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The greatest number of packages of baked items that can be sold is 10
<em><u>Solution:</u></em>
Given that there are 50 cupcakes and 160 cookies to be sold at the school bake sale
To find: greatest number of packages of baked items that can be sold
Each package has the same number of cupcakes and same number of cookies with none left over
So, we have to find the greatest common factor of 50 and 160
The greatest number that is a factor of two (or more) other numbers.
When we find all the factors of two or more numbers, and some factors are the same ("common"), then the largest of those common factors is the Greatest Common Factor.
<em><u>Greatest common factor of 50 and 160:</u></em>
The factors of 50 are: 1, 2, 5, 10, 25, 50
The factors of 160 are: 1, 2, 4, 5, 8, 10, 16, 20, 32, 40, 80, 160
Then the greatest common factor is 10
So the greatest number of packages sold is 10
<em><u>Each package will contain:</u></em>
Given that there 50 cupcakes and 160 cookies
<em><u>Number of cupcakes in 1 package:</u></em>
<u><em>Number of cookies in 1 package:</em></u>
So each package has the same number of cupcakes and same number of cookies with none left over
All the theorems are proved as follows.
<h3>What is a Triangle ?</h3>A triangle is a polygon with three sides , three vertices and three angles.
1. The Triangle sum Theorem
According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.
To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.
We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.
180° - ∠A + 180° - ∠B + 180° - ∠C = 360°
- ∠A - ∠B - ∠C = 360° - (180°+ 180°+ 180°) = -180°
∠A + ∠B + ∠C = 180°
(Hence Proved)
2. Isosceles Triangle Theorem
Considering an isosceles triangle ΔABC
with AB = AC, we have by sine rule;
as AB = AC
sin B = sin C
angle B = angle C
3.Converse of the Isosceles theorem
Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;
as ∠B= ∠C ,
AB = AC
4. Midsegment of a triangle theorem
It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.
Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC
In ΔABC and ΔADF
∠A ≅ ∠A
BA = 2 × DA, BC = 2 × FA
Hence;
ΔABC ~ ΔADF (SAS similarity)
BA/DA = BC/FA = DF/AC = 2
Hence AC = 2×DF
5.Concurrency of Medians Theorem
A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and the point of intersection is inside the triangle known as Centroid .
Consider a triangle ABC , X,Y and Z are the midpoints of the sides
Since the medians bisect the segment AB into AZ + ZB
BC into BX + XB
AC into AY + YC
Where:
AZ = ZB
BX = XB
AY = YC
AZ/ZB = BX/XB = AY/YC = 1
AZ/ZB × BX/XB × AY/YC = 1 and
the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).
To know more about Triangle
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