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The diagram below is a quadrilateral ABCD in which CD = 8 cm, BC = 11cm, AC = 13cm, AB = 4cm and angle ADC = 80°.

Mathematics

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4 parts
a

62.7∘\text{62.7}^\circ

b

11.7 cm\text{11.7 cm}

c

16.7∘\text{16.7}^\circ

d

66.7 cm2\text{66.7 cm}^2

Step-by-step solution

9 steps

Haya wmiriam395, twende na swali hili la jiometri.

Part a· 2 steps

Kokotoa ukubwa wa pembe ACD.

Ili kupata ∠ACD\angle ACD, tutatumia sheria ya Sine katika △ADC\triangle ADC.
Tunajua CD=8CD = 8 cm, AC=13AC = 13 cm, na ∠ADC=80∘\angle ADC = 80^\circ.

  1. Tumia sheria ya Sine kupata ∠CAD\angle CAD.

    sin⁡(∠CAD)CD=sin⁡(∠ADC)AC\frac{\sin(\angle CAD)}{CD} = \frac{\sin(\angle ADC)}{AC} sin⁡(∠CAD)8=sin⁡80∘13\frac{\sin(\angle CAD)}{8} = \frac{\sin 80^\circ}{13} sin⁡(∠CAD)=8sin⁡80∘13\sin(\angle CAD) = \frac{8 \sin 80^\circ}{13} sin⁡(∠CAD)=8×0.984813\sin(\angle CAD) = \frac{8 \times 0.9848}{13} sin⁡(∠CAD)≈0.6060\sin(\angle CAD) \approx 0.6060 ∠CAD=arcsin⁡(0.6060)≈37.3∘\angle CAD = \arcsin(0.6060) \approx 37.3^\circ
  2. Tumia jumla ya pembe katika pembetatu kupata ∠ACD\angle ACD.

    Jumla ya pembe katika pembetatu ni 180∘180^\circ.

    ∠ACD=180∘−∠ADC−∠CAD\angle ACD = 180^\circ - \angle ADC - \angle CAD ∠ACD=180∘−80∘−37.3∘\angle ACD = 180^\circ - 80^\circ - 37.3^\circ ∠ACD=62.7∘\angle ACD = 62.7^\circ

    Ukubwa wa pembe ACD ni 62.7∘\boxed{\text{62.7}^\circ}.

Part b

11.7 cm\text{11.7 cm}

Kokotoa urefu wa upande AD.

Tutatumia sheria ya Cosine katika △ADC\triangle ADC.
Tunajua CD=8CD = 8 cm, AC=13AC = 13 cm, na ∠ADC=80∘\angle ADC = 80^\circ.

  1. Tumia sheria ya Cosine kupata urefu wa AD.

    AC2=AD2+CD2−2(AD)(CD)cos⁡(∠ADC)AC^2 = AD^2 + CD^2 - 2(AD)(CD)\cos(\angle ADC) 132=AD2+82−2(AD)(8)cos⁡(80∘)13^2 = AD^2 + 8^2 - 2(AD)(8)\cos(80^\circ) 169=AD2+64−16AD(0.1736)169 = AD^2 + 64 - 16 AD (0.1736) 169=AD2+64−2.7776AD169 = AD^2 + 64 - 2.7776 AD
  2. Panga upya mlinganyo kuwa mlinganyo wa quadratic na utatue kwa AD.

    AD2−2.7776AD+64−169=0AD^2 - 2.7776 AD + 64 - 169 = 0 AD2−2.7776AD−105=0AD^2 - 2.7776 AD - 105 = 0

    Tumia formula ya quadratic AD=−b±b2−4ac2aAD = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}:

    AD=2.7776±(−2.7776)2−4(1)(−105)2(1)AD = \frac{2.7776 \pm \sqrt{(-2.7776)^2 - 4(1)(-105)}}{2(1)} AD=2.7776±7.715+4202AD = \frac{2.7776 \pm \sqrt{7.715 + 420}}{2} AD=2.7776±427.7152AD = \frac{2.7776 \pm \sqrt{427.715}}{2} AD=2.7776±20.6812AD = \frac{2.7776 \pm 20.681}{2}

    Kwa kuwa urefu hauwezi kuwa hasi, tunachukua thamani chanya:

    AD=2.7776+20.6812AD = \frac{2.7776 + 20.681}{2} AD=23.45862AD = \frac{23.4586}{2} AD≈11.7293 cmAD \approx 11.7293 \text{ cm}

    Urefu wa upande AD ni 11.7 cm\boxed{\text{11.7 cm}}.

