We read your photo as:

Calculate the perimeter of the shaded part of the circle below. Correct your answer into 2 d.p

Mathematics

ScanSolve AI Answer

Answer

112 cm2\text{}112 \text{ cm}^2\text{}

Step-by-step solution

3 steps

Hapa kuna suluhisho la swali la 5:

Swali la 5: Tafuta eneo la sehemu isiyo na kivuli kwenye mchoro hapa chini (Tumia π=227\pi = \frac{22}{7}).

Mchoro unaonyesha mraba wenye upande wa 14 cm14 \text{ cm}. Sehemu zenye kivuli ni maumbo mawili yanayofanana. Ili kupata eneo la sehemu isiyo na kivuli, tutatoa eneo la sehemu mbili zenye kivuli kutoka eneo la mraba.

Kila sehemu yenye kivuli inaweza kuhesabiwa kwa kutumia mbinu ya "eneo la sehemu ya duara" (segment of a circle).
Hata hivyo, kuna njia rahisi zaidi kwa aina hii ya mchoro.
Eneo la sehemu isiyo na kivuli ni sawa na eneo la mraba ukiondoa eneo la sehemu mbili zenye kivuli.
Kila sehemu yenye kivuli inaundwa na eneo la sekta ya duara (quarter circle) ukitoa eneo la pembetatu.
Hebu tuchukulie mraba ABCD.
Eneo la mraba ni S2S^2.
Eneo la robo duara (quarter circle) lenye radius SS ni 14πS2\frac{1}{4} \pi S^2.
Eneo la pembetatu ya kulia (right-angled triangle) ndani ya robo duara ni 12S2\frac{1}{2} S^2.
Eneo la sehemu moja yenye kivuli (segment) ni Asegment=Aquarter_circle−Atriangle=14πS2−12S2A_{segment} = A_{quarter\_circle} - A_{triangle} = \frac{1}{4} \pi S^2 - \frac{1}{2} S^2.
Kuna sehemu mbili zenye kivuli, kwa hiyo eneo la jumla la kivuli ni 2×(14πS2−12S2)=12πS2−S22 \times (\frac{1}{4} \pi S^2 - \frac{1}{2} S^2) = \frac{1}{2} \pi S^2 - S^2.
Eneo lisilo na kivuli ni Asquare−Ashaded=S2−(12πS2−S2)=S2−12πS2+S2=2S2−12πS2A_{square} - A_{shaded} = S^2 - (\frac{1}{2} \pi S^2 - S^2) = S^2 - \frac{1}{2} \pi S^2 + S^2 = 2S^2 - \frac{1}{2} \pi S^2.

Njia nyingine rahisi:
Eneo la sehemu isiyo na kivuli linaweza kupatikana kwa kutoa eneo la sehemu mbili za duara kutoka eneo la mraba.
Fikiria robo duara mbili, moja ikiwa na kituo kwenye kona ya juu kushoto na nyingine kwenye kona ya chini kulia.
Eneo la robo duara moja ni Aquarter_circle=14πr2A_{quarter\_circle} = \frac{1}{4} \pi r^2.
Eneo la sehemu isiyo na kivuli ni 2×(Eneo la robo duara)−(Eneo la mraba)2 \times (\text{Eneo la robo duara}) - (\text{Eneo la mraba}).
Hii ni kwa sababu eneo la mraba linaweza kugawanywa katika sehemu nne: sehemu mbili zenye kivuli, na sehemu mbili zisizo na kivuli.
Eneo la sehemu isiyo na kivuli ni Aunshaded=2×Aquarter_circle−AsquareA_{unshaded} = 2 \times A_{quarter\_circle} - A_{square}.
Hii ni njia ya kawaida ya kutatua matatizo ya aina hii.

  1. Tafuta eneo la mraba.

    Upande wa mraba S=14 cmS = 14 \text{ cm}.

    Amraba=S2=(14 cm)2=196 cm2A_{mraba} = S^2 = (14 \text{ cm})^2 = 196 \text{ cm}^2
  2. Tafuta eneo la robo duara moja.

    Radius ya robo duara ni sawa na upande wa mraba, r=14 cmr = 14 \text{ cm}.

    Arobo_duara=14πr2=14×227×(14 cm)2A_{robo\_duara} = \frac{1}{4} \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times (14 \text{ cm})^2 Arobo_duara=14×227×196 cm2A_{robo\_duara} = \frac{1}{4} \times \frac{22}{7} \times 196 \text{ cm}^2 Arobo_duara=14×22×28 cm2A_{robo\_duara} = \frac{1}{4} \times 22 \times 28 \text{ cm}^2 Arobo_duara=22×7 cm2=154 cm2A_{robo\_duara} = 22 \times 7 \text{ cm}^2 = 154 \text{ cm}^2
  3. Tafuta eneo la sehemu isiyo na kivuli.

    Eneo la sehemu isiyo na kivuli ni jumla ya eneo la robo duara mbili ukitoa eneo la mraba.

    Aisiyo_na_kivuli=(2×Arobo_duara)−AmrabaA_{isiyo\_na\_kivuli} = (2 \times A_{robo\_duara}) - A_{mraba} Aisiyo_na_kivuli=(2×154 cm2)−196 cm2A_{isiyo\_na\_kivuli} = (2 \times 154 \text{ cm}^2) - 196 \text{ cm}^2 Aisiyo_na_kivuli=308 cm2−196 cm2A_{isiyo\_na\_kivuli} = 308 \text{ cm}^2 - 196 \text{ cm}^2 Aisiyo_na_kivuli=112 cm2A_{isiyo\_na\_kivuli} = 112 \text{ cm}^2

    Eneo la sehemu isiyo na kivuli ni 112 cm2\text{}112 \text{ cm}^2\text{}

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp