∆ABC is an isosceles triangle. The length of CB is 12 feet 4 inches and the congruent sides are each 3/4 this length.

Question

∆ABC is an isosceles triangle. The length of CB is 12 feet 4 inches and the congruent sides are

each 3/4 this length.

2. What is the perimeter of ∆ABC?

a. 31 ft. 4 in.

b. 21 ft. 7 in.

c. 30 ft. 10 in.

d. 18 ft. 6 in.

3. In ∆DEF, DE and DF are each 6 feet 3 inches long. This length is 0.75 times the length of

FE. What is the perimeter of ∆DEF?

a. 12 ft. 4 in.

b. 17 ft. 2 in.

c. 14 ft. 7 in.

d. 20 ft. 10 in.

4. ∆JKL is an isosceles triangle with JL ≅ KL. If JK is three more than x, KL is 17 less than four

times x, and JL is 45 less than six times x, find x and the measure of each side.

e. 39, 39, 17

f. 17, 15, 17

g. 17,17,39

h. 42,42,42​

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Katherine 2 weeks 2021-10-14T03:24:29+00:00 1 Answer 0

Answers ( )

    0
    2021-10-14T03:26:19+00:00

    Answer:

    Answer for 2nd is option c, for 3rd is option d, for 4th is option e

    Step-by-step explanation:

    As we know 1 ft.=12 in.

    1. In ΔABC

       ∴ The congruent sides are AB and AC respectively

    • CB =12 ft. 4 in.=148 in.
    • AB=\frac{3}{4}CB =111 in. =9 ft. 3 in.
    • AC=\frac{3}{4}CB =111 in. =9 ft. 3 in.

        ∵  Perimeter of ΔABC =AB+AC+CB

                                            =9 ft. 3 in. + 9 ft. 3 in. +12 ft. 4 in.

                                            =30 ft. 10 in.

        2. In ΔDEF

        ∴ The congruent sides are DE and DF respectively

    • DE =  6 ft. 3 in. =75 in.
    • DF =  6 ft. 3 in. =75 in.
    • Let the length of FE is equal to x
    • 0.75FE =DE =DF
    • 0.75x = 6 ft. 3 in. =75 in.
    • x =100 in. =8 ft. 4 in.

       ∵ Perimeter of ΔDEF =DE+DF+FE

                                            = 6 ft. 3 in. +6 ft. 3 in. +8 ft. 4 in.

                                            = 20 ft. 10 in.

        3. In ΔJKL

        ∴ The congruent sides are JL and KL respectively

    • JK = x+3
    • KL =4x-17
    • JL  =6x-45
    • JL≅KL
    • 4x-17 =6x-45  . . . . . . . . . . . . . . . . . . . . . . . (1)
    • Subracting 4x from both sides from eq 1
    • -17 =2x-45
    • Adding 45 on both the sides
    • 28 =2x
    • Dividing by 2 on both sides
    • 14 =x
    • JK = 14+3 =17
    • KL = 4×14-17 =39
    • JL = 6×14-45 =39

       ∵ The dimensions of the ΔJKL are 39,39 and 17.

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