Part c

16.7∘\text{16.7}^\circ

Kokotoa ukubwa wa pembe ACB.

Tutatumia sheria ya Cosine katika △ABC\triangle ABC.
Tunajua AB=4AB = 4 cm, BC=11BC = 11 cm, na AC=13AC = 13 cm.

  1. Tumia sheria ya Cosine kupata ∠ACB\angle ACB.

    AB2=BC2+AC2−2(BC)(AC)cos⁡(∠ACB)AB^2 = BC^2 + AC^2 - 2(BC)(AC)\cos(\angle ACB) 42=112+132−2(11)(13)cos⁡(∠ACB)4^2 = 11^2 + 13^2 - 2(11)(13)\cos(\angle ACB) 16=121+169−286cos⁡(∠ACB)16 = 121 + 169 - 286 \cos(\angle ACB) 16=290−286cos⁡(∠ACB)16 = 290 - 286 \cos(\angle ACB)
  2. Tatua kwa cos⁡(∠ACB)\cos(\angle ACB) na kisha ∠ACB\angle ACB.

    286cos⁡(∠ACB)=290−16286 \cos(\angle ACB) = 290 - 16 286cos⁡(∠ACB)=274286 \cos(\angle ACB) = 274 cos⁡(∠ACB)=274286\cos(\angle ACB) = \frac{274}{286} cos⁡(∠ACB)≈0.9580\cos(\angle ACB) \approx 0.9580 ∠ACB=arccos⁡(0.9580)≈16.7∘\angle ACB = \arccos(0.9580) \approx 16.7^\circ

    Ukubwa wa pembe ACB ni 16.7∘\boxed{\text{16.7}^\circ}.

Part d

66.7 cm2\text{66.7 cm}^2

Kokotoa eneo la quadrilateral ABCD.

Eneo la quadrilateral ABCD ni jumla ya eneo la △ADC\triangle ADC na eneo la △ABC\triangle ABC.

  1. Kokotoa eneo la △ADC\triangle ADC.

    Tunajua CD=8CD = 8 cm, AD≈11.7293AD \approx 11.7293 cm, na ∠ADC=80∘\angle ADC = 80^\circ.

    Eneo(△ADC)=12×CD×AD×sin⁡(∠ADC)\text{Eneo}(\triangle ADC) = \frac{1}{2} \times CD \times AD \times \sin(\angle ADC) Eneo(△ADC)=12×8×11.7293×sin⁡(80∘)\text{Eneo}(\triangle ADC) = \frac{1}{2} \times 8 \times 11.7293 \times \sin(80^\circ) Eneo(△ADC)=4×11.7293×0.9848\text{Eneo}(\triangle ADC) = 4 \times 11.7293 \times 0.9848 Eneo(△ADC)≈46.20 cm2\text{Eneo}(\triangle ADC) \approx 46.20 \text{ cm}^2
  2. Kokotoa eneo la △ABC\triangle ABC.

    Tunajua BC=11BC = 11 cm, AC=13AC = 13 cm, na ∠ACB≈16.664∘\angle ACB \approx 16.664^\circ.

    Eneo(△ABC)=12×BC×AC×sin⁡(∠ACB)\text{Eneo}(\triangle ABC) = \frac{1}{2} \times BC \times AC \times \sin(\angle ACB) Eneo(△ABC)=12×11×13×sin⁡(16.664∘)\text{Eneo}(\triangle ABC) = \frac{1}{2} \times 11 \times 13 \times \sin(16.664^\circ) Eneo(△ABC)=12×143×0.2866\text{Eneo}(\triangle ABC) = \frac{1}{2} \times 143 \times 0.2866 Eneo(△ABC)≈20.49 cm2\text{Eneo}(\triangle ABC) \approx 20.49 \text{ cm}^2
  3. Jumlisha maeneo ya pembetatu mbili.

    Eneo(ABCD)=Eneo(△ADC)+Eneo(△ABC)\text{Eneo(ABCD)} = \text{Eneo}(\triangle ADC) + \text{Eneo}(\triangle ABC) Eneo(ABCD)=46.20+20.49\text{Eneo(ABCD)} = 46.20 + 20.49 Eneo(ABCD)=66.69 cm2\text{Eneo(ABCD)} = 66.69 \text{ cm}^2

    Eneo la quadrilateral ABCD ni 66.7 cm2\boxed{\text{66.7 cm}^2}.

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