Test of Mathematics Solution Subjective 166 -The Grazing Field

Test of Mathematics at the 10+2 Level

This is a Test of Mathematics Solution Subjective 166 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.


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Problem

A cow is grazing with a rope around its neck and the other end of the rope is tied to a pol. The length of the rope is 10 metres. There are two boundary walls perpendicular to each other, one at a distance of 5 metres to the east of the pole and another at a distance of $ 5 \sqrt{2}$ metres to the north of the pole. Find the area the cow can graze on.


Solution

First we need to analyse the question and draw a diagram. The figure for the problem would look something like this

Untitled

We have a circle with centre P and radius 10m. Let AC and AB be the boundary walls present. The striped region shows the area available for grazing. Now if we see carefully we can see that area of the shaded region is nothing but the sum of the areas of the rectangle AEPC, the two right angled triangles EPB and DPC and the sector BPC.

PB = PC = radius of the circle = 10m

PE = $ 5\sqrt{2}$m

AS $ PE^2 + EB^2 = PB^2$      (Pythagorean property of a right angled triangle)

From here, we get $ EB = 5\sqrt{2}$m

Similarly, as PD = 5m, PC = 10m

$ CD = 5\sqrt{3}$m              (pythagorean property)

Therefore,

Area of rectangle AEPC = $ (5\sqrt{2} *5) m^2 = 25\sqrt{2}m^2$

Area of triangle PEB = $\frac{1}{2} * 5\sqrt{2} * 5\sqrt{2} = 25m^2$

Area of triangle PDC = $ \frac{1}{2} * 5 * 5\sqrt{3} = \frac{25\sqrt{3}}{2} m^2$

Area of sector BPC = $ \pi(10)^2 * \frac {\angle BPC}{360^0}$

Now $ \angle BPC = 360^0 - (\angle BPE + \angle CPD + \angle EPD)$

$ => \angle BPC = 360^0 - (tan^{-1}\frac{BE}{EP} + tan^{-1}\frac{CD}{DP} + 90^0)$

$ => \angle BPC = 360^0 - (tan^{-1}\frac{5\sqrt{2}}{5\sqrt{2}} + tan^{-1}\frac{5\sqrt{3}}{5} + 90^0)$

$ => \angle BPC = 360^0 - (tan^{-1}1 + tan^{-1}\sqrt{3} + 90^0)$

$ => \angle BPC = 360^0 - (45^0 + 60^0 + 90^0)$

$ => \angle BPC = 165^0$

Therefore area of sector BPC = $ \pi(10)^2 * \frac {165^0}{360^0} = \pi * 100 * \frac{11}{24} = \frac{275}{6}\pi$

Thus Total area of the Grazing Field = $ (25\sqrt{2} + 25 + \frac{25\sqrt{3}}{2} + \frac{275}{6}\pi) m^2$

or to simplify Total Area = $ \frac{25}{6}(6\sqrt{2} + 6 + 3\sqrt{3} + 11) m^2$

Problem based on Cylinder | AMC 10A, 2015 | Question 9

Try this beautiful problem from Mensuration: Problem based on Cylinder from AMC 10A, 2015.

Cylinder - AMC-10A, 2015- Problem 9


Two right circular cylinders have the same volume. The radius of the second cylinder is $10 \%$ more than the radius of the first. What is the relationship between the heights of the two cylinders?

  • (A) The second height is $10 \%$ less than the first.
  • (B) The first height is $10 \%$ more than the second.
  • (C) The second height is $21 \%$ less than the first.
  • (D) The first height is $21 \%$ more than the second.
  • (E) The second height is $80 \%$ of the first.

Key Concepts


Mensuration

Cylinder

Check the Answer


Answer: (D) The first height is $21 \%$ more than the second.

AMC-10A (2015) Problem 9

Pre College Mathematics

Try with Hints


Let the radius of the first cylinder be $r_{1}$ and the radius of the second cylinder be $r_{2}$. Also, let the height of the first cylinder be $h_{1}$ and the height of the second cylinder be $h_{2}$.

Can you now finish the problem ..........

According to the problem,

$r_{2}=\frac{11 r_{1}}{10}$
$\pi r_{1}^{2} h_{1}=\pi r_{2}^{2} h_{2}$

can you finish the problem........

$r_{1}^{2} h_{1}=\frac{121 r_{1}^{2}}{100} h_{2} \Rightarrow h_{1}=\frac{121 h_{2}}{100}$

Therefore the Possible answer will be (D) The first height is $21 \%$ more than the second.

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Problem on Cylinder | AMC-10A, 2004 | Problem 11

Try this beautiful problem from AMC 10A, 2004 based on Mensuration: Cylinder

Problem on Cylinder - AMC-10A, 2004- Problem 11


A company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by \(25\%\) without altering the volume, by what percent must the height be decreased?

  • \(16\)
  • \(18\)
  • \(20\)
  • \(36\)
  • \(25\)

Key Concepts


Mensuration

Cylinder

Percentage

Check the Answer


Answer: \(36\)

AMC-10A (2004) Problem 11

Pre College Mathematics

Try with Hints


Let the radius of the jar be \(x\) and height be \(h\).then the volume (V) of the jar be\(V\)= \(\pi (x)^2 h\). Diameter of the jar increase \(25 \)% Therefore new radius will be \(x +\frac{x}{4}=\frac{5x}{4}\) .Now the given condition is "after increase the volume remain unchange".Let new height will be \(h_1\).Can you find out the new height....?

can you finish the problem........

Let new height will be \(H\).Therefore the volume will be \(\pi (\frac{5x}{4})^2 H\).Since Volume remain unchange......

\(\pi (x)^2 h\)=\(\pi (\frac{5x}{4})^2 H\) \(\Rightarrow H=\frac{16h}{25}\).

height decrease =\(h-\frac{16h}{25}=\frac{9h}{25}\).can you find out the decrease percentage?

can you finish the problem........

Decrease Percentage=\( \frac {\frac {9h}{25}}{h} \times 100=36\)%

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Mensuration : Area of Triangle , AMC 8 2015 Problem 21

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

Knowledge Graph


area of triangle- knowledge graph

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

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