Problem of the Week Problem C
Locate the Fourth Vertex
Quadrilateral BDFH is constructed so that each vertex is on a different side of
square ACEG. Vertex B is on side AC so that AB = 4 cm and BC = 6 cm.
Vertex F is on EG so that EF = 3 cm and FG = 7 cm. Vertex H is on GA so
that GH = 4 cm and HA= 6 cm. The area of quadrilateral BDFH is 47 cmº.
The fourth vertex of quadrilateral BDFH, labelled D, is located on side CE so
that the lengths of CD and DE are both positive integers.
Determine the lengths of CD and DE.

CD is 6 cm and DE is 4 cm.

Step-by-step explanation:

Well it‘s the correct answer and I got it right so good luck out there bois!

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

i need the options

step-by-step explanation: