*Solution:* Let O be the center of the circle and we draw the segment OP, where P belongs to the side BC. Using power at a point C, we have: PC² = CF × CE = 4×9 → *PC=6.* Now, we draw the segment OT, where T belongs to AD, note that AEOT and EBPO are squares whose sides are equal to the radius R of the circle. Therefore, AE=EB=BP=R. By Pythagoras in ∆BEC: EC² = BE² + BC² = BE² + (BP+ PC)² 9² = R² + (6+R)² = R²+36+12R+R² 2R² + 12R = 81 - 36 = 45 2R(R + 6) = 45 . note that, AB = 2R e BC = R + 6, consequently, *AB × BC = 45* , This is the area of rectangle ABCD.
Very nice, slick answer. Mine was more clunky. If BC is r + x, then x^2 = 4x9 = 36 so x =6. This was encouraging, so I used triangle EBC to give r^2 + (6+r)^2 = 9^2. This gave a slightly nasty answer of r = (3rt14 - 6)/2, so r+6 =(3rt14 + 6)/2, and then 2r x (r+6) = 45. Oh well!
It's messy to get a value of r, sure enough. But it turns out you don't need it!! Using your variables, r=radius & x=6: A(rectangle) = height * width = (2r)(r+6) = 2r^2 + 12r From your quadratic equation: r^2 + (r+6)^2 = 81 r^2 + (r^2 + 12r + 36) = 81 Collecting like-terms and subtracting 36 to RHS: 2r^2 + 12r = 45 and the LHS is EXACTLY the expression we got for area of the rectangle! A(rectangle) = 45 Done!!
Let the point of tangency of BC and the circle be point M. Then, applying the tangent secant theorem, (CM)² = (CF)(CE) = (4)(4 + 5) = 36 and CM = 6. Let the radius of the circle be r. Then, EB = MB = r. Apply the Pythagorean theorem to ΔBCE: (EB)² + (BC)² = (9)², r² + (r + 6)² = 81, 2r² + 12r - 45 = 0. The positive root, simplified, is r = √(31.5) - 3. AB = 2r = 2(√(31.5) - 3). BC = r + 6 = √(31.5) + 3. Area of ABCD = (AB)(BC) = 2(√(31.5) - 3)(√(31.5) + 3) = 2(31.5 - 9) = 2(22.5) = 45, as Math Booster also found. Checking the comments, I find that RAG981 may have solved it the same way.
Thanks for your comments! I think this is the more straightforward way to solve the problem, but the equations include 2 radicals. Luckily, we get a product of form (a - b)(a + b) = a² - b², so the square roots go away. Math Booster's solution avoids the radicals also, but, in my opinion, it takes a lot of creativity to find his solution!
Well you probably made some unintentional spelling mistakes while writing the comment, but that's exactly the method I used to do it, just use the tangent secant theorem to find length from C to bottom tangency point, then the rest of the rectangle length would be equal to the radius (can be proved by drawing some lines to complete a square and prove it's a square EONB but I won't bother getting into it here) as well as EB which is also r, and using pythagoras' theorem we can find r, multiply 2r by r+6 and we get the area
The area is 45 units square. Also I have noticed something that I should have thought about: delta EFP and delta EQC are similar because these triangles are NOT opposite to one another. This is different from one of the last two problems on this channel which showed HL congruence. I hope that this counts as a sufficient reason being that shows why and how similar triangles correspond to the area of the square.
Drop perpendicular from O onto chord EF. The little triangle formed is similar to triangle EBC. Let NC = x. Then (r+x)/9 = 2.5/r. This gives r^2 + rx =22.5 and 2r^2 + 2rx = 45. But this is the area since the rectangle has sides (r+x) and 2r.
АD=BC=a , AB=DC=2r . Area ADCD=2ar , theorem about tangent and secant lines coming from one point : (a-r)*2=CExCF , a*2+r*2-2ar=(5+4)4 (1) . From a triangle BCE according to the Pythagorean theorem - CE*2=r*2+a*2 , r*2+a*2=9*2 , substituting into eguation (1) - 9*2-2ar=9x4 , 2ar=81-36=45 . Area ABCD=45 .
Exactly!!! Very simple, without complications !! Congratulations !! My solution : cos α = 5/2R = b/(5+4) A = b.h= b.2R= 5.(5+4) A = 45 cm² ( Solved √ ) Exactly the same !!
r:9=x:4...x=FH=4r/9..(Hprooezione di F)...risulta arcsin(5r/9/5)=arccos(5/2r)..r^2=81/2-9√14..b(base)=√(81-(81/2-9√14))=√(81/2+9√14)..Arett=b*2r=2√(40,5^2-81*14)=2*22,5=45
Too clumsy, 4*(4+5)= tangent line sq, so tangent line =6. Let r is rad, so rectangle area =(6+r)*2r= 12r+ (2r) sq. Use Gougu theorem, r sq + (6+r) sq =81, expand, 12r+ (2r) sq -45=0. NO NEED TO SOLVE THE EQ, ang the 1st 2 terms is the area of rectangle, so =45
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Uma bela solução! Parabéns! 🎉🎉🎉🎉
*Solution:*
Let O be the center of the circle and we draw the segment OP, where P belongs to the side BC. Using power at a point C, we have:
PC² = CF × CE = 4×9 → *PC=6.*
Now, we draw the segment OT, where T belongs to AD, note that AEOT and EBPO are squares whose sides are equal to the radius R of the circle. Therefore, AE=EB=BP=R.
By Pythagoras in ∆BEC:
EC² = BE² + BC² = BE² + (BP+ PC)²
9² = R² + (6+R)² = R²+36+12R+R²
2R² + 12R = 81 - 36 = 45
2R(R + 6) = 45 .
note that, AB = 2R e BC = R + 6, consequently,
*AB × BC = 45* , This is the area of rectangle ABCD.
Very nice, slick answer. Mine was more clunky. If BC is r + x, then x^2 = 4x9 = 36 so x =6. This was encouraging, so I used triangle EBC to give r^2 + (6+r)^2 = 9^2. This gave a slightly nasty answer of r = (3rt14 - 6)/2, so r+6 =(3rt14 + 6)/2, and then 2r x (r+6) = 45. Oh well!
It's messy to get a value of r, sure enough. But it turns out you don't need it!!
Using your variables, r=radius & x=6:
A(rectangle) = height * width
= (2r)(r+6)
= 2r^2 + 12r
From your quadratic equation:
r^2 + (r+6)^2 = 81
r^2 + (r^2 + 12r + 36) = 81
Collecting like-terms and subtracting 36 to RHS:
2r^2 + 12r = 45
and the LHS is EXACTLY the expression we got for area of the rectangle!
A(rectangle) = 45
Done!!
Let the point of tangency of BC and the circle be point M. Then, applying the tangent secant theorem, (CM)² = (CF)(CE) = (4)(4 + 5) = 36 and CM = 6. Let the radius of the circle be r. Then, EB = MB = r. Apply the Pythagorean theorem to ΔBCE: (EB)² + (BC)² = (9)², r² + (r + 6)² = 81, 2r² + 12r - 45 = 0. The positive root, simplified, is r = √(31.5) - 3. AB = 2r = 2(√(31.5) - 3). BC = r + 6 = √(31.5) + 3. Area of ABCD = (AB)(BC) = 2(√(31.5) - 3)(√(31.5) + 3) = 2(31.5 - 9) = 2(22.5) = 45, as Math Booster also found.
Checking the comments, I find that RAG981 may have solved it the same way.
cos α = 5/2R = b/(5+4)
A = b.h= b.2R= 5.(5+4)
A = 45 cm² ( Solved √ )
I solved it the same way you did. It was time consuming.
Good thinking, nevertheless
Thanks for your comments! I think this is the more straightforward way to solve the problem, but the equations include 2 radicals. Luckily, we get a product of form (a - b)(a + b) = a² - b², so the square roots go away. Math Booster's solution avoids the radicals also, but, in my opinion, it takes a lot of creativity to find his solution!
