. Describe a way you can use graham crackers to demonstrate the division problem . Include the result in your description.
1 answer:
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See proof below
Step-by-step explanation:
Assume triangle ABC to have vertices at;
A(2,-1), B(2,-7) and C(6,-7)
D is midpoint of BC, thus D is at (4,-7)
The P and Q, are lying on side AB and AC, hence assume P is at (2,-4) and Q is at (4,-4) such at DP is parallel to QA
Plot the points on a graph tool and join the points to view the sketch.
To prove area of triangle CPQ is 1/4 area of ABC will be;
Find area ABC and CPQ then compare the areas.
Apply the distance formula to find the length of sides of the triangles then find the areas.
The distance formula is;
Length of side AB from the sketch is;
Length of side BC will be;
Thus area of triangle ABC will be;
1/2 *base length*height ------because it is a right-triangle
1/2*4*6=12 square units
Find the lengths of all sides of triangle CPQ
Length of side PQ is half that of side BC thus PQ=2 units
Length of side PC is;
Length of side QC will be;
QC= √13 = 3.6 units
Find area of triangle CPQ given all sides by applying the Heron's formula for area of triangle which is;
A=√s(s-a)(s-b)(s-c) where;
A=area of the triangle
s= half the perimeter of the triangle
a=side PQ = 2 units
b=side PC = 5 units
c= side QC = 3.6 units
Finding the perimeter of triangle CPQ will be;
P=sum of all sides
P=2+5+3.6 =10.6 units
s=10.6/2 = 5.3
Area of the triangle CPQ will be;
A=3.0 (1 decimal place)
Compare the areas;
Area of triangle ABC=12 square units
Area of triangle CPQ = 3 square units
Area of triangle CPQ / Area of triangle ABC = 3/12 =1/4
Thus you have proved that area of triangle CPQ is 1/4 th area of triangle ABC because 1/4 *12 =3
Learn More
Area of a triangle ;brainly.com/question/14869984
The Heron's formula : brainly.com/question/10713495
Keywords: midpoint, triangle, sides, parallel, prove , area, equal
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