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Triangle Problem | AMC 10B, 2013 | Problem 16

Try this beautiful problem from Geometry from AMC-10B, 2013, Problem-16, based on triangle

Triangle | AMC-10B, 2013 | Problem 16

In triangle $ABC$, medians $AD$ and $CE$ intersect at $P$, $PE=1.5$, $PD=2$, and $DE=2.5$. What is the area of $AEDC$?

• $12$
• $13.5$
• $15.5$

Key Concepts

Geometry

Triangle

Pythagorean

Answer:$13.5$

AMC-10B, 2013 problem 16

Pre College Mathematics

Try with Hints

We have to find out the area of AEDC which is divided in four triangles i.e$\triangle APC$,$\triangle APE$, $\triangle PED$, $\triangle CPD$

Now if we find out area of four triangle then we can find out the required area AEDC. Now in the question they supply the information only one triangle i.e $\triangle PED$ such that $PE=1.5$, $PD=2$, and $DE=2.5$ .If we look very carefully the given lengths of the $\triangle PED$ then $( 1.5 )^2 +(2)^2=(2.5)^2$ $\Rightarrow$ $(PE)^2 +(PD)^2=(DE)^2$ i,e $\triangle PED$ is a right angle triangle and $\angle DEP =90^{\circ}$ .so we can easily find out the area of the $\triangle PED$ using formula $\frac{1}{2} \times base \times height$ . To find out the area of other three triangles, we must need the lengths of the sides.can you find out the length of the sides of other three triangles…

Can you now finish the problem ……….

To find out the lengths of the sides of other three triangles:

Given that AD and CD are the medians of the given triangle and they intersects at the point P. .we know the fact that the centroid ($P$) divides each median in a $2:1$ ratio .Therefore AP:PD =2:1 & CP:PE=2:1. Now $PE=1.5$ and $PD=2$ .Therefore AP=4 and CP=3.

And also the angles i.e ($\angle APC ,\angle CPD, \angle APE$) are all right angles as $\angle DPE= 90^{\circ}$

can you finish the problem……..

Area of four triangles :

Area of the $\triangle APC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PC$ = $\frac{1}{2} \times 4 \times 3$ =6

Area of the $\triangle DPC$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times CP \times PD$ = $\frac{1}{2} \times 3 \times 2$ =3

Area of the $\triangle PDE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times PD \times DE$ = $\frac{1}{2} \times 2 \times 1.5$=1.5

Area of the $\triangle APE$ =$\frac{1}{2} \times base \times height$ = $\frac{1}{2} \times AP \times PE$ = $\frac{1}{2} \times 4 \times 1.5 =6$

Total area of ACDE=area of ($\triangle APC$+$\triangle APE$+ $\triangle PED$+ $\triangle CPD$)=$(6+3+1.5+6)=17.5$ sq.unit

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Finding side of Triangle | PRMO-2014 | Problem 15

Try this beautiful problem from PRMO, 2014 based on Finding side of Triangle.

Finding side of Triangle | PRMO | Problem 15

Let XOY be a triangle with angle XOY=90 degrees. Let M and N be the midpoints of the legs OX and OY, respectively. Suppose that XN=19 and YM=22. what is XY?

• $28$
• $26$
• $30$

Key Concepts

Geometry

Triangle

Pythagoras

Answer:$26$

PRMO-2014, Problem 15

Pre College Mathematics

Try with Hints

Given that $\angle XOY=90^{\circ}$ .Let M and N be the midpoints of the legs OX and OY, respectively, and that XN=19 and YM=22. Now $\triangle XON$ & $\triangle MOY$ are Right angle Triangle. Use Pythagoras theorem …….

Can you now finish the problem ……….

Let $XM=MO=p$ and $ON=NY=q$.Now using Pythagoras theorm on $\triangle XON$ & $\triangle MOY$ we have…

$OX^2 +ON^2=XN^2$ $\Rightarrow 4p^2 +q^2=19^2$ $\Rightarrow 4p^2 +q^2=361$………..(1) and $OM^2 +OY^2=MY^2$ $\Rightarrow p^2 +4q^2=22^2$ $\Rightarrow p^2 +4q^2=484$……(2)

Now Adding (1)+(2)=$(4p^2 +q^2=361)$+$(p^2 +4q^2=484$ $\Rightarrow 5(p^2+q^2)=845$ $\Rightarrow (p^2+q^2)=169$ $\Rightarrow 4(p^2+q^2)=676$ $\Rightarrow (OX)^2+(OY)^2=(26)^2$ $\Rightarrow (XY)^2=(26)^2$ $\Rightarrow XY=26$.

