Hexagons
There is a square ABCD, a square EFGD, and a rectangle HIJD. Points J and G lie on side CD, with DJ less than DG, and points H and E lie on side DA, with DH less than DE. We also know that DJ equals GC. Hexagon ABCGFE has a perimeter of 96 cm, hexagon EFGJIH has a perimeter of 60 cm, and rectangle HIJD has a perimeter of 28 cm. Determine the area of hexagon EFGJIH.
Final Answer:
You need to know the following knowledge to solve this word math problem:
planimetricsbasic operations and conceptsthemes, topicsGrade of the word problem
We encourage you to watch this tutorial video on this math problem: video1
Related math problems and questions:
- Hexagon - MO
The picture shows the ABCD square, the EFGD square, and the HIJD rectangle. Points J and G lie on the side CD and are true |DJ| - The rectangle 5
The rectangle OABC has one vertex at O, the center of a circle, and a second vertex, A, 2 cm from the edge of the circle, as shown. The vertex A is also 7 cm from C. The points B and C lie on the circumference of the circle. a. What is the radius? b. Find - Quadrilateral calc
The square ABCD is given. The midpoint of AB is E, the midpoint of BC is F, CD is G, and the midpoint of DA is H. Join AF, BG, CH, and DE. Inside the square (approximately in the middle), the intersections of these line segments form a quadrilateral. Calc - Circumscribed by triangle
Inside the rectangle ABCD, the points E and F lie so that the line segments EA, ED, EF, FB, and FC are congruent. Side AB is 22 cm long, and the circle circumscribed by triangle AFD has a radius of 10 cm. Determine the length of side BC. - Quadrilateral perimeter angles
Quadrilateral ABCD has side lengths AB=13 cm, CD=3 cm, AD=4 cm. Angles ACB and ADC are right angles. Calculate the perimeter of quadrilateral ABCD. - Trapezoid thirds
The ABCD trapezoid has parallel sides AB and CD. The E point lies on the AB side. The segment DE divides the trapezoid into two parts with the same area. Find the length of the AE line segment. - Trapezoid angles
Given an isosceles trapezoid ABCD, in which | AB | = 2 | BC | = 2 | CD | = 2 | DA | holds. On its side BC, the point K is such that | BK | = 2 | KC |; on its CD side, the point L is such that | CL | = 2 | LD |, and on its DA side, the point M is such that
