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Mathematics 15 Online
OpenStudy (sanra123):

can any one help plz

OpenStudy (sanra123):

ganeshie8 (ganeshie8):

for part a : find the area of square find the area of triangle add them both

ganeshie8 (ganeshie8):

area of square = (4x-4)^2

ganeshie8 (ganeshie8):

area of triangle = 1/2 (4x-4)(2x-2)

ganeshie8 (ganeshie8):

Area of pentagon = (4x-4)^2 + 1/2(4x-4)(2x-2)

ganeshie8 (ganeshie8):

simplify

OpenStudy (sanra123):

ok..why do we add the 1/2

ganeshie8 (ganeshie8):

good question :)

ganeshie8 (ganeshie8):

maybe first factor out 4

ganeshie8 (ganeshie8):

\((4x-4)^2 + \frac{1}{2}(4x-4)(2x-2)\) \(4^2(x-1)^2 + \frac{1}{2}4(x-1)2(x-1)\) \(16(x-1)^2 + 4(x-1)(x-1)\) \(16(x-1)^2 + 4(x-1)^2\) \(20(x-1)^2 \)

ganeshie8 (ganeshie8):

thats the final simplified expression

ganeshie8 (ganeshie8):

see if that looks okay

ganeshie8 (ganeshie8):

next, think about part b

OpenStudy (sanra123):

oh thanks now i knew why 1/2

ganeshie8 (ganeshie8):

good :)

OpenStudy (sanra123):

we start multiplying the answer and see if we get the expression

OpenStudy (sanra123):

@ganeshie8

OpenStudy (sanra123):

area of triangle = 1/2 (4x-4)(2x-2) hi sorry but how did u to this for the triangle ?

ganeshie8 (ganeshie8):

area of triangle = 1/2(b)(h)

OpenStudy (sanra123):

isn't the answer have to be (16+4) (x-1)

OpenStudy (sanra123):

@ganeshie8

ganeshie8 (ganeshie8):

how ?

ganeshie8 (ganeshie8):

we're done wid part a right ?

OpenStudy (sanra123):

by grouping

ganeshie8 (ganeshie8):

it shoudl be (16+4)(x-1)^2 okay ?

ganeshie8 (ganeshie8):

when u add u wud get : 20(x-1)^2

ganeshie8 (ganeshie8):

thats for part a

OpenStudy (sanra123):

oh ok...

OpenStudy (sanra123):

ya now part be we will be checking how is it the right answer

OpenStudy (sanra123):

b

ganeshie8 (ganeshie8):

for part b, u may do below :- plugin x = 2, and see wat u get for area using the simplified formula : 20(x-1)^2

ganeshie8 (ganeshie8):

next, when x = 2, find out area of square and triangle separately. and add them.

ganeshie8 (ganeshie8):

u should get the same answer by using both mehtods..

ganeshie8 (ganeshie8):

Area of pentagon using simplified formula, when x = 2 :- A = 20(2-1)^2 = 20

OpenStudy (sanra123):

why =20

ganeshie8 (ganeshie8):

Area of square = (4*2-4)(4*2-4) = (4)(4) = 16 Area of triangle = 1/2(4*2-4)(2*2-2) = 1/2(4)(2) = 4 adding them gives : 16+4 = 20 so we just verified that the formula is correct for x = 2

OpenStudy (sanra123):

how did u solve for square

ganeshie8 (ganeshie8):

Area of square = (4x-4)(4x-4)

ganeshie8 (ganeshie8):

plugin x = 2 above

OpenStudy (sanra123):

4(2)-4 4(2)-4

OpenStudy (sanra123):

then here how do we get 16

ganeshie8 (ganeshie8):

whats 4(2) - 4 ?

