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ISI MStat 2016 Problem 1 | Area bounded by the curves | PSB Sample

This is a beautiful problem from ISI MStat 2016 Problem 1, PSB Sample, based on area bounded by the curves. We provide a detailed solution with the prerequisites mentioned explicitly.

Problem- ISI MStat 2016 Problem 1

In the diagram below, $L(x)$ is a straight line that intersects the graph of a polynomial $P(x)$ of degree 2 at the points $A=(-1,0)$ and $B=(5,12) .$ The area of the shaded region is 36 square units. Obtain the expression for $P(x)$.

Prerequisites

• Area bounded by the curve
• Polynomials of degree 2
• Area of a triangle

Solution

Let $P(x)=ax^2 +bx + c$ as given $P(x)$ is of degree 2 .

Now from the figure we can see that $L(x)$ intersect $P(x)$ at points $A=(-1,0)$ and $B=(5,12) .$

Hence we have $P(-1)=0$ and $P(5)=12$ , which gives ,

$a-b+c=0$ —(1) and $25a+5b +c =12$ —-(2)

Then ,

See from Fig-1 we can say that Area of the shaded region = (Area bounded by the curve P(x) and x-axis )- (Area of the triangle ABC) – (Area bounded by the curve P(x) , x=5 and x=L )

= $\int^{L}_{-1} P(x)\,dx – \frac{1}{2} \times (5+1) \times 12 -\int^{5}_{L} P(x)\,dx$

=$\int^{L}_{-1} P(x)\,dx – \int^{5}_{L} P(x)\,dx$ -36

=$\int^{5}_{-1} P(x)\,dx$ -36

Again it is given that area of the shaded region is 36 square units.

So, $\int^{5}_{-1} P(x)\,dx$ -36 =36 $\Rightarrow$ $\int^{5}_{-1} P(x)\,dx$ =$2 \times 36$

$\int^{5}_{-1} (ax^2+bx+c) \,dx = 2 \times 36$ . After integration we get ,

$7a + 2b +c =12$ —(3)

Now we have three equations and three unknows

$a-b+c=0$

$25a+5b +c =12$

$7a + 2b +c =12$

Solving this three equations by elimination and substitution we get ,

$a=-1 , b=6 , c=7$

Therefore , the expression for $P(x)$ is $P(x)= -x^2+6x+7$ .

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Problem on Area of Triangle | SMO, 2010 | Problem 32

Try this beautiful problem from Singapore Mathematics Olympiad based on area of triangle.

Problem – Area of Triangle (SMO Entrance)

Given that ABCD is a square . Points E and F lie on the side BC and CD respectively, such that BE = $\frac {1}{3} AB$ = CF. G is the intersection of BF and DE . If

$\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$ is in its lowest term find the value of m+n.

• 12
• 19
• 21
• 23

Key Concepts

2D – Geometry

Area of Triangle

Challenges and Thrills – Pre – College Mathematics

Try with Hints

Here is the 1st hint for them who got stuck in this problem:

At 1st we will join the points BD and CG .

Then to proceed with this sum we will assume the length of AB to be 1. The area of $\triangle {BGE}$ and $\triangle {FGC}$ are a and b respectively.

Try to find the area of $\triangle {EGC}$ and $\triangle {DGF}$……………………..

Now for the 2nd hint let’s start from the previous hint:

So the area of $\triangle {EGC}$ and $\triangle {DGF}$ are 2a and 2b .From the given value in the question we can say the area of $\triangle {BFC} = \frac {1}{3}$(as BE = CF = 1/3 AB\).

We can again write 3a + b = $\frac {1}{6}$

Similarly 3b + 2a = area of the $\triangle DEC = \frac {1}{3}$.

Now solve this two equation and find a and b…………..

In the last hint :

2a + 3b = 1/3……………………(1)

3a + b = 1/6……………(2)

so in $(1) \times 3$ and in $(2) \times 2$

6a + 9b = 1

6a + 2b = 1/3

so , 7 b = 2/3 (subtracting above 2 equation)

b = $\frac {2}{21}$ and a = $\frac {1}{42}$

So,$\frac {Area of ABGD}{Area of ABCD} = 1-3(a+b) = 1- \frac {15}{42} = \frac {9}{14}$

Comparing the given values from the question , $\frac {Area of ABGD}{Area of ABCD} = \frac {m}{n}$

m = 9 and n = 14

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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).

Categories

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= $BD$=$\frac{4 \sqrt3}{3}$

Categories

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) .

Categories

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

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 $ABCD$?
Source of the problem
American Mathematical Contest 2019, AMC 8  Problem 2

2D Geometry (Area of Rectangle)

3/10
Suggested Book
Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!
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.
So find the longer side which is 5*2 = 10 feet.
the area of the identical rectangles is 5*10 = 50 square feet
Thus at end we have 3 identical rectangles formed a bigger rectangle……Try to solve …   50*3 = 150 square feet

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External Tangent | AMC 10A, 2018 | Problem 15

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Dice Problem | AMC 10A, 2014| Problem No 17

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Problem on Curve | AMC 10A, 2018 | Problem 21

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Right-angled Triangle | AMC 10A, 2018 | Problem No 16

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Length of the crease | AMC 10A, 2018 | Problem No 13

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Right-angled shaped field | AMC 10A, 2018 | Problem No 23

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Area of region | AMC 10B, 2016| Problem No 21

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Coin Toss Problem | AMC 10A, 2017| Problem No 18

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Categories

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

The last digit of  $9!+3^{9966}$ is

(A) 3

(B) 9

(C) 7

(D) 1

Source of the problem
Indian Statistical Institute (ISI) 2012 Problem 8.

Number system

4/10
Suggested Book
Challenges and Thrills in Pre College Mathematics

Excursion Of Mathematics

Do you really need a hint? Try it first!
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.
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.
$3^{1}$ 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.

9966 when divided by 4 gives 2 as remainder.

So, in $3^{9964} 3^{2}$ , 9 will be the last digit.

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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

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

Source of the problem
American Mathematical Contest 2019, AMC 10A  Problem 17
Combinatorics
4/10
Suggested Book

Do you really need a hint? Try it first!
Arroanging 8 Cubes is same as arranging 9 cubes and removing 1 cube from the arrangment. It will be same as arranging 8 cubes out of 9 on top of one another.
If we want to arrange the 9 things then the total no of ways are $9!$.
But 2 cubes are red (or similiar type), 3 Blue and 4 green so we have to divide the total arrangment by $2!$ , $3!$ and $4!$.

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Problem on Curve | AMC 10A, 2018 | Problem 21

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Area of region | AMC 10B, 2016| Problem No 21

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