Answer is 45 It forms a pentagon with Angles A=C=E 120 degree D=B=90 degree AB=AC=CE=ED= r (radius) BD= r. sqrt(3) All 5 Triangles formed are congruent ABC , COB, DOB,COD ,ECD So 9×5=45 I mostly found out measure of arcs & with them the angles & figured it out!
hello! my answer is 45: ABOC is clearly a rhombus, since AB=AC=BO=CO, so diagonals are perpendicular. So BAC is congruent to BOC, with < BAC= < BCA = 2 * ABO = 2 *60 = 120 since ABO and ACO are equalateral triangle, and
After constructing the diagram, it is clear that the area of the polygon can be (inelegantly) described as the sum of the areas of ∆ABC, ∆BOC, ∆BOD, ∆COD, and ∆CED. AB = BO => AB = radius (r). CO = r = BO, therefore CO = AB = AC. If B and C are equal distances from A (by the definition of an isosceles triangle), it follows that they would also be equal distances from the point D on the opposite side of the circle from A (BD = CD). From this, it is clear that ∆BOD is congruent to ∆COD, by SSS congruence (BD = CD, OD = OD, and BO = CO). Angle BAC subtends the arc BDC, and since B and C are equal distances from D, this implies that the arc BDC = 2 * the arc BD, which is contained by angle BOD. From this, and by the definition of inscribed angles, it is clear that
Hello, The answer is 45 square units. First I constructed the diagram. I realised that the area of ABDEC can be composed as the sum of 5 smaller triangles based around O; ∆AOB, ∆BOD, ∆DOE, ∆EOC and ∆COE. Because AB=BO we can easily derive that all triangles are congruent and equilateral except ∆BOD. Next, we see that ∆BOD is congruent to ∆COD which has an area of 9. Since ∆DOE and ∆EOC are congruent and equailateral when the rhombus is split into two by line DC it creates two equal areas. Since one of these areas is ∆COD which has area 9 the entire rhombus has area 18. If we instead split the rhombus with line OE one of the slices must have area 9 because half the rhombus has area 9. Since half of the rhombus is either ∆DOE or ∆EOC then all the other congruent triangles must have the same area. Since all 5 triangles have an area of 9 the total area must be 5x9=45 square units.
the answer is 45 because AB=AO=AC=OC=OD and there are infact 5 congruent triangles (ABC, BOC, BOD, COD and CED) with the area of 9 so just multiply by 5 and you get 45 units ^2.
Hi, I am from Brazil and your channel made me want to learn more about mathematics but I am only at highschool level so you have any tips at where to start?
Hi, LoLMistic! I love your enthusiasm for mathematics, and I am glad my channel helped you foster some of it. Perhaps the best resource I know for studying mathematics (and various other subjects) is www.khanacademy.org/. They have thousands of videos (all free!) covering mathematical topics from basic arithmetic to multivariable calculus. The teacher, Salman Khan, is certainly one of the best teachers I have known in my life. I hope you continue to explore the beautiful realm of mathematics! =)
Answer is 45
It forms a pentagon
with Angles
A=C=E 120 degree
D=B=90 degree
AB=AC=CE=ED= r (radius)
BD= r. sqrt(3)
All 5 Triangles formed are congruent
ABC , COB, DOB,COD ,ECD
So 9×5=45
I mostly found out measure of arcs & with them the angles & figured it out!
Btw I love your videos !
I get to learn all sorts of new techniques so thanks allot!
hello! my answer is 45:
ABOC is clearly a rhombus, since AB=AC=BO=CO, so diagonals are perpendicular. So BAC is congruent to BOC, with < BAC= < BCA = 2 * ABO = 2 *60 = 120 since ABO and ACO are equalateral triangle, and
After constructing the diagram, it is clear that the area of the polygon can be (inelegantly) described as the sum of the areas of ∆ABC, ∆BOC, ∆BOD, ∆COD, and ∆CED. AB = BO => AB = radius (r). CO = r = BO, therefore CO = AB = AC. If B and C are equal distances from A (by the definition of an isosceles triangle), it follows that they would also be equal distances from the point D on the opposite side of the circle from A (BD = CD). From this, it is clear that ∆BOD is congruent to ∆COD, by SSS congruence (BD = CD, OD = OD, and BO = CO). Angle BAC subtends the arc BDC, and since B and C are equal distances from D, this implies that the arc BDC = 2 * the arc BD, which is contained by angle BOD. From this, and by the definition of inscribed angles, it is clear that
Hello,
The answer is 45 square units.
First I constructed the diagram. I realised that the area of ABDEC can be composed as the sum of 5 smaller triangles based around O; ∆AOB, ∆BOD, ∆DOE, ∆EOC and ∆COE. Because AB=BO we can easily derive that all triangles are congruent and equilateral except ∆BOD. Next, we see that ∆BOD is congruent to ∆COD which has an area of 9. Since ∆DOE and ∆EOC are congruent and equailateral when the rhombus is split into two by line DC it creates two equal areas. Since one of these areas is ∆COD which has area 9 the entire rhombus has area 18. If we instead split the rhombus with line OE one of the slices must have area 9 because half the rhombus has area 9. Since half of the rhombus is either ∆DOE or ∆EOC then all the other congruent triangles must have the same area. Since all 5 triangles have an area of 9 the total area must be 5x9=45 square units.
Using five congruent triangles with the area of nine, the answer is 9*5=45.
BO has to be outside of triangle ABC and therefore form rhombus ABOC where all 4 sides have the equal length of the radius.
45 finally an easy one
5 isosceles triangles resulting 9*5=45
Answer is 45. There are 5 congruent triangles and since one of them has area 9 so total area of figure is 45.
Easy geometry...
45 sq units
the answer is 45 because AB=AO=AC=OC=OD and there are infact 5 congruent triangles (ABC, BOC, BOD, COD and CED) with the area of 9 so just multiply by 5 and you get 45 units ^2.
45.
I'm not sure how I'm supposed to answer it but I got 45
It's 45 it was easier this time.
Hi, I am from Brazil and your channel made me want to learn more about mathematics but I am only at highschool level so you have any tips at where to start?
Hi, LoLMistic! I love your enthusiasm for mathematics, and I am glad my channel helped you foster some of it. Perhaps the best resource I know for studying mathematics (and various other subjects) is www.khanacademy.org/. They have thousands of videos (all free!) covering mathematical topics from basic arithmetic to multivariable calculus. The teacher, Salman Khan, is certainly one of the best teachers I have known in my life. I hope you continue to explore the beautiful realm of mathematics! =)
LetsSolveMathProblems does this site also have Physics videos?
Chaitanya Paranjape yup