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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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Angles of Star | AMC 8, 2000 | Problem 24

Try this beautiful problem from Geometry from AMC-8, 2000, Problem-24, based on angles of Star

Angles of Star | AMC-8, 2000 | Problem 24

If $\angle A = 20^\circ$ and $\angle AFG =\angle AGF$, then $\angle B+\angle D =$

• $90$
• $70$
• $80$

Key Concepts

Geometry

Star

Triangle

Answer:$80$

AMC-8, 2000 problem 24

Pre College Mathematics

Try with Hints

Find the $\angle AFG$

Can you now finish the problem ……….

sum of the angles of a Triangle is $180^\circ$

can you finish the problem……..

we know that the sum of the angles of a Triangle is $180^\circ$

In the $\triangle AGF$ we have,$(\angle A +\angle AGF +\angle AFG) =180^\circ$

$\Rightarrow 20^\circ +2\angle AFG=180^\circ$(as $\angle A =20^\circ$ & $\angle AFG=\angle AGF$)

$\Rightarrow \angle AFG=80^\circ$ i.e $\angle EFD=\angle 80^\circ$

So the $\angle BFD=\frac{360^\circ -80^\circ-80^\circ}{2}=100^\circ$

Now in the $\triangle BFD$,$(\angle BFD +\angle B +\angle D$)=$180^\circ$

$\Rightarrow \angle B +\angle D=180^\circ -100^\circ$

$\Rightarrow \angle B +\angle D=80^\circ$