.find the area of the blue portion.

ifada123:
.find the area of the blue portion.

The area of the blue portion is Agege because that is the only area I know where structures can be jam-packed like this.


*spits

LuciferYaweh:


The area of the blue portion is Agege because that is the only area I know where structures can be jam-packed like this.


*spits

What ?
My friend if u can't do it, wait for others to do it, before u are speaking a language I can't understand

ifada123:
.find the area of the blue portion.

myanswer is 2489 sq. unit

dejt4u:
answer is 3257 sq. unit

No sir try again sir

dejt4u:
answer is 3257 sq. unit

Show your workings let me detect your error sir

loading

dejt4u:
loading

Your error

The breadth is not the sum of diamter of one big circle and one small circle,
It is the sum of one small circle and and unknown value which you have to calculate for.
Try it out

ifada123:

Your error

The breadth is not the sum of diamter of one big circle and one small circle,
It is the sum of one small circle and and unknown value which you have to calculate for.
Try it out

yea.. I noticed that too. my final answer is 2489 sq. unit

ifada123:

Your error

The breadth is not the sum of diamter of one big circle and one small circle,
It is the sum of one small circle and and unknown value which you have to calculate for.
Try it out

yea.. I noticed that too. my final answer is 2489 sq. unit

The area of the blue region A_B is :
A_B = (768)[ (7 + 30√2)^½ + 2 ] - 1408π ≈ 2511.9783 square unit (4 d.p)

Confirm. I can show my working if you want.
Cc. ifada123

dejt4u:

yea.. I noticed that too. my final answer is 2489 sq. unit

Correct √√√√√√√√√√ (10/10)

You can show your workings for others to learn from it

Martinez39:
The area of the blue region A_B is :
A_B = (768)[ (7 + 30√2)^½ + 2 ] - 1408π ≈ 2511.9783 square unit (4 d.p)

Confirm. I can show my working if you want.
Cc. ifada123

Show workings sir. Is like you approximated too early
But very close, show workings let me detect your error sir.

MY WORKINGS.
Lengh=4xR=4×24=96

Breadth=√(16+24)^2-24^2) +16 +24=32+16+24=72

Area of rectangle =72×96=6912

Area of blue =6192-2×22/7 24^2-22/7 x16^2

A=2488.64unit^2

ifada123:

Correct √√√√√√√√√√ (10/10)

You can show your workings for others to learn from it

I've attached the image to my first comment

ifada123:

Show workings sir. Is like you approximated too early
But very close, show workings let me detect your error sir.

Okay.

1) The area of the blue region is the area of the overall rectangle minus the sum of the areas of the circle. The sum of the areas, A_c, of the circle is :
A_c = 2[ π(24)²] + π(16)² = 1408π unit²

2) The length of one side of the rectangle is 96 units. The length of the other side is (x + 16) units. x is the size of one of the legs of right triangle whose vertices are:
a) The centre of the circle of radius 16 units,
b) the mid-point of the upper horizontal side of the overall rectangle, and
c) any of the two vertices of the rectangle close to the two big circles.

3) repeatedly using the Pythagorean Theorem, we have.
x² + 48² = (40 + 24√2)²
x² = (40 + 24√2)² - 48² = 448 + 1920√2
x = (448 + 1920√2)^1/2
x = 8(7 + 30√2)^1/2

4) The area of the overall rectangle A_R is :
A_R = 96 × (x + 16)
= 768[ (7 + 30√2)^1/2 + 2 ]

5) Area of the blue portion A_B is given by A_R - A_c

A_B = 768[ (7 + 30√2)^1/2 + 2 ] - 1408π ≈ 2511.9783 unit²

Cc. Ifada123, I am yet to see an error in my working. If you can see any, spot it out.

Martinez39:
Okay.

1) The area of the blue region is the area of the overall rectangle minus the sum of the areas of the circle. The sum of the areas, A_c, of the circle is :
A_c = 2[ π(24)²] + π(16)² = 1408π unit²

2) The length of one side of the rectangle is 96 units. The length of the other side is (x + 16) units. x is the size of one of the legs of right triangle whose vertices are:
a) The centre of the circle of radius 16 units,
b) the mid-point of the upper horizontal side of the overall rectangle, and
c) any of the two vertices of the rectangle close to the two big circles.

3) repeatedly using the Pythagorean Theorem, we have.
x² + 48² = (40 + 24√2)²
x² = (40 + 24√2)² - 48² = 448 + 1920√2
x = (448 + 1920√2)^1/2
x = 8(7 + 30√2)^1/2

4) The area of the overall rectangle A_R is :
A_R = 96 × (x + 16)
= 768[ (7 + 30√2)^1/2 + 2 ]

5) Area of the blue portion A_B is given by A_R - A_c

A_B = 768[ (7 + 30√2)^1/2 + 2 ] - 1408π ≈ 2511.9783 unit²

Cc. Ifada123, I am yet to see an error in my working. If you can see any, spot it out.

Your x value is not correct,

ifada123:

Your x value is not correct,

Oh yes. I have seen it.

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