Can you use the asa postulate or the aas theorem to prove the triangles congruent? by aas only neither apply by asa only both apply

(C) by ASA only

Step-by-step explanation:

Let us name the given figure, there are two triangles that areΔABO and ΔCOD.

Thus, from ΔABO and ΔCOD, we have

∠ABO=∠DCO(Given in the question)

BO=DO( given in question)

∠AOB=∠COD(Vertically opposite angles)

Therefore, by ASA rule,

ΔABO ≅ΔCOD

Therefore, only ASA rule can be applied on the given figure.

Hence, option C is correct.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

(2x + 16) + (x) = 180

step-by-step explanation:

given,

abcd is a quadrilateral inscribed in a circle

angle a is labeled x plus 16, (x + 16)

angle b is labeled x degrees, (x)

angle c is labeled 6x minus 4, (6x – 4)

angle d is labeled 2x plus 16, (2x + 16)

what equation can be used to solve for angle b?

we know that the sum of the opposite angles of a simple quadrilateral inscribed in a circle is 180 degrees.

angle b is opposite angle d

therefore, angle b + angle d = 180

(x) + (2x + 16) = 180 

same as (2x + 16) + (x) = 180