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Loci Using a ruler and a pair of compasses only, a. Construct a triangle ABC such that angle ABC = 135°C, AB = 8.2cm and BC = 9.6cm b. Given that D is a position equidistant from both AB and BC and also from B and C i. Locate D ii. Find the area of triangle DBC.

Mathematics

ScanSolve AI Answer

1.a) Kujenga pembetatu ABC
Huu ni ujenzi wa pembetatu kwa kutumia pande mbili na pembe moja (SAS).

  • Hatua ya 1: Chora mstari wa urefu wa AB=8.2 cmAB = 8.2 \text{ cm}.
  • Hatua ya 2: Katika ncha BB, jenga pembe ya 135∘135^\circ.
  • Hatua ya 3: Kwenye mkono wa pembe ya 135∘135^\circ, pima 9.6 cm9.6 \text{ cm} kutoka BB ili kupata ncha CC.
  • Hatua ya 4: Unganisha ncha AA na CC kukamilisha pembetatu ABCABC.

1.b) i) Kupata ncha D

  • Hatua ya 1: Kwa kuwa DD iko umbali sawa kutoka ABAB na BCBC, DD iko kwenye bisector ya pembe ya ∠ABC\angle ABC. Jenga bisector ya pembe ya ∠ABC\angle ABC.
  • Hatua ya 2: Kwa kuwa DD iko umbali sawa kutoka BB na CC, DD iko kwenye bisector ya perpendicular ya mstari BCBC. Jenga bisector ya perpendicular ya BCBC.
  • Hatua ya 3: Ncha DD ni mahali ambapo bisector ya pembe ya ∠ABC\angle ABC na bisector ya perpendicular ya BCBC zinakutana.

1.b) ii) Kupata eneo la pembetatu DBC

Tunajua BC=9.6 cmBC = 9.6 \text{ cm}.
Kwa kuwa DD iko kwenye bisector ya pembe ya ∠ABC\angle ABC, ∠DBC=12∠ABC\angle DBC = \frac{1}{2} \angle ABC.

∠DBC=12×135∘=67.5∘\angle DBC = \frac{1}{2} \times 135^\circ = 67.5^\circ

Kwa kuwa DD iko kwenye bisector ya perpendicular ya BCBC, ikiwa MM ni katikati ya BCBC, basi DMDM ni perpendicular kwa BCBC. Hivyo, DMDM ni urefu wa pembetatu DBCDBC na BM=12BCBM = \frac{1}{2} BC.

BM=12×9.6 cm=4.8 cmBM = \frac{1}{2} \times 9.6 \text{ cm} = 4.8 \text{ cm}

Katika pembetatu ya kulia △DBM\triangle DBM:

tan⁡(∠DBM)=DMBM\tan(\angle DBM) = \frac{DM}{BM} tan⁡(67.5∘)=DM4.8 cm\tan(67.5^\circ) = \frac{DM}{4.8 \text{ cm}} DM=4.8×tan⁡(67.5∘)DM = 4.8 \times \tan(67.5^\circ)

Tunajua tan⁡(67.5∘)=2+1≈2.4142\tan(67.5^\circ) = \sqrt{2} + 1 \approx 2.4142.

DM=4.8×(2+1) cmDM = 4.8 \times (\sqrt{2} + 1) \text{ cm} DM≈4.8×2.4142 cmDM \approx 4.8 \times 2.4142 \text{ cm} DM≈11.588 cmDM \approx 11.588 \text{ cm}

Eneo la pembetatu DBCDBC ni:

Eneo=12×base×height\text{Eneo} = \frac{1}{2} \times \text{base} \times \text{height} Eneo=12×BC×DM\text{Eneo} = \frac{1}{2} \times BC \times DM Eneo=12×9.6 cm×(4.8×(2+1)) cm\text{Eneo} = \frac{1}{2} \times 9.6 \text{ cm} \times (4.8 \times (\sqrt{2} + 1)) \text{ cm} Eneo=4.8×4.8×(2+1) cm2\text{Eneo} = 4.8 \times 4.8 \times (\sqrt{2} + 1) \text{ cm}^2 Eneo=23.04×(2+1) cm2\text{Eneo} = 23.04 \times (\sqrt{2} + 1) \text{ cm}^2 Eneo≈23.04×2.41421356 cm2\text{Eneo} \approx 23.04 \times 2.41421356 \text{ cm}^2 Eneo≈55.6197 cm2\text{Eneo} \approx 55.6197 \text{ cm}^2

Kwa usahihi wa sehemu mbili za desimali:

Eneo≈55.62 cm2\text{Eneo} \approx 55.62 \text{ cm}^2

Eneo la pembetatu DBCDBC ni 55.62 cm2\boxed{\text{55.62 cm}^2}.

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