There is a paper of rectangle as shown in a figure.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
Solution
A figure is a figure of the rectangle is being folded so that the vertex A and C may overlap.
A fold is set to EF and the intersection of EF and AC is set to G.
The triangles with which paper has not overlapped are △ABE and △ADF.
Since △C(D) F was turned up to be △ADF, find the area of △ABE and △C(D) F.
The area is calculated as (rectangular area) - (area of a quadrangle AECF).
As for a quadrangle AECF, AF and EC, AE and FC are parallel and since the diagonal line crosses vertically, it is a rhombus.
The area is determined by EC × AB.
△GEC and △ABC are homothetic and BC : AC = GC : EC = 20 : 25 = 4 : 5.
EC = GC × 5/4 = 25/2 × 5/4 = 125/8 =15.625cm.
The length of vertical side, horizontal side and diagonal line are 15 cm, 20 cm and 25 cm respectively.
This paper is folded so that A and C may overlap.
Find the sum total of a triangular area with which paper has not overlapped in this case.
Answer
65.625 cm2
A figure is a figure of the rectangle is being folded so that the vertex A and C may overlap.
A fold is set to EF and the intersection of EF and AC is set to G.
The triangles with which paper has not overlapped are △ABE and △ADF.
Since △C(D) F was turned up to be △ADF, find the area of △ABE and △C(D) F.
The area is calculated as (rectangular area) - (area of a quadrangle AECF).
As for a quadrangle AECF, AF and EC, AE and FC are parallel and since the diagonal line crosses vertically, it is a rhombus.
The area is determined by EC × AB.
△GEC and △ABC are homothetic and BC : AC = GC : EC = 20 : 25 = 4 : 5.
EC = GC × 5/4 = 25/2 × 5/4 = 125/8 =15.625cm.
Therefore, as for the area of rhombus AECF, it is 15.625 × 15 = 234.375cm2.
The area of rectangles is 15 × 20 = 300cm2.
The area to find is 300 - 234.375 = 65.625cm2.