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Ratio Of Two Triangles | AMC-10A, 2004 | Problem 20

Try this beautiful problem from AMC 10A, 2004 based on Geometry: Ratio Of Two Triangles

Ratio Of Two Triangles – AMC-10A, 2004- Problem 20

Points $E$ and $F$ are located on square $ABCD$ so that $\triangle BEF$ is equilateral. What is the ratio of the area of $\triangle DEF$ to that of $\triangle ABE$

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $2$
• $1+\sqrt 3$

Key Concepts

Square

Triangle

Geometry

Answer: $2$

AMC-10A (2002) Problem 20

Pre College Mathematics

Try with Hints

We have to find out the ratio of the areas of two Triangles $\triangle DEF$ and $\triangle ABE$.Let us take the side length of $AD$=$1$ & $DE=x$,therefore $AE=1-x$

Now in the $\triangle ABE$ & $\triangle BCF$ ,

$AB=BC$ and $BE=BF$.using Pythagoras theorm we may say that $AE=FC$.Therefore $\triangle ABE \cong \triangle CEF$.So $AE=FC$ $\Rightarrow DE=DF$.Therefore the $\triangle DEF$ is  an isosceles right triangle. Can you find out the area of isosceles right triangle $\triangle DEF$

Can you now finish the problem ……….

Length of $DE=DF=x$.Then the the side length of $EF=X \sqrt 2$

Therefore the area of $\triangle DEF= \frac{1}{2} \times x \times x=\frac{x^2}{2}$ and area of $\triangle ABE$=$\frac{1}{2} \times 1 \times (1-x) = \frac{1-x}{2}$.Now from the Pythagoras theorm $(1-x)^2 +1 =2x^2 \Rightarrow x^2=2-2x=2(1-x)$

can you finish the problem……..

The ratio of the area of $\triangle DEF$ to that of $\triangle ABE$ is $\frac{\frac{x^2}{2}}{\frac{(1-x)}{2}}$=$\frac{x^2}{1-x}$=$\frac {2(1-x)}{(1-x)}=2$

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Length of a Tangent | AMC-10A, 2004 | Problem 22

Try this beautiful problem from American Mathematics Competitions, AMC 10A, 2004 based on Geometry: Length of a Tangent.

Length of a Tangent – AMC-10A, 2004- Problem 22

Square $ABCD$ has a side length $2$. A Semicircle with diameter $AB$ is constructed inside the square, and the tangent  to the semicircle from $C$ intersects side $AD$at $E$. What is the length of $CE$?

• $\frac{4}{3}$
• $\frac{3}{2}$
• $\sqrt 3$
• $\frac{5}{2}$
• $1+\sqrt 3$

Key Concepts

Square

Semi-circle

Geometry

Answer: $\frac{5}{2}$

AMC-10A (2004) Problem 22

Pre College Mathematics

Try with Hints

We have to find out length of $CE$.Now $CE$ is a tangent of inscribed the semi circle .Given that length of the side is $2$.Let $AE=x$.Therefore $DE=2-x$. Now $CE$ is the tangent of the semi-circle.Can you find out the length of $CE$?

Can you now finish the problem ……….

Since $EC$ is tangent,$\triangle COF$ $\cong$ $\triangle BOC$ and $\triangle EOF$ $\cong$ $\triangle AOE$ (By R-H-S law).Therefore $FC=2$ & $EC=x$.Can you find out the length of $EC$?

can you finish the problem……..

Now the $\triangle EDC$ is a Right-angle triangle……..

Therefore $ED^2+ DC^2=EC^2$ $\Rightarrow (2-x)^2 + 2^2=(2+x)^2$ $\Rightarrow x=\frac{1}{2}$

Hence $EC=EF+FC=2+\frac{1}{2}=\frac{5}{2}$

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Radius of a Semi Circle -AMC 8, 2017 – Problem 22

Try this beautiful problem from Geometry based on the radius of a semi circle and tangent of a circle.

AMC-8(2017) – Geometry (Problem 22)

In the right triangle ABC,AC=12,BC=5 and angle C is a right angle . A semicircle is inscribed in the triangle as shown.what is the radius of the semi circle?

• $\frac{7}{6}$
• $\frac{10}{3}$
• $\frac{9}{8}$

Key Concepts

Geometry

congruency

similarity

Answer:$\frac{10}{3}$

AMC-8(2017)

Pre College Mathematics

Try with Hints

Here O is the center of the semi circle. Join o and D(where D is the point where the circle is tangent to the triangle ) and Join OB.

Can you now finish the problem ……….

Now the $\triangle ODB$and $\triangle OCB$ are congruent

can you finish the problem……..

Let x be the radius of the semi circle

Now the $\triangle ODB$ and $\triangle OCB$ we have

OD=OC

OB=OB

$\angle ODB$=$\angle OCB$= 90 degree`

so $\triangle ODB$ and $\triangle OCB$ are congruent (by RHS)

BD=BC=5

And also $\triangle ODA$ and $\triangle BCA$ are similar….

$\frac{8}{12}$=$\frac{x}{5}$

i.e x =$\frac{10}{3}$