Find the side length of CB and side AB

Find the side length of CB and side AB


[tex]Find the side length of CB and side AB[/tex]

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  1. A) The length of the mid-segment  is 6.25 cm

    B) The length of AT = 33 units

    C) The value of x is 3

    Step-by-step explanation:

    A) Find the length of the mid segment of an equilateral triangle with side lengths of 12.5 cm

    Solution:

    A mid segment of a triangle is a segment connecting the midpoints of two sides of a triangle. This segment has two special properties. It is always parallel to the third side, and the length of the mid segment is half the length of the third side.

    This is an equilateral triangle with side lengths 12.5cm

    The length of the mid segment = 1/2 the length of the third side.

    The length of the mid segment = 1/2 * 12.5

    The length of the mid segment = 6.25 cm

    B) Given that UT is the perpendicular bisector of AB, where T is on AB, find the length of AT given AT = 3x +6 and TB = 42 - x

    Solution:

    UT is the perpendicular bisector of AB

    T lies on AB

    AT = BT

    3x+6 = 42-x

    Combine the like terms:

    3x+x=42-6

    4x= 36

    Divide both sides by 4

    4x/4 = 36/4

    x= 9

    Now plug the value of x in AT= 3x+6.

    =3(9)+6

    =27+6

    =33

    The length of AT = 33 units.

    C) Given angle ABC has angle bisector for BD, where AB = CB, find the value of x if AD = 5x + 10 and DC = 28 - x.

    Solution:

    In Δ ABC

    AB = BC

    Δ ABC is an isosceles triangle

    BD bisects angle ABC

    AC is the opposite side of the vertex B

    BD bisects the side AC at D

    AD=CD

    AD=5x+10

    CD= 28-x

    Equate both the equations:

    5x+10 = 28-x

    Combine the like terms:

    5x+x=28-10

    6x=18

    Divide both sides by 6

    6x/6 = 18/6

    x= 3

    Thus the value of x is 3

  2. a) The length of the mid-segment  is 6.25 cm

    b) The length of AT = 33 units

    c) The value of x is 3

    Step-by-step explanation:

    a)

    * Lets explain the mid-segment of a triangle

    - A mid-segment of a triangle is a segment connecting the midpoints

     of two sides of a triangle

    - This segment has two special properties

    # It is parallel to the third side

    # The length of the mid-segment is half the length of the third side

    ∵ The triangle is equilateral triangle

    ∴ All sides are equal in length

    ∵ the side lengths = 12.5 cm

    ∵ The length of the mid-segment = 1/2 the length of the third side

    ∴ The length of the mid-segment = 1/2 × 12.5 = 6.25 cm

    * The length of the mid-segment  is 6.25 cm

    b)

    ∵ UT is a perpendicular bisector of AB

    ∵ T lies on AB

    ∴ T is the mid-point of AB

    ∵ AT = BT

    ∵ AT = 3x + 6

    ∵ BT = 42 - x

    - Equate AT and BT

    ∴ 3x + 6 = 42 - x

    - Add x to both sides

    ∴ 4x + 6 = 42

    - Subtract 6 from both sides

    ∴ 4x = 36

    - Divide both sides by 4

    ∴ x = 9

    ∵ AT = 3x + 6

    - Substitute x by 9

    ∴ AT = 3(9) + 6 = 27 + 6 = 33

    * The length of AT = 33 units

    c)

    - In Δ ABC

    ∵ AB = BC

    ∴ Δ ABC is an isosceles triangle

    ∵ BD bisects angle ABC

    - In the isosceles Δ the bisector of the vertex angle bisects the base

     of the triangle which is opposite to the vertex angle

    ∵ AC is the opposite side of the vertex B

    ∴ BD bisects the side AC at D

    ∴ AD = CD

    ∵ AD = 5x + 10

    ∵ CD = 28 - x

    ∴ 5x + 10 = 28 - x

    - Add x to both sides

    ∴ 6x + 10 = 28

    - Subtract 10 from both sides

    ∴ 6x = 18

    - Divide both sides by 6

    ∴ x = 3

    * The value of x is 3

  3. 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=[tex]\frac{3}{4}[/tex]CB =111 in. =9 ft. 3 in.AC=[tex]\frac{3}{4}[/tex]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 x0.75FE =DE =DF0.75x = 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-45JL≅KL4x-17 =6x-45  . . . . . . . . . . . . . . . . . . . . . . . (1)Subracting 4x from both sides from eq 1-17 =2x-45Adding 45 on both the sides 28 =2xDividing by 2 on both sides14 =xJK = 14+3 =17KL = 4×14-17 =39JL = 6×14-45 =39

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

  4. length=19

    Step-by-step explanation:

    top is length and right side is breadth.

    perimeter = 2(l+b)

    74 = 2[(4z-1)+(3z+3)]

    74/2 = 4z-1+3z+3

    37 = 7z+2

    7z= 37-2

    7z = 35

    z = 35/7

    z = 5

    .: length = top = 4z - 1= 4(5) - 1

    = 20-1=19

    .: length= 19

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