Maths Quiz: Tafuta eneo lililotiwa kivuli

[SMU]

Hahahah! itakuwa swaafi sana maana imeninyima raha kwenye majukwaa mengine nitaacha kubwabwaja!!

Nimeshaibadilisha mkuu! Pamoja sana.

 

[WA-UKENYENGE]

Nimeshaibadilisha mkuu! Pamoja sana.

Pamoja sana mkuu!! Mwenzio ishanitoa jasho hapa! Hahaha:nerd:

 

[simplemind]

Bila simplifying assumption below problem is unsolvable

- ABC is right angle triangle
- AC is diameter of large circle
-BC is radius small circle.

What a man believes may be ascertained not from his creed but from assumptions on which he habitually acts.

 

[simplemind]

Find the area enclosed by line y=x and curve y=x^2 from x=0 to the point where curve intersects the line. (5 marks)

 

[Jodoki Kalimilo]

Hapana mkuu, tunaanzia kwenye umbo kama hilo then we finally come up with structure like this!!

Kumbe nyie ndio mungu alivyomaliza kuumba ulimwengu aliwakabidhi nchi na akawapa kazi ya uumbaji (ENGINEERS) wa baadhi ya vitu kwa niaba yake?

 

[RedDevil]

Kumbe nyie ndio mungu alivyomaliza kuumba ulimwengu aliwakabidhi nchi na akawapa kazi ya uumbaji (ENGINEERS) wa baadhi ya vitu kwa niaba yake?

Engineers believe are small Gods.

 

[RedDevil]

Bila simplifying assumption below problem is unsolvable

- ABC is right angle triangle
- AC is diameter of large circle
-BC is radius small circle.

What a man believes may be ascertained not from his creed but from assumptions on which he habitually acts.

What if ABC is not a right angled triangle i.e /_ABC is not equal to 90 degree? Cosine and sine rules are suitable to obtain the length AC.

 

[Jodoki Kalimilo]

Engineers believe are small Gods.

Ni kweli mdau ila siku hizi wanasiasa wanawaingilia mpaka mnaingia katika mkumbo wa kuchakachua BOQ na kuumba vitu below standard

 

[mwanamabadiliko]

Aise kama alivosema jembe, assumption ni muhimu. Kwa kuangalia tu hio ni right angled triangle, in engineering we make lots of assuptions. Pia std 7 tulikuwa tunachorewa maumbo then tunasov, huwa hawastet

 

[simplemind]

What if ABC is not a right angled triangle i.e /_ABC is not equal to 90 degree? Cosine and sine rules are suitable to obtain the length AC.

sine rule au cosine rule itatumika vipi, all angles of triangle ABC unknown/not given.

 

[NingaR]

MUNGU ni MUNGU na Engineers are nothing like MUNGU

 

[WA-UKENYENGE]

Find the area enclosed by line y=x and curve y=x^2 from x=0 to the point where curve intersects the line. (5 marks)

Mkuu unatutoa kwenye maada!
Anyway! Enclosed area = integral { x - x^2} dx; Limits should be --> x1 = 0, x2 = 1 after solving the resulting quadric equation x-x^2 = 0. Enclosed area = 1/6 [units sqr].

 

[fortunho]

1.Find the area ya Nusu Duara(AC)
2.Find the area of Triangle(ABC)
3.Total Area=AC+ABC
4.Find Area of Quarter Circle(BC)
5.Shaded Area=Total-BC

good boy,i like that

 

[QUALIFIED]

Wakuu! Assumptions ni mhimu kuweka, Mfano: The triangle ABC is not stated to be a right, however most of the attempters are babtizing the triangle ABC to be the right angled!!

nyie ndo wale ukipewa 1+1=? Chuoni unaanza mbwela nying kisa una formulae nying kichwan kijana hapo hakuna assumption piga triangle ka right angled not other wise ambapo ni_:
(area of triangle+area of sem-circle)-area of chord
kuku ni kuku,jogoo jina tu

 

[WA-UKENYENGE]

Solution 1:

• Assumption 1: Angle ABC = 90[SUP]o[/SUP], Length BD = BC, then BCD is a quarter-circle with radius 42 [cm].
• Area of triangle ABC = (1/2)x42x56 = 1176 [cm[SUP]2[/SUP]]
• Assumption 2: Let point E be the mid-point between point A and point C, then the structure ACA is a semi-circle with radius, R = (1/2)x AC=35 [cm]
• Area of the semi-circle = 1924.2255 [cm[SUP]2[/SUP]]
• Area of quarter-circle BCD = 1385.44 [cm[SUP]2[/SUP]]
• Area of the shaded region = (Area of ΔABC) + (Area of Semi-circle ACA) - (Area of quarter-circle) = 1176 +1924.22-1385.44 = 1714.79 [cm[SUP]2[/SUP]]

 

[Super Sub Steve]

big up lakini eneo la nusudua na robo duara nadhani formula zake umeoverlook

Find area of the whole figure which is a triangle (56*42*0.5) and a semi circle (35*35*22/7) the total will be 3101 cm square. From this minus the unshaded part which is a quarter of a circle (22/7*42*42*0.5) which is 1386 cm squire. You will remain with 1715 cm squire.

 

[Lughe]

Vijana swali simple sana hii!!
1)Eneo la triangle ABC
½×56×42=1176cm²
2)Eneo la semicircle AC
½×Π×r²
r=AB/2
By Pythagoras theorem
AB=70cm
r=70/2=35cm
:-area=½×Π×35²=1825cm²
3)Eneo la Robo duara BC
¼×Π×42²=1386cm²

Turudi kwenye swari sasa
SHADED AREA=(Eneo la triangle ABC+Eneo la semicircle AC)-(Eneo la robo duara BC)
=>Numerically
Required area
=[(1176+1925)-1386]cm²
=(3101-1386)cm²
=1715cm²
:-The shaded area is 1715cm²

Kaka unakwama wapi?
Hakuna nusu duara hapo.

it sounds if we say;

Hapo kuna robo duara mbili tu lenye radius 56 na 42.
So Tol area=quarter circle equal to area of quater cirlce using radius56
Un Shaded area=area of quarter circle using radius 42.
Shaded area=Tol area-Un shaded area.

 

[NingaR]

Kaka unakwama wapi?
Hakuna nusu duara hapo.

it sounds if we say;

Hapo kuna robo duara mbili tu lenye radius 56 na 42.
So Tol area=quarter circle equal to area of quater cirlce using radius56
Un Shaded area=area of quarter circle using radius 42.
Shaded area=Tol area-Un shaded area.

Mkuu hebu angalia hata uwiano wa hayo marobo duara mawili unaona yana fanana kweli???? hai hitaji miwani kuona kua yanatofautiana tena sana just angalia vizuri ujionee mwenyewe,
angalia hiyo robo duara kubwa na uniambie hizo radius mbili zinazo bound curved part zinalingana?? maana inatakiwa katika robo duara any distance from center to a curved surface must be equal (AU UMESAHAU HILO?)

 

[mkupuo]

find the area of the shaded region!!

ebo nitafute eneo lililowekwa kivuli wakati naliona kwa macho yangu! Huwa tunatafuta vitu vilivyopotea bana!

 

[Preta]

uwiii... sweetlady.....lete ile bikari yangu hapa.....kuna hesabu.....