Blessings has cherries
2 answers:
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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.
- In ΔABC
∴ The congruent sides are AB and AC respectively
- CB =12 ft. 4 in.=148 in.
- AB=
CB =111 in. =9 ft. 3 in.
- AC=
∵ <em>Perimeter of ΔABC</em> =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.
∵ <em>Perimeter of ΔDEF</em> =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
∵ <em>The dimensions of the ΔJKL are 39,39 and 17.</em>