder_version=”4.2.2″ title_level=”h2″ background_color=”#00457a” custom_button_one=”on” button_one_text_color=”#44580e” button_one_bg_color=”#ffffff” button_one_border_color=”#ffffff” button_one_border_radius=”5px”]
x 
. Now fill in the tables with positive integers so that the sum of the rows, columns, and diagonals are equal. Does there exist such a rectangle?
. There are 
. See the picture below.
. See the picture below.
. Therefore the number of rows and columns must be equal.
