\(L=\frac{90}{360}\times2\pi(12)=\boxed{6\pi\text{ cm}}\).

\(A_A=\frac14\pi t^2\) and \(A_B=\frac{135}{360}\pi(2t)^2=\frac32\pi t^2\). Therefore \(k=\frac{A_B}{A_A}=\frac{3/2}{1/4}=\boxed{6}\).

The perimeter is \(12\times6=72\text{ cm}\). With length \(L\), \(2L+2(6)=72\). Thus \(L=\frac{72-12}{2}=\boxed{30\text{ cm}}\).

The major angle is \(360^\circ-50^\circ=310^\circ\). Its area is \(\frac{310}{360}\pi(6)^2=\boxed{31\pi\text{ cm}^2}\).

Include the semicircular arc and the straight diameter: \(P=\frac12\pi(15.8)+15.8=40.61858\ldots\), giving \(\boxed{40.6\text{ cm}}\) to 3 significant figures.

The trapezium area is \(\frac12(8.1+12.3)(6.2)=63.24\text{ cm}^2\). The triangle has hypotenuse \(8.5\text{ cm}\) and one leg \(5.1\text{ cm}\). Its other leg is \(\sqrt{8.5^2-5.1^2}=6.8\text{ cm}\), so its area is \(\frac12(5.1)(6.8)=17.34\text{ cm}^2\).

Total area \(=63.24+17.34=80.58\text{ cm}^2\), giving \(\boxed{80.6\text{ cm}^2}\).

Opposite sides are equal, so \(P=2(12+9)=\boxed{42\text{ cm}}\).

The parallel sides are 15 cm and 9 cm; their perpendicular separation is 8 cm. \(A=\frac12(15+9)(8)=\boxed{96\text{ cm}^2}\).

The chord lies \(3-2.5=0.5\text{ m}\) below the centre. Drop a perpendicular from O to the chord. Half the minor central angle satisfies \(\cos(\theta/2)=0.5/2.5\), so \(\theta=2\cos^{-1}(0.2)=156.926081\ldots^\circ\).

The minor segment area is its sector area minus its triangle area: \(A_{\rm minor}=\frac{\theta}{360}\pi(2.5)^2-\frac12(2.5)^2\sin\theta\). The major segment area is \(\pi(2.5)^2-A_{\rm minor}=12.300708\ldots\text{ m}^2\).

Volume \(=12.300708\ldots(800)=9840.567135\ldots\), giving \(\boxed{9840\text{ m}^3}\) to 3 significant figures.

Arc length \(=\frac{45}{360}(2\pi\times18)=\frac92\pi\), so \(\boxed{n=9/2}\).

The major angle is \(360^\circ-40^\circ=320^\circ\). Arc length \(=\frac{320}{360}(2\pi\times6)=\boxed{\frac{32\pi}3\text{ cm}}\).

The major angle is \(360^\circ-126^\circ=234^\circ\). Perimeter \(=2(7.5)+\frac{234}{360}(2\pi\times7.5)=45.630528\ldots\), giving \(\boxed{45.6\text{ cm}}\).

\(A=\frac12(10+12)(8)=\boxed{88\text{ cm}^2}\).

The sector radius is \((2/3)(12)=8\text{ cm}\) and its angle is \(60^\circ\). Shaded area \(=\frac12(12)(12)\sin60^\circ-\frac{60}{360}\pi(8)^2=28.843507\ldots\), giving \(\boxed{28.8\text{ cm}^2}\).

Each trapezium has parallel sides x and 11.4 and perpendicular separation 7 cm. Total area \(=2\times\frac12(x+11.4)\times7\). Thus \(7(x+11.4)=130.2\), so \(x+11.4=18.6\), giving \(\boxed{x=7.2\text{ cm}}\).

\(\frac12(8)h=12\), giving \(\boxed{h=3\text{ cm}}\).

The two radii contribute 18 cm, so arc length is \(2\pi\text{ cm}\). The full circumference is \(18\pi\), so the sector is \(1/9\) of the circle. Its area is \(\frac19\pi(9)^2=\boxed{9\pi\text{ cm}^2}\).

(a) \(\frac{60}{360}\pi(12)^2=\boxed{24\pi\text{ cm}^2}\). (b) The major angle is 300°, so its arc length is \(\frac{300}{360}(2\pi)(12)=\boxed{20\pi\text{ cm}}\).

Two separate rectangles have total perimeter \(2[2(19+5)]=96\text{ cm}\). The joined 5 cm edge is counted twice and becomes internal. Thus \(P=96-2(5)=\boxed{86\text{ cm}}\).

