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## Area of Trapezoid | AMC 10A, 2018 | Problem 9

Try this beautiful problem from AMC 10A, 2018 based on area of trapezoid.

## Area of Trapezoid | AMC 10A

All of the triangles in the diagram below are similar to isosceles triangle ABC , in which AB = AC . Each of the 7 smallest triangles has area 1 and $\triangle {ABC}$ has area 40.  What is the area of trapezoid DBCE ?

• 24
• 25
• 26
• 20

### Key Concepts

2D – Geometry

Isosceles Triangle

Area of Triangle

American Mathematics Competition, 2018 Problem 9

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

We can try this problem using this hint :

Let the base length of the small triangle be  x :

Then, there is a triangle ADE encompassing the 7 small triangles and sharing the top angle with a base length of 4x.

Try the rest of the sum ………………………..

Continuing from the last hint:

Because the area is proportional to the square of the side, let the base BC be $\sqrt {40}x$. Then triangle ADE  has an area of 16. So the area is 40 – 16 = 24(Answer).

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## What is Area of triangle?

The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is $A = 1/2 × b × h$, where $b$ is the base and $h$ is the height of the given triangle. Here we have to find the side of triangle.

## Try the problem

The area of $\triangle (EBD)$ is one third of the area of $\triangle (ABC)$. Segment  is perpendicular to segment $(AB)$. What is $(BD)$?

2011 AMC 10B-Problem 9

Geometry

6 out of 10

Challenges and Thrills in Pre College Mathematics

## Use some hints

Notice that here $\triangle ABC$ and $\triangle BDE$ are similar.

Therefore $DE$=$\frac{3}{4} BD$

Now we have to find the area.

We know this is = $\frac{1}{2}$ $\times$ base $\times$ height.

Now using this formula can you find the area of $\triangle ABC$ and $\triangle BDE$?

Now $\triangle ABC$ = $\frac{1}{2} \times 3 \times 4$ =$6$ and $\triangle BDE$= $\frac{1}{2} \times DE \times BD$ = $\frac{1}{2}$* $\frac{3}{4} \times BD$ $\times$ $BD$ = $\frac{3}{8} BD^2$

Again we know the area of $\triangle BDE$ is one third of the area of $\triangle ABC$ .

Therefore, $\frac{3}{8} BD^2$ = $6$ $\times$ $\frac{1}{3}$

$9*BD^2$=$48$ , or,$BD^2$=$\frac{48}{9}$ , or, $BD^2$=$\frac{16}{3}$

so, the answer is= $BD$=$\frac{4 \sqrt3}{3}$

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## What is Area of Triangle ?

The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × h,  where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral.

## Try This Problem from AMC 10A – 2019 -Problem No.7

Two lines with slopes $\frac{1}{2}$ and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?

A) 4 B) $4\sqrt 2$ C) 6 D) 8 E) $6 \sqrt 2$

American Mathematics Competition 10 (AMC 10A), 2019, Problem Number – 7

Area of Triangle

6 out of 10

Problems in Plane Geometry by Sharygin

## Use some hints

If you need a hint to start this sum use this

Lets try to find the slop – intercept form of all three lines : (x,y) = (2,2) and y =

$\frac{x}{2}+b$ implies $2 = \frac{2}{2}+b = 1+b$. So, b = 1 . While y = 2x + c implies 2 = 2.2 + c So, c = -2 And again x+y = 10 implies y = -x + 10.

Thus the lines are $y = \frac {x}{2} + 1$ , y = 2x – 2 and y = -x + 10 . Now we find the intersection points between each of the lines with y = -x + 10 , which are (6,4) and (4,6) .

In the last hint we can apply the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle where the base is $2\sqrt 2$ and the height $3 \sqrt 2$, whose area is 6 .The answer is 6 (c) .

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## What is Area of a triangle?

Area of a triangle $=\frac{1}{2}\times BASE \times HEIGHT$

## Try the problem

In the given figure hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas $18$ and $32$ respectively, $\triangle JBK$ is equilateral and $FE=BC$. What is the area of $\triangle KBC$?

$\textbf{(A) }6\sqrt{2}\quad\textbf{(B) }9\quad\textbf{(C) }12\quad\textbf{(D) }9\sqrt{2}\quad\textbf{(E) }32$.

AMC 8 2015 Problem 21

Area of Squares and triangles.

6 out of 10

Challenges and Thrills of Pre college Mathematics

## Use some hints

Can you find the lengths of one side of the squares $ABJI$ and $FEHG$ ??

If you can, then try to use the given conditions to find out the lengths of two sides of $\triangle KBC$

Area of a square =$(\textbf{Side of the square})^2$

Then Side of a square =$\sqrt{(\textbf{Area of the square})}$

Then you can easily find the lengths of a side of the squares $ABJI$ and $FEHG$ which is $3\sqrt{2}$ and $4\sqrt{2}$ respectively.

