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Construct a trapezium ABCD in which AB = 10cm, BC = 8cm, AD = 7cm, angle ABC = 60° and angle BAD is 90°.

Mathematics

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Haya, twende tukasuluhishe swali la 4.

Kwa kuwa mimi ni AI, siwezi kuchora au kufanya ujenzi halisi. Hata hivyo, nitaeleza hatua za ujenzi na maelezo ya loci kama inavyohitajika.

4(a) Construct a trapezium ABCD

Hatua za ujenzi:

  1. Chora mstari ABAB wenye urefu wa 10 cm10 \text{ cm}.
  2. Kwenye ncha AA, jenga pembe ya 90∘90^\circ (∠BAD \angle BAD).
  3. Kwenye mstari wa pembe ya 90∘90^\circ kutoka AA, pima na uweke alama DD ili AD=7 cmAD = 7 \text{ cm}.
  4. Kwenye ncha BB, jenga pembe ya 60∘60^\circ (∠ABC \angle ABC).
  5. Kwenye mstari wa pembe ya 60∘60^\circ kutoka BB, pima na uweke alama CC ili BC=8 cmBC = 8 \text{ cm}.
  6. Unganisha ncha CC na DD kukamilisha trapezi.

4(b) Measure and write the length of CD.

Ili kupata urefu wa CDCD bila kuchora, tunaweza kutumia jiometria ya kuratibu.
Tuchukulie AA iko kwenye asili (0,0)(0,0).
Hivyo, A=(0,0)A = (0,0) na B=(10,0)B = (10,0).
Kwa kuwa ∠BAD=90∘\angle BAD = 90^\circ na AD=7 cmAD = 7 \text{ cm}, kuratibu za DD ni (0,7)(0,7).
Kwa CC, tunajua BC=8 cmBC = 8 \text{ cm} na ∠ABC=60∘\angle ABC = 60^\circ.
Kuratibu za CC zinaweza kupatikana kwa kutumia trigonometry kutoka BB:

xC=xB+BCcos⁡(180∘−60∘)=10+8cos⁡(120∘)=10+8(−12)=10−4=6x_C = x_B + BC \cos(180^\circ - 60^\circ) = 10 + 8 \cos(120^\circ) = 10 + 8(-\frac{1}{2}) = 10 - 4 = 6 yC=yB+BCsin⁡(180∘−60∘)=0+8sin⁡(120∘)=8(32)=43y_C = y_B + BC \sin(180^\circ - 60^\circ) = 0 + 8 \sin(120^\circ) = 8(\frac{\sqrt{3}}{2}) = 4\sqrt{3}

Hivyo, C=(6,43)C = (6, 4\sqrt{3}).
Sasa, tunaweza kutumia formula ya umbali kati ya D(0,7)D(0,7) na C(6,43)C(6, 4\sqrt{3}):

CD=(xC−xD)2+(yC−yD)2CD = \sqrt{(x_C - x_D)^2 + (y_C - y_D)^2} CD=(6−0)2+(43−7)2CD = \sqrt{(6 - 0)^2 + (4\sqrt{3} - 7)^2} CD=62+(43)2−2(43)(7)+72CD = \sqrt{6^2 + (4\sqrt{3})^2 - 2(4\sqrt{3})(7) + 7^2} CD=36+(16×3)−563+49CD = \sqrt{36 + (16 \times 3) - 56\sqrt{3} + 49} CD=36+48−563+49CD = \sqrt{36 + 48 - 56\sqrt{3} + 49} CD=133−563CD = \sqrt{133 - 56\sqrt{3}}

Kukokotoa thamani ya namba:

CD≈133−56×1.73205CD \approx \sqrt{133 - 56 \times 1.73205} CD≈133−96.9948CD \approx \sqrt{133 - 96.9948} CD≈36.0052CD \approx \sqrt{36.0052} CD≈6.00 cmCD \approx 6.00 \text{ cm}

Urefu wa CDCD ni takriban 6.0 cm\boxed{\text{6.0 cm}}.

4(c) Within the trapezium ABCD, draw the locus of points which are

  • (i) 2cm from CD: Hii ni mstari sambamba na CDCD, umbali wa 2 cm2 \text{ cm} kutoka CDCD, na ndani ya trapezi ABCDABCD.
  • (ii) 9cm from B: Hii ni sehemu ya duara (arc) yenye kitovu BB na rediasi 9 cm9 \text{ cm}, ndani ya trapezi ABCDABCD.
  • (iii) equidistant from AB and AD: Hii ni mstari unaogawanya pembe BADBAD katika sehemu mbili sawa (angle bisector). Kwa kuwa ∠BAD=90∘\angle BAD = 90^\circ, mstari huu utakuwa kwenye 45∘45^\circ kutoka ABAB (na ADAD), ukianzia AA, na ndani ya trapezi ABCDABCD.

4(d) A point E within the trapezium ABCD, is such that it is greater than or equal to 2cm from CD, less than or equal to 9cm from B and nearer to AD than AB. Indicate, by shading, the region in which E must lie.

Eneo ambalo EE lazima liwe ndani yake linafafanuliwa na masharti matatu:

  1. Greater than or equal to 2cm from CD: Hii inamaanisha EE iko kwenye upande wa mstari wa locus kutoka 4(c)(i) ambao uko mbali zaidi na CDCD.
  2. Less than or equal to 9cm from B: Hii inamaanisha EE iko ndani au kwenye arc ya locus kutoka 4(c)(ii).
  3. Nearer to AD than AB: Hii inamaanisha EE iko kwenye upande wa mstari wa locus kutoka 4(c)(iii) ambao uko karibu zaidi na ADAD.

Ili kuonyesha eneo hili kwa kivuli, ungetakiwa kuchora loci zote tatu kwanza. Kisha, eneo la kivuli litakuwa makutano ya maeneo yote matatu yanayokidhi masharti yaliyotajwa hapo juu.

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