OpenStudy (sanra123):

4

ganeshie8 (ganeshie8):

yes

OpenStudy (sanra123):

then 4*4 =16

ganeshie8 (ganeshie8):

yup

OpenStudy (sanra123):

then with the triangle im confused in solving it

ganeshie8 (ganeshie8):

write down Area of triangle formula first

OpenStudy (sanra123):

1/2(4*2-4)(2*2-2)

ganeshie8 (ganeshie8):

Area of triangle = 1/2(4x-4)(2x-2)

ganeshie8 (ganeshie8):

plugin x = 2

OpenStudy (sanra123):

1/2 4(2)-4 2 (2)-2

OpenStudy (sanra123):

@ganeshie8

ganeshie8 (ganeshie8):

yes, simplify it

ganeshie8 (ganeshie8):

1/2 [4(2)-4 ] [2 (2)-2]

OpenStudy (sanra123):

4 +2

ganeshie8 (ganeshie8):

nope, 1/2 [4(2)-4 ] [2 (2)-2] first u need to work the thing inside parenthesis

ganeshie8 (ganeshie8):

1/2 [4(2)-4 ] [2 (2)-2] ^^^^^ ^^^^^

OpenStudy (sanra123):

oh ok

ganeshie8 (ganeshie8):

1/2[8 - 4] [4 - 2]

ganeshie8 (ganeshie8):

1/2[4] [2]

ganeshie8 (ganeshie8):

4

OpenStudy (sanra123):

ohh ok

OpenStudy (sanra123):

lets do c

OpenStudy (sanra123):

@ganeshie8

ganeshie8 (ganeshie8):

for part c, start by writing out the area of triangle formula again

ganeshie8 (ganeshie8):

Area of triangle = \(\large \frac{1}{2}(4x-4)(2x-2)\)

ganeshie8 (ganeshie8):

right ?

OpenStudy (sanra123):

yes

ganeshie8 (ganeshie8):

\(\large \frac{1}{2}(4x-4)(2x-2)\) \(\large \frac{1}{2}4(x-1)2(x-1)\) \(\large 4(x-1)^2\)

ganeshie8 (ganeshie8):

right ?

OpenStudy (sanra123):

yes

ganeshie8 (ganeshie8):

\(\large 4(x-1)^2\) take 4 inside the square, it becoems : \(\large (2(x-1))^2\) \(\large (2x-2)^2\)

ganeshie8 (ganeshie8):

clearly thats the area of a square of side (2x-2)

ganeshie8 (ganeshie8):

so, for part c, the required area of square is : \(\large (2x-2)^2\)

OpenStudy (sanra123):

then d

ganeshie8 (ganeshie8):

expand the square

ganeshie8 (ganeshie8):

u knw below identity (a + b)^2 = a^2 + 2ab + b^2 right ?

ganeshie8 (ganeshie8):

use it

ganeshie8 (ganeshie8):

\((2x-2)^2 = ?\)

OpenStudy (sanra123):

yes

OpenStudy (sanra123):

(2x+2)(2x-2)

OpenStudy (sanra123):

?

ganeshie8 (ganeshie8):

nope, use the identity...

ganeshie8 (ganeshie8):

\((2x-2)^2 = (2x)^2 - 2(2x)(2) + 2^2\)

ganeshie8 (ganeshie8):

\( = 4x^2 - 8x + 4\)

ganeshie8 (ganeshie8):

what type of polynomial it is ?

OpenStudy (sanra123):

what is it?

ganeshie8 (ganeshie8):

how many terms it has

ganeshie8 (ganeshie8):

?

ganeshie8 (ganeshie8):

\(4x^2-8x+4\)

ganeshie8 (ganeshie8):

3 terms right ?

ganeshie8 (ganeshie8):

so its called a "trinomial"

OpenStudy (sanra123):

i really didnt understand how did u expand that

ganeshie8 (ganeshie8):

do u knw below formula ? \((a+b)^2 = a^2 + 2ab + b^2\)

OpenStudy (sanra123):

yes

OpenStudy (sanra123):

oh k

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