Then by the given condition $\overline{FE}=\overline{BC}=4\sqrt{2}$

Since $\triangle JBK$ is an equilateral triangle then all of its sides are equal.

and $\overline{JB}=\overline{BK}=3\sqrt{2}$

Now as you know that $\triangle JBK$ is equilateral and the hexagon $ABCDEF$ is equiangular. Can you find out the measure of $\angle KBC$ ??

From the figure we can clearly see $\angle JBA + \angle ABC + \angle KBC + \angle KBJ = 360^{\circ}$

$\angle KBC = 360^{\circ}-90^{\circ}-120^{\circ}-60^{\circ}$ [Since $\angle JBA=90^{\circ}$ (an angle of a square) $\angle ABC=120^{\circ}$ (an angle of an equiangular hexagon) and $\angle JBK= 60^{\circ}$(an angle of an equilateral triangle)]

i.e., $\angle KBC= 90^{\circ}$

Then $\triangle KBC$ is a right angle triangle

Then the base and height of $\triangle KBC$ are $BC$ and $KB$

So the area of $\triangle KBC=\frac12 \times KB \times BC = \frac12 \times 4\sqrt{2} \times 3\sqrt{2} = 12$

Categories

## What is the Area of Triangle ?

The area of a Triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × hwhere b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral. Also the area of Quadrilateral is defined as the half of the product of the length of the diagonals

## Try the problem from AMC 10 (2020)

Triangle AMC is isosceles with AM = AC. Medians $\overline {MV}$ and $\overline {CU}$ are perpendicular to each other, and $MV = CU =12$ . What is the area of $\triangle {AMC}$ ?

American Mathematics Competition 10 (AMC 10A), {2020}, {12}

Geometry – Area of Triangle

4 out of 10

Challenges and Thrills of Pre – College Mathematics

## Use some hints

We can imagine the portion $UVCM$ to be a quadrilateral having perpendicular diagonals .So its area can be found as half of the product of the length of the diagonals .

Again : – $\triangle AUV$ has $\frac {1}{4}$ of the triangle

$AMC$ by similarity.

So, $UVVM = \frac {3}{4} AMC$

$\frac {1}{2} .12.12 = \frac {3}{4} AMC$

$72 = \frac {3}{4} AMC$

$AMC = 96$

Categories

## Competency in Focus: 2D Geometry (Area of Rectangle)

This problem from American Mathematics contest (AMC 8, 2019) is based on calculation of area of  Rectangle .

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]Three identical rectangles are put together to form rectangle $ABCD$, as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is 5 feet, what is the area in square feet of rectangle $ABCD$? [/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” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” _builder_version=”4.2.2″ open=”off”]American Mathematical Contest 2019, AMC 8  Problem 2[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”on” _builder_version=”4.2.2″]

### 2D Geometry (Area of Rectangle)

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]The length of the smaller side of each rectangle is given 5 feet. So using the diagram we can find  that the larger side of the small rectangle is 2 times the length of the smaller side.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]So find the longer side which is 5*2 = 10 feet.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″] the area of the identical rectangles is 5*10 = 50 square feet[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]  Thus at end we have 3 identical rectangles formed a bigger rectangle……Try to solve …   50*3 = 150 square feet [/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”]

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## Competency in Focus: Number System

This problem from Indian Statistical Institute (ISI Entrance 2012) is based on Number System. It includes finding the remainder when a number is divided by another digit.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]The last digit of  $9!+3^{9966}$ is (A) 3 (B) 9 (C) 7 (D) 1[/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”]Indian Statistical Institute (ISI) 2012 Problem 8.[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” open=”off” _builder_version=”4.2.2″]

### Number system

[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.2.2″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.2.2″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.2.2″]Unit digit of the whoe expression will be sum of unit digit of the first term and unit digit of the second term. So if the first term gives last digit 5 and 2nd terms gives 2 then unit digit of whole expresion is (5+2) or 7.[/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.2.2″]when we expand $9!$ there will be 5 and 2 in between that when multiplied will give 10 as a factor so the term $9!$ will have $0!$ as last digit.[/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.2.2″]$3^{1}$ has last digit as 3 $3^{2}$ has last digit as 9 $3^{3}$ has last digit as 7 $3^{4}$ has last digit as 1 $3^{5}$ has last digit as 3  and the pattern repeats with the power(indxe) at an interval of 4.[/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.2.2″]9966 when divided by 4 gives 2 as remainder. So, in $3^{9964} 3^{2}$ , 9 will be the last digit.[/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″ saved_tabs=”all” locked=”off”][et_pb_fullwidth_header title=”I.S.I. & C.M.I. Program” button_one_text=”Learn more” button_one_url=”https://www.cheenta.com/isicmientrance/” header_image_url=”https://www.cheenta.com/wp-content/uploads/2018/03/ISI.png” _builder_version=”4.2.2″ title_level=”h2″ title_font=”Acme||||||||” background_color=”#220e58″]

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## Competency in Focus: Combinatorics

This problem is based on Combinatorics from American Mathematics contest (AMC 10A, 2019). It includes arrangement of  $n$  items out of which  $r$ items are similar.

## Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.2.2″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” box_shadow_style=”preset2″]A child builds towers using identically shaped cubes of different color. How many different towers with a height 8 cubes can the child build with 2 red cubes, 3 blue cubes, and 4 green cubes? (One cube will be left out.) (A) 24 (B) 288 (C) 312 (D) 1260 (E) 40320[/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” open=”on” _builder_version=”4.2.2″]American Mathematical Contest 2019, AMC 10A  Problem 17[/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.2.2″ open=”off”]Combinatorics[/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.1″ open=”off”]4/10[/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.2.2″ open=”off”]

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Categories

## Finding Side of Triangle

The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × h, where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral. This problem is based on finding side of a triangle by its area.

## Try the Problem

This problem is from AMC 10A, 2013.

Square $ABCD$ has a side length $10$. The point $E$ is on $\overline{BC}$ and the area of $\triangle ABE$ is $40$. What is $\overline{BE}?$

$\textbf{(A)}\quad 4\quad \textbf{(B)}\quad 5\quad \textbf{(C)}\quad 6\quad \textbf{(D)}\quad 7\quad \textbf{(E)}\quad 8\quad$

AMC 10A, 2013 Problem 3

Area of Triangle

4/10

Challenges and Thrills in Pre College Mathematics

Excursion of Mathematics

## Use some hints

Given, Square  ABCD  has side length 10.

So, AB = 10.

Now, we know area of a triangle =$\frac{(\text{height})\times(\text{base})}{2}$. Try to use this here

So we have the area of $\triangle ABE$ is equal to $\frac{AB\times BE}{2}$. Plugging in $AB=10$. What we get?

We get $10\times BE=80 \Rightarrow BE=8$

Categories

# What are we learning ?

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” hover_enabled=”0″ box_shadow_style=”preset2″]Competency in Focus: Area of triangles This problem from American Mathematics contest (AMC 10A, 2013) is based on calculation of area of triangles . [/et_pb_text][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]

# Next understand the problem

[/et_pb_text][et_pb_text _builder_version=”4.1″ text_font=”Raleway||||||||” text_font_size=”20px” text_letter_spacing=”1px” text_line_height=”1.5em” background_color=”#f4f4f4″ custom_margin=”10px||10px” custom_padding=”10px|20px|10px|20px” hover_enabled=”0″ box_shadow_style=”preset2″]Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$? $[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$ $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ [/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=”4.1″ toggle_font=”||||||||” body_font=”Raleway||||||||” text_orientation=”center” custom_margin=”10px||10px” hover_enabled=”0″][et_pb_accordion_item title=”Source of the problem” open=”on” _builder_version=”4.1″ hover_enabled=”0″]American Mathematical Contest 2013, AMC 10A  Problem 3 [/et_pb_accordion_item][et_pb_accordion_item title=”Key Competency” _builder_version=”4.1″ hover_enabled=”0″ open=”off”]Area of triangles [/et_pb_accordion_item][et_pb_accordion_item title=”Difficulty Level” _builder_version=”4.1″ hover_enabled=”0″ open=”off”]4/10 [/et_pb_accordion_item][et_pb_accordion_item title=”Suggested Book” _builder_version=”4.0.9″ open=”off”]Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

[/et_pb_text][et_pb_tabs _builder_version=”4.1″ hover_enabled=”0″][et_pb_tab title=”HINT 0″ _builder_version=”4.0.9″]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title=”HINT 1″ _builder_version=”4.1″ hover_enabled=”0″]Given Square $ABCD$ has side length $10$.  So, $AB=10$. [/et_pb_tab][et_pb_tab title=”HINT 2″ _builder_version=”4.1″ hover_enabled=”0″]Now, we know area of a triangle =$\frac{(height).(base)}{2}$. Try to use this here . [/et_pb_tab][et_pb_tab title=”HINT 3″ _builder_version=”4.1″ hover_enabled=”0″]So , we have the area of $\triangle ABE$ is equal to $\frac{AB(BE)}{2}$Plugging in $AB=10$ , what we get ? [/et_pb_tab][et_pb_tab title=”HINT 4″ _builder_version=”4.1″ hover_enabled=”0″]we get $80 = 10BE$. Dividing, we find that $BE=\boxed{\textbf{(E) }8}$ [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version=”3.27.4″ text_font=”Raleway|300|||||||” text_text_color=”#ffffff” header_font=”Raleway|300|||||||” header_text_color=”#e2e2e2″ background_color=”#0c71c3″ min_height=”12px” custom_margin=”50px||50px” custom_padding=”20px|20px|20px|20px” border_radii=”on|5px|5px|5px|5px” box_shadow_style=”preset3″]