Question

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

Answer 1

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.