Huu hapa ni ufumbuzi wa maswali yako:

Mathematics

ScanSolve AI Answer

Answers

3 parts
a

7.5 cm\text{7.5 cm}

b

72.9∘72.9^\circ

c

29.4∘29.4^\circ

Step-by-step solution

6 steps

Huu hapa ni ufumbuzi wa maswali yako:

Kutokana na mchoro na taarifa ulizotoa, vipimo vya mchemraba ni:
Urefu (GH) = 8 cm
Upana (FG) = 4.5 cm
Kimo (CG) = 6 cm

Part a· 2 steps

Kokotoa urefu wa FC

  1. Tambua pembetatu sahihi.

    FC ni ulalo kwenye uso wa mbele FGCB. Pembetatu FGC ni pembetatu ya kulia yenye pembe ya 90∘90^\circ kwenye G.

  2. Tumia Theorem ya Pythagoras.

    FC2=FG2+GC2FC^2 = FG^2 + GC^2 FC2=(4.5)2+(6)2FC^2 = (4.5)^2 + (6)^2 FC2=20.25+36FC^2 = 20.25 + 36 FC2=56.25FC^2 = 56.25 FC=56.25FC = \sqrt{56.25} FC=7.5 cmFC = 7.5 \text{ cm}

    Urefu wa FC ni 7.5 cm\boxed{\text{7.5 cm}}.

Part b

72.9∘72.9^\circ

Kokotoa ukubwa wa pembe kati ya mistari FC na FH

  1. Kokotoa urefu wa FH na CH.

    FH ni ulalo kwenye msingi EFGH. Pembetatu FGH ni pembetatu ya kulia yenye pembe ya 90∘90^\circ kwenye G.

    FH2=FG2+GH2FH^2 = FG^2 + GH^2 FH2=(4.5)2+(8)2FH^2 = (4.5)^2 + (8)^2 FH2=20.25+64FH^2 = 20.25 + 64 FH2=84.25FH^2 = 84.25 FH=84.25 cmFH = \sqrt{84.25} \text{ cm}

    CH ni ulalo kwenye uso wa nyuma CGHD. Pembetatu CGH ni pembetatu ya kulia yenye pembe ya 90∘90^\circ kwenye G.

    CH2=CG2+GH2CH^2 = CG^2 + GH^2 CH2=(6)2+(8)2CH^2 = (6)^2 + (8)^2 CH2=36+64CH^2 = 36 + 64 CH2=100CH^2 = 100 CH=100CH = \sqrt{100} CH=10 cmCH = 10 \text{ cm}

  2. Tumia Kanuni ya Kosine (Cosine Rule) katika pembetatu FCH.

    Tunataka kupata pembe ∠CFH\angle CFH.

    CH2=FC2+FH2−2(FC)(FH)cos⁡(∠CFH)CH^2 = FC^2 + FH^2 - 2(FC)(FH)\cos(\angle CFH) 102=(7.5)2+(84.25)2−2(7.5)(84.25)cos⁡(∠CFH)10^2 = (7.5)^2 + (\sqrt{84.25})^2 - 2(7.5)(\sqrt{84.25})\cos(\angle CFH) 100=56.25+84.25−1584.25cos⁡(∠CFH)100 = 56.25 + 84.25 - 15\sqrt{84.25}\cos(\angle CFH) 100=140.5−1584.25cos⁡(∠CFH)100 = 140.5 - 15\sqrt{84.25}\cos(\angle CFH) 1584.25cos⁡(∠CFH)=140.5−10015\sqrt{84.25}\cos(\angle CFH) = 140.5 - 100 1584.25cos⁡(∠CFH)=40.515\sqrt{84.25}\cos(\angle CFH) = 40.5 cos⁡(∠CFH)=40.51584.25\cos(\angle CFH) = \frac{40.5}{15\sqrt{84.25}} cos⁡(∠CFH)≈40.515×9.1787\cos(\angle CFH) \approx \frac{40.5}{15 \times 9.1787} cos⁡(∠CFH)≈40.5137.6805\cos(\angle CFH) \approx \frac{40.5}{137.6805} cos⁡(∠CFH)≈0.29415\cos(\angle CFH) \approx 0.29415 ∠CFH=arccos⁡(0.29415)\angle CFH = \arccos(0.29415) ∠CFH≈72.9∘\angle CFH \approx 72.9^\circ

    Ukubwa wa pembe kati ya mistari FC na FH ni 72.9∘\boxed{72.9^\circ}.

Part c

29.4∘29.4^\circ

Kokotoa ukubwa wa pembe kati ya mistari AB na FH

  1. Tambua uhusiano kati ya mistari.

    Mstari AB ni sambamba na mstari GH (AB ∥\parallel GH).
    Kwa hiyo, pembe kati ya AB na FH ni sawa na pembe kati ya GH na FH, ambayo ni ∠GHF\angle GHF.

  2. Tumia trigonometry katika pembetatu FGH.

    Pembetatu FGH ni pembetatu ya kulia yenye pembe ya 90∘90^\circ kwenye G.
    Tunataka kupata ∠GHF\angle GHF.
    Upande wa pili wa ∠GHF\angle GHF ni FG = 4.5 cm.
    Upande wa karibu wa ∠GHF\angle GHF ni GH = 8 cm.

    tan⁡(∠GHF)=upande wa piliupande wa karibu=FGGH\tan(\angle GHF) = \frac{\text{upande wa pili}}{\text{upande wa karibu}} = \frac{FG}{GH} tan⁡(∠GHF)=4.58\tan(\angle GHF) = \frac{4.5}{8} tan⁡(∠GHF)=0.5625\tan(\angle GHF) = 0.5625 ∠GHF=arctan⁡(0.5625)\angle GHF = \arctan(0.5625) ∠GHF≈29.4∘\angle GHF \approx 29.4^\circ

    Ukubwa wa pembe kati ya mistari AB na FH ni 29.4∘\boxed{29.4^\circ}.

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

More Mathematics Questions

QHuu hapa ni utatuzi wa maswali yako:QHaya hapa majibu ya maswali yaliyosalia:QHaya hapa majibu ya maswali yako ya Kiswahili.QHapa kuna majibu kamili kulingana na kazi yako:QHuu hapa utatuzi wa swali la 35:
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.