Menjana Kecemerlangan 4 (Soalan 5 – 7) – Buku Teks Matematik Tingkatan 2 Bab 4 (Poligon)


Soalan 5:
(a) Hitung nilai bagi x + y dalam rajah di bawah.

(b) Rajah menunjukkan logo berbentuk pentagon sekata. FED ialah garis lurus.
Hitung nilai x + y.


(c) Dalam rajah di bawah, HIJKL ialah sebuah pentagon. KJM ialah garis lurus. Hitung nilai a + b + c + d.



Penyelesaian:
(a)

$$ \begin{aligned} r+150^{\circ} & =180^{\circ} \\ r & =180^{\circ}-150^{\circ} \\ r & =30^{\circ} \end{aligned} $$
$$ \begin{aligned} s+65^{\circ} & =180^{\circ} \\ s & =180^{\circ}-65^{\circ} \\ s & =115^{\circ} \end{aligned} $$


$$ \begin{aligned} &\text { Hasil tambah sudut pedalaman suatu poligon }=(n-2) \times 180^{\circ}\\ &\begin{aligned} x+s+y+r & =(4-2) \times 180^{\circ} \\ x+115^{\circ}+y+30^{\circ} & =2 \times 180^{\circ} \\ x+y+145^{\circ} & =360^{\circ} \\ x+y & =360^{\circ}-145^{\circ} \\ x+y & =215^{\circ} \end{aligned} \end{aligned} $$


(b)
$$ \begin{aligned} \text { Sudut peluaran poligon sekata } & =\frac{360^{\circ}}{n} \\ y & =\frac{360^{\circ}}{5} \\ & =72^{\circ} \end{aligned} $$
$$ \begin{aligned} \text { Sudut pedalaman poligon sekata } & =\frac{(n-2) \times 180^{\circ}}{n} \\ x & =\frac{(5-2) \times 180^{\circ}}{5} \\ & =\frac{3 \times 180^{\circ}}{5} \\ & =\frac{540^{\circ}}{5} \\ & =108^{\circ} \end{aligned} $$
$$ \begin{aligned} x+y & =108^{\circ}+72^{\circ} \\ & =180^{\circ} \end{aligned} $$

(c)
$$ \begin{aligned} \angle K J I+65^{\circ} & =180^{\circ} \\ \angle K J I & =180^{\circ}-65^{\circ} \\ & =115^{\circ} \end{aligned} $$
$$ \begin{aligned} &\text { Hasil tambah sudut pedalaman suatu poligon }=(n-2) \times 180^{\circ}\\ &\begin{aligned} a+b+c+d+\angle K J I & =3 \times 180^{\circ} \\ a+b+c+d+115^{\circ} & =540^{\circ} \\ a+b+c+d & =540^{\circ}-115^{\circ} \\ a+b+c+d & =425^{\circ} \end{aligned} \end{aligned} $$


Soalan 7:
Hasil tambah semua sudut pedalaman sebuah poligon sekata ialah 2 700°. Nyatakan bilangan sisi poligon itu.

Penyelesaian:
$$ \begin{aligned} &\text { Hasil tambah sudut pedalaman suatu poligon }=(n-2) \times 180^{\circ}\\ &\begin{aligned} 2700^{\circ} & =(n-2) \times 180^{\circ} \\ n-2 & =\frac{2700^{\circ}}{180^{\circ}} \\ n-2 & =15 \\ n & =15+2 \\ n & =17 \text { sisi } \end{aligned} \end{aligned} $$


Soalan 8:
Dalam rajah di bawah, hitung nilai p + q.


Penyelesaian:



$$ r=70^{\circ}=\text { sudut sepadan } 70^{\circ} $$
$$ \begin{aligned} & s=180^{\circ}-70^{\circ} \\ & s=110^{\circ} \end{aligned} $$
$$ \begin{aligned} t+98^{\circ} & =360^{\circ} \\ t & =360^{\circ}-98^{\circ} \\ t & =262^{\circ} \end{aligned} $$
$$ \begin{aligned} u & =180^{\circ}-80^{\circ} \\ u & =100^{\circ} \end{aligned} $$


$$ \begin{aligned} &\text { Hasil tambah sudut pedalaman suatu poligon }=(n-2) \times 180^{\circ}\\ &\begin{aligned} p+q+100^{\circ}+110^{\circ}+92^{\circ}+262^{\circ}+60^{\circ} & =(7-2) \times 180^{\circ} \\ p+q+624^{\circ} & =900^{\circ} \\ p+q & =900^{\circ}-624^{\circ} \\ p+q & =276^{\circ} \end{aligned} \end{aligned} $$

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