Well you probably made some unintentional spelling mistakes while writing the comment, but that's exactly the method I used to do it, just use the tangent secant theorem to find length from C to bottom tangency point, then the rest of the rectangle length would be equal to the radius (can be proved by drawing some lines to complete a square and prove it's a square EONB but I won't bother getting into it here) as well as EB which is also r, and using pythagoras' theorem we can find r, multiply 2r by r+6 and we get the area
The area is 45 units square. Also I have noticed something that I should have thought about: delta EFP and delta EQC are similar because these triangles are NOT opposite to one another. This is different from one of the last two problems on this channel which showed HL congruence. I hope that this counts as a sufficient reason being that shows why and how similar triangles correspond to the area of the square.
Similarity of triangles:
cos α = 5/2R = b/(5+4)
A = b.h= b.2R= 5.(5+4)
A = 45 cm² ( Solved √ )
This was the easiest and shortest solition. I strongly recommend pinning ur comment
PERFECT 👍
@nexen1041
Thanks
Drop perpendicular from O onto chord EF. The little triangle formed is similar to triangle EBC. Let NC = x. Then (r+x)/9 = 2.5/r. This gives r^2 + rx =22.5 and 2r^2 + 2rx = 45. But this is the area since the rectangle has sides (r+x) and 2r.
АD=BC=a , AB=DC=2r . Area ADCD=2ar , theorem about tangent and secant lines coming from one point : (a-r)*2=CExCF , a*2+r*2-2ar=(5+4)4 (1) . From a triangle BCE according to the Pythagorean theorem - CE*2=r*2+a*2 , r*2+a*2=9*2 , substituting into eguation (1) - 9*2-2ar=9x4 , 2ar=81-36=45 . Area ABCD=45 .
A=AB*BC=2r*x. r^2+x^2=81. (x-r)^2=9*4=36. (x-r)^2=x^2 + r^2 - 2r*x = 81 -A = 36.
A = 45.
🎉🎉🎉 A mesma solução que eu encontrei 🎉🎉🎉
puissance du point C par rapport au cercle= 4(4+5)=36= OQ²
BC=6+r et BC² =EC²-r²=(6+r)²=9²-r²
2r²+12r+36=81
2r(r+6)=45= surface du rectangle
(5)^2 (4)^2rm={25+16}=41 {60°A60°B+60°C+60°D}=360°ABCD/41= 90ABCD 3^30 3^3^10 1^3^2^5 1^3^2^1 3^2 (ABCD ➖ 3ABCD+2)
NC=6... r^2+(r+6)^2=81... 2r^2+12r=45... S=2r(r+6)=2r^2+12r=45
Uau! Que questão bonita. Eu encontrei uma solução um pouco mais rápida que a sua. Parabéns pela escolha! 👏👏👏 Brasil 9 de novembro de 2024.
S(ABCD)=45 square units
2r:9=5:base
2r=height
Base *height =9*5=45
Exactly!!! Very simple, without complications !!
Congratulations !!
My solution :
cos α = 5/2R = b/(5+4)
A = b.h= b.2R= 5.(5+4)
A = 45 cm² ( Solved √ )
Exactly the same !!
@ thank you 🙏 Yes I know you like the shorter the better
45
r:9=x:4...x=FH=4r/9..(Hprooezione di F)...risulta arcsin(5r/9/5)=arccos(5/2r)..r^2=81/2-9√14..b(base)=√(81-(81/2-9√14))=√(81/2+9√14)..Arett=b*2r=2√(40,5^2-81*14)=2*22,5=45
Too clumsy,
4*(4+5)= tangent line sq, so tangent line =6. Let r is rad, so rectangle area =(6+r)*2r= 12r+ (2r) sq. Use Gougu theorem, r sq + (6+r) sq =81, expand, 12r+ (2r) sq -45=0. NO NEED TO SOLVE THE EQ, ang the 1st 2 terms is the area of rectangle, so =45
Is it given that the point E is touching the tangent?