₦airaland Forum

LAST SHAPE FOR TODAY

Find the area of the polygon,
Or tell us the regularity your observe.

Anyone?

pls sir mr shape,i have been following ur threads, I am a mathematics and education student in unn but all this shape questions are quite strange to me in what way can u assist a young mathematician

Need2learnmuch:
pls sir mr shape,i have been following ur threads, I am a mathematics and education student in unn but all this shape questions are quite strange to me in what way can u assist a young mathematician

Follow me on niaraland and always attempt shape questions and ask questions were you are not clear,
Shape is an aspect of maths for people that can see beyond the obvious, and u can't get that from books, you can only get that from practice, soon u will start seeing connection and apply maths basic rules to make u vet unknown,
Feel free to tag me in any shape u encounter or find difficult to solve, I will assist u in understanding how to see connections,
No book can help u.

Anyone?

ifada123:
Anyone?

First I can see an isosceles triangle

Is AD the diameter of the circle?

Justinakusu:
Is AD the diameter of the circle?

Sure,
It is a line that pass through the center of the circle and divide it into two equal parts

oyekanwahab:
First I can see an isosceles triangle

OK go on sir

ifada123:
OK go on sir

The area of the polygon is 1110 unit

oyekanwahab:
The area of the polygon is 1110 unit

We are ask to look for the radius not area sir

Anyone?

Anyone

Should I bring solution

23 cm. Not really good at Maths.

I see a kite and triangle

femi4:
I see a kite and triangle

OK good try it out find the radius

Academicwizman:
23 cm. Not really good at Maths.

You tried for attempting,
This is why your solution is not correct,.
That angle is not 90 since it was not subtended by a diameter
Try again sir

ifada123:
LAST SHAPE FOR TODAY

Find the area of the polygon,
Or tell us the regularity your observe.

Here is my solution.

r=25 and Area= 1536 or r=18 and Area =250√11=829.1562. Two possible solutions.

ifada123:
You tried for attempting,
This is why your solution is not correct,.
That angle is not 90 since it was not subtended by a diameter
Try again sir

Ok sire. Can't remember all those theorems again. It's been ages.

Lets go

k²=30² + 14² - 2*30*14 cos a

k² = 1096 - 840 cos a

But 180 + 2b =2a

a = b+90°

Hence k² =1096 - 840 (cosb cos90 - sinb sin90)

K² = 1096 + 840 sinb


And sin b =30/2r

k² = 1096 + 840 *30/2r
= 1096 + 12600/r

Also 4r² = k² + 900

4r² - 900 = 1096 + 12600/r

4r³ -1996r - 12600 = 0

r³ - 499r - 3150 = 0

r = 25, -7 or - 18

R = 25 (The positive value)

femi4:
Lets go

k²=30² + 14² - 2*30*14 cos a

k² = 1096 - 840 cos a

But 180 + 2b =2a

a = b+90°

Hence k² =1096 - 840 (cosb cos90 - sinb sin90)

K² = 1096 + 840 sinb


And sin b =30/2r

k² = 1096 + 840 *30/2r
= 1096 + 12600/r

Also 4r² = k² + 900

4r² - 900 = 1096 + 12600/r

4r³ -1996r - 12600 = 0

r³ - 499r - 3150 = 0

r = 25, -7 or - 18

R = 25 (The positive value)

Please how did you get 180 + 2b =2a?

Mitocyan:
Please how did you get 180 + 2b =2a?

angle subtended at the center (the obtuse angle) = twice angle subtended at the circumference

You guys killed this shape,
Thumbs up.
Look forward to my next shape.
Today's shape

I like the attitude you guys gave this shape,
We are thinking smarter now.
Soon none of my shapes will ever go unsolved.

femi4:
angle subtended at the center (the obtuse angle) = twice angle subtended at the circumference

Ok makes sense