Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given parabola is

[tex]\sf \: y = - {x}^{2} + 4x + c \\ \\ [/tex]

can be rewritten as

[tex]\sf \: y = - ( {x}^{2} - 4x) + c \\ \\ [/tex]

[tex]\sf \: y = - ( {x}^{2} - 4x + 4 - 4) + c \\ \\ [/tex]

[tex]\sf \: y = - ( {x}^{2} - 4x + 4) + 4 + c \\ \\ [/tex]

[tex]\sf \: y = - {(x - 2)}^{2} + 4 + c \\ \\ [/tex]

[tex]\sf \: y - (4 + c)= - {(x - 2)}^{2} - - - (1) \\ \\ [/tex]

So,

[tex]\sf \:  \implies \: Vertex \: of \: parabola \: = \: (2, \: 4 + c) \\ \\ [/tex]

As it is given that Point B is also the maximum point of the parabola (vertex).

So, Coordinates of B = (2, 4 + c).

[tex]\sf \:  \implies \: AB = 4 + c \: units \\ \\ [/tex]

Now, We have to find the point of intersection with x- axis of the given parabola.

On x - axis, y = 0

So, Substitute this value in equation (1), we get

[tex]\sf \: 0 - (4 + c)= - {(x - 2)}^{2} \\ \\ [/tex]

[tex]\sf \: 4 + c= {(x - 2)}^{2} \\ \\ [/tex]

[tex]\sf \: x - 2 = \: \pm \: \sqrt{4 + c} \\ \\ [/tex]

[tex]\sf \:\sf \:  \implies \: x = 2\: \pm \: \sqrt{4 + c} \\ \\ [/tex]

So,

[tex]\sf \:  \implies \: Coordinates \: of \: C = (2 + \sqrt{4 + c}, \: 0) \\ \\ [/tex]

[tex]\sf \:  \implies \: OC = 2 + \sqrt{4 + c} \: units \\ \\ [/tex]

So,

[tex]\sf \: AC=OC-OA \\ \\ [/tex]

[tex]\sf \: AC=2 + \sqrt{4 + c} -2\\ \\ [/tex]

[tex]\sf \: \sf \:  \implies \: AC=\sqrt{4 + c} \: units\\ \\ [/tex]

Now, Further given that

[tex]\sf \: Area_{(\triangle ABC)} \: = \: 32 \\ \\ [/tex]

[tex]\sf \: \frac{1}{2} \times AB \times AC \: = \: 32 \\ \\ [/tex]

[tex]\sf \: (4 + c) \times \sqrt{4 + c} \: = \: 64 \\ \\ [/tex]

[tex]\sf \: (4 + c)^{ \frac{3}{2} } \: = \: {4}^{3} \\ \\ [/tex]

[tex]\sf \: \sqrt{4 + c} = 4 \\ \\ [/tex]

[tex]\sf \: 4 + c = 16 \\ \\ [/tex]

[tex]\sf \:  \implies \: c = 12 \\ \\ [/tex]

Answer:

hey I like your best answer ♡

you are so good brain I am your follower please help me when I need and you also deserve brainliest mark