Try this beautiful problem from Geometry: Area of a triangle
Triangle - AMC-10A, 2006- Problem 21
A circle of radius 1 is tangent to a circle of radius 2 . The sides of $\triangle A B C$ are tangent to the circles as shown, and the sides $\overline{A B}$ and $\overline{A C}$ are congruent. What is the area of $\triangle A B C ?$
,
i
$15 \sqrt{2}$
$\frac{35}{2} $
$\frac{64}{3}$
$16 \sqrt{2}$
\(24\)
Key Concepts
Geometry
Circle
Triangle
Check the Answer
Answer: $16 \sqrt{2}$
AMC-10A (2006) Problem 21
Pre College Mathematics
Try with Hints
Given that there are two circle of radius 1 is tangent to a circle of radius 2.we have to find out the area of the \(\triangle ABC\).Now draw a perpendicular line \(AF\) on \(BC\).Clearly it will pass through two centers \(O_1\) and \(O_2\). and $\overline{A B}$ and $\overline{A C}$ are congruent i.e \(\triangle ABC\) is an Isosceles triangle. Therefore \(BF=FC\)
So if we can find out \(AF\) and \(BC\) then we can find out the area of the \(\triangle ABC\).can you find out \(AF\) and \(BC\)?
Can you now finish the problem ..........
Now clearly $\triangle A D O_{1} \sim \triangle A E O_{2} \sim \triangle A F C$ ( as \(O_1D\) and \(O_2E\) are perpendicular on \(AC\) , R-H-S law )
From Similarity we can say that , $\frac{A O_{1}}{A O_{2}}=\frac{D O_{1}}{E O_{2}} \Rightarrow \frac{A O_{1}}{A O_{1}+3}=\frac{1}{2} \Longrightarrow A O_{1}=3$
By the Pythagorean Theorem we have that $A D=\sqrt{3^{2}-1^{2}}=\sqrt{8}$
Again from $\triangle A D O_{1} \sim \triangle A F C$ $\frac{A D}{A F}=\frac{D O_{1}}{C F} \Longrightarrow \frac{2 \sqrt{2}}{8}=\frac{1}{C F} \Rightarrow C F=2 \sqrt{2}$
can you finish the problem........
The area of the triangle is $\frac{1}{2} \cdot A F \cdot B C=\frac{1}{2} \cdot A F \cdot(2 \cdot C F)=A F \cdot C F=8(2 \sqrt{2})$=\(16\sqrt2\)
Circles with centers $A$ and $B$ have radii 3 and 8 , respectively. A common internal tangent intersects the circles at $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at $E,$ and $A E=5 .$ What is $C D ?$
,
i
$13$
$\frac{44}{3} $
$\sqrt{221}$
$\sqrt{255}$
\(\frac{55}{3}\)
Key Concepts
Geometry
Circle
Tangents
Check the Answer
Answer: $ \frac{44}{3}$
AMC-10 (2006) Problem 23
Pre College Mathematics
Try with Hints
Given that Circles with centers $A$ and $B$ have radii 3 and 8 and $A E=5 .$.we have to find out \(CD\).So join \(BC\) and \(AD\).then clearly \(\triangle BCE\) and \(\triangle ADE\) are Right-Triangle(as \(CD\) is the common tangent ).Now \(\triangle BCE\) and \(\triangle ADE\) are similar.Can you proof \(\triangle BCE\) and \(\triangle ADE\)?
Can you now finish the problem ..........
$\angle A E D$ and $\angle B E C$ are vertical angles so they are congruent, as are angles $\angle A D E$ and $\angle B C E$ (both are right angles because the radius and tangent line at a point on a circle are always perpendicular). Thus, $\triangle A C E \sim \triangle B D E$.
By the Pythagorean Theorem, line segment \(DE=4\)
Therefore from the similarity we can say that \(\frac{D E}{A D}=\frac{C E}{B C} \Rightarrow \frac{4}{3}=\frac{C E}{8}\) .
Try this beautiful problem from Geometry based on Area of a Triangle Using similarity
Area of Triangle - AMC-8, 2018 - Problem 20
In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.
What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?
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
Check the Answer
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....
Try this beautiful problem from Geometry based on the radius and tangent of a circle.
SMO 2013 - Geometry (Problem 25)
As shown in the figure below ,circles $C_1 $and$ C_2$ of radius 360 are tangent to each other , and both tangent to the straight line l.if the circle$ C_3$ is tangent to $C_1$ ,$C_2$ and l ,and circle$ C_4 $is tangent to$ C_1$,$C_3$ and l ,find the radius of$ C_4$
30
35
40
Key Concepts
Geometry
Pythagoras theorm
Distance Formula
Check the Answer
Answer:40
SMO -Math Olympiad-2013
Pre College Mathematics
Try with Hints
Let R be the radius of $C_3$
$C_2E$ =360-R
$C_3E=360$
$C_2C_3$=360+R
Using pythagoras theorm ....
$ (360-R)^2+360^2=(360+R)^2$
i.e R=90
Can you now finish the problem ..........
Let the radius of$ C_4$ be r
then use the distacce formula and tangent property........
