Question

rajah 3 menunjukkan keratan rentas pinu hadapan sebuah bangsa di mana abcd adalah segi empat tepat dan oapb adalah sebuah sektor berpusat di o dengan jejari r m.
diagram 3 shows the cross section of a front door of a barn such that abcd is a rectangle and oapb is a sector with centre o of radius r m.
(a) diberi bahawa panjang ab ialah 8 m dan pq adalah pembahagi dua sama serenjang ab. given that the length ab is 8 m and pq is a perpendicular bisector ab. given that the length ab is 8 m and pq is 2 m.
guna : π = 3.142
(i) hitung nilai r, dalam m,
calculate the value of r, in m,
2 markah
2 marks
(b) seterusnya, cari
hence, find
(i) perimeter, dalam m,
the perimeter, in m,
(ii) luas, dalam m²,
the area, in m²,
permukaan melengkung bangsa seperti yang berlorek dalam rajah 3.
of the curved surface as shaded in diagram 3.
6 markah
6 marks

Explanation:

Step1: Find the value of \( r \)

Since \( PQ \) is the perpendicular bisector of \( AB \), and \( AB = 8\mathrm{m}\), \( AQ=\frac{AB}{2}=4\mathrm{m}\). Also, \( OQ = r - PQ\), and \( PQ = 2\mathrm{m}\), so \( OQ=r - 2\).
By the Pythagorean theorem in right - triangle \( OAQ\): \(OA^{2}=AQ^{2}+OQ^{2}\). Since \( OA=r\), we have \(r^{2}=4^{2}+(r - 2)^{2}\).
Expand \((r - 2)^{2}=r^{2}-4r + 4\). Then \(r^{2}=16+r^{2}-4r + 4\).
Subtract \(r^{2}\) from both sides: \(0=20-4r\). Solve for \(r\): \(4r = 20\), so \(r = 5\mathrm{m}\).

Step2: Calculate the perimeter of the shaded region

The perimeter of the shaded region consists of the arc length \(l\) of the sector and two radii \(OA\) and \(OB\).
The central angle \(\theta\) of the sector: In right - triangle \(OAQ\), \(\sin\angle AOQ=\frac{AQ}{OA}=\frac{4}{5}\), \(\angle AOQ=\sin^{-1}(\frac{4}{5})\approx53.13^{\circ}\), and the central angle of the sector \(\alpha = 2\angle AOQ\approx106.26^{\circ}\).
The arc length formula is \(l=\frac{\alpha}{360}\times2\pi r\). Substitute \(\alpha = 106.26^{\circ}\) and \(r = 5\mathrm{m}\), \(l=\frac{106.26}{360}\times2\times3.14\times5\approx9.2\mathrm{m}\).
The perimeter \(P=l + 2r\). Substitute \(l\approx9.2\mathrm{m}\) and \(r = 5\mathrm{m}\), \(P\approx9.2+10=19.2\mathrm{m}\).

Step3: Calculate the area of the shaded region

The area of the sector \(A_{s}=\frac{\alpha}{360}\times\pi r^{2}\). Substitute \(\alpha = 106.26^{\circ}\) and \(r = 5\mathrm{m}\), \(A_{s}=\frac{106.26}{360}\times3.14\times5^{2}\approx23.1\mathrm{m}^{2}\).

Answer:

(a) \(r = 5\mathrm{m}\)
(b)(i) The perimeter is approximately \(19.2\mathrm{m}\)
(b)(ii) The area is approximately \(23.1\mathrm{m}^{2}\)