The perimeter of triangle ABC is 138 cm and BC || DE. Is triangle ABC = triangle ADE? Which conjecture supports your conclusion? please help me it’s due in 30 minutes!!

Answer

yes triangle ABC is equal to triangle ADE

Step-by-step explanation:

x=46 cm, I got this after simplifying the equation of triangle ABC, x+4+x-4+x=138 cm, so this equation simplifies to 3x as +4-4 cancels out and there're 3 x's, so divide 138 by 3 and you get 46 cm, so AB is 50 cm, BC is 42 cm, and AC is 46 cm, this means EA is equal to EC, now here come a property of equality, EC attaches 2 parallel lines, as does DB, this means these 2 lines are transversals, so a number of angles are equal as well, corresponding angles are congruent, a.k.a equal to each other along with alternate interior angles, so since line DB AND line EC are transversals, we can deduce angle ADE is congruent to angle ABC, and angle DEA is equal to ACB, this mean we have 2 congruent angles, and 1 congruent set of sides, this mean that the 2 triangles are congruent due to Angle Angle Side, if your teacher doesn't use these terms, look it up, but the answer in short is yes, they are congruent or equal

the smallest possible measurement of the width is 14 feet

step-by-step explanation:

let's assume length of parallelogram is 'l'

width of parallelogram is 'w'

the perimeter of a parallelogram must be no less than 40 feet

so, we get

we are given

length of parallelogram is 6 feet

so,

l=6

we can plug it

now, we can solve for w

subtract both sides by 12

divide both sides by 2

and we get

so,

the smallest possible measurement of the width is 14 feet