(a) \(2(6.51+x)=2\pi(4)\), so \(x=4\pi-6.51=6.056370\ldots=6.06\text{ m}\).

(b) The height is \(6.06\cos36^\circ\), since the 36° angle is measured from the vertical. Area \(=6.51(6.06)\cos36^\circ\). Cost \(=6.51(6.06)\cos36^\circ(18)=574.491705\ldots\), giving \(\boxed{\$574}\) to 3 significant figures.

Use the perpendicular height 3.8 cm: \(A=\frac12(5.3+8.7)(3.8)=\boxed{26.6\text{ cm}^2}\).

(a) The remaining angle is 310°. Perimeter includes the major arc and two radii: \(P=(310/360)(2\pi\times12)+24=\boxed{88.9\text{ cm}}\). (b) Remaining cross-section area \(=(310/360)\pi(12)^2\). Thus \(h=3510/[(310/360)\pi(12)^2]=\boxed{9.01\text{ cm}}\).

The shaded angle is 360−48=312°. Area \(=\frac{312}{360}\pi(9)^2=\boxed{221\text{ cm}^2}\) to 3 significant figures.

(a) Radius=6.4 cm. Area \(=\frac12(12.8)^2+\frac12\pi(6.4)^2=\boxed{146\text{ cm}^2}\) to 3 significant figures. (b) KL=12.8√2. The joining diameter JL is internal, so perimeter \(=12.8+12.8\sqrt2+6.4\pi=\boxed{51.0\text{ cm}}\).

(a) Both arcs and two radial sides are on the perimeter: \(\frac{60}{360}2\pi(17+10)+2(17-10)=9\pi+14\). Thus \(\boxed{a=9,b=14}\). (b)(i) Divide into 6 equilateral triangles, each area½s²sin 60°. \(127.3=3s^2\sin60^\circ\), so \(s=\sqrt{127.3/(3\sin60^\circ)}=6.999\ldots=\boxed{7.0\text{ cm}}\). (ii)(a) \(127.3(10)=\boxed{1273\text{ cm}^3}\). (ii)(b) Two ends plus six rectangles: \(2(127.3)+6(7)(10)=674.6\approx\boxed{675\text{ cm}^2}\).

(a) Arc lengths are proportional to their central angles, with minor:major=3:7. Thus \(x=360\times3/(3+7)=\boxed{108}\). (b)(i) \(\frac12r^2\sin y=\frac12\frac{y}{360}\pi r^2\). Cancel r²/2 and multiply by 360: \(\boxed{360\sin y=\pi y}\). (ii) Calculate using degree mode and π.

(iii) At 108.6 the two sides are closest; their difference changes sign just above 108.6. To 1 decimal place, \(\boxed{y=108.6^\circ}\).

\(\frac12(x+9.5)(6)=42\). Thus \(x+9.5=14\), so \(\boxed{x=4.5\text{ cm}}\).

The side squared is 72. The diagonal is the diameter d, so \(d^2=72+72=144\). Thus r=d/2=6 and circle area \(=\boxed{36\pi\text{ cm}^2}\).

(a)(i) \((2x+3)(x+1)=190\), giving \(2x^2+5x-187=0\). Factor: \((2x-17)(x+11)=0\). Reject x=−11 because lengths would be negative; thus x=8.5. (ii) Sides=20 and 9.5 cm, so \(P=2(20+9.5)=\boxed{59\text{ cm}}\).

(b)(i) Segment area=sector−triangle: \(r^2\bigl(\frac{50\pi}{360}-\frac12\sin50^\circ\bigr)=30\). Therefore \(r=\sqrt{\frac{30}{50\pi/360-\frac12\sin50^\circ}}=23.72\ldots\text{ cm}\), giving 23.7 cm. (ii) Perimeter=arc+chord=\(\frac{50}{360}(2\pi r)+2r\sin25^\circ=\boxed{40.8\text{ cm}}\), using the unrounded radius.

(a)(i) \(\tfrac12(9+13)(8)=\boxed{88\text{ cm}^2}\). (ii)(a) \(\tfrac12[(y+4)+(y+1)](y+2)=264\). Multiply by 2 and expand: \(2y^2+9y+10=528\), hence \(2y^2+9y-518=0\). (b) \((2y+37)(y-14)=0\). Roots−18.5 and 14; lengths must be positive, so \(\boxed{y=14}\). (b) Perimeter=\(8+8+\frac{75}{360}(2\pi8)=\boxed{26.5\text{ cm}}\). (c) \(OA^2=5^2+5^2=50\). Sector area=\(\frac{90}{360}\pi(50)=25\pi/2\). Two triangles contribute \(2(\tfrac12)(5)(5)=25\). Total=\(\boxed{25+25\pi/2\text{ cm}^2}\).

Bisect the isosceles triangle formed by the two radii. Half the central angle is \(\sin^{-1}(3/9)\).

\(\theta=2\sin^{-1}(1/3)=0.6796738\text{ radians}\). Arc length \(=r\theta=9(0.6796738)=6.12\text{ cm}\).

Outer radius \(OB=3.2+2.6=5.8\).

\(A=\frac{140}{360}\pi(5.8^2-3.2^2)=9.1\pi\), so \(k=9.1\).

(a)(i) \(\frac12(24)(8)=96\text{ cm}^2\). (ii) \(\tan^{-1}(8/24)=18.4^\circ\).

(b) \(\frac12\pi(6)^2(11)=198\pi=622\text{ cm}^3\).

(c)(i) The removed part is a \(4\times4\) square plus three quarters of the circle: \(16+\frac34\pi(4)^2=16+12\pi\).

Remaining area \(20(15)-16-12\pi=246.300888\ldots\approx246\text{ cm}^2\).

(ii) Straight edges total \((20-4)+(15-4)+20+15=62\). Arc length \(\frac34(2\pi\times4)=6\pi\). Total \(62+6\pi=80.84956\ldots=80.8\text{ cm}\).

(a)(i) Each sector is a quarter circle of radius 11 cm. Area \(=11^2+2\left(\frac14\pi11^2\right)=311.066\ldots=311\text{ cm}^2\).

(ii) Straight edges total \(4(11)=44\) cm and the two arcs total \(\pi(11)\). Perimeter \(=44+11\pi=78.6\text{ cm}\).

(b) Projection of \(AG\) onto the base is \(AC\), with \(AC=7\sqrt2\). \(\tan\theta=7/(7\sqrt2)=1/\sqrt2\), so \(\theta=35.3^\circ\).

Rhombus area \(R=13.6^2\sin41^\circ\). One sector area \(S=\frac{41}{360}\pi13.6^2\). The unshaded lens is the overlap of the sectors, area \(2S-R\).

Shaded area \(=R-(2S-R)=2R-2S=2(13.6)^2\left(\sin41^\circ-\frac{41\pi}{360}\right)=110\text{ cm}^2\).

(a)(i) \(\sin\angle BOC=2/6\), so \(\angle BOC=19.4712^\circ\approx19.5^\circ\). (ii) Minor \(\angle AOB=135^\circ+19.4712^\circ\). Major angle is \(360^\circ-135^\circ-19.4712^\circ\). Sector area \(=\frac{360-135-19.4712}{360}\pi6^2=64.6\text{ cm}^2\).

(iii) \(OC=\sqrt{6^2-2^2}=\sqrt{32}\), \(OD=2\sqrt{32}\). Cosine rule \(AD^2=6^2+(2\sqrt{32})^2-2(6)(2\sqrt{32})\cos135^\circ=260\). Thus \(AD=\sqrt{260}=16.1\text{ cm}\).

(iv) Triangle \(OBC\) has area \(\frac12(2)\sqrt{32}=\sqrt{32}\). Triangle \(AOD\) has area \(\frac12(6)(2\sqrt{32})\sin135^\circ=24\). Total, using unrounded sector area, \(64.56877+\sqrt{32}+24=94.2\text{ cm}^2\).

(b) Areas scale by the square of length factor: \(160(20/8)^2=1000\text{ cm}^2\).

\(2\pi(4.7)=29.5\text{ cm}\).

Rectangle area \(11(4)=44\). Sector radius 4 cm and angle \(\theta=\tan^{-1}(11/4)=70.0169^\circ\).

\(A_{\rm sector}=\frac{\theta}{360}\pi4^2\). Shaded percentage \(100\frac{44-A_{\rm sector}}{44}=77.8\%\).

Sector radius \(k/2\) and angle 45°. Sector area \(\frac{45}{360}\pi(k/2)^2=\pi k^2/32\). Shaded percentage \(100(1-\pi/32)=90.2\%\).

(a) \((12-2)180/12=150\)°. (b)(i) Central angle 30°. Bisect triangle OAB: \(r=3/\sin15^\circ=11.5911\ldots=11.6\text{ cm}\). (ii)(a) \(2\pi r=72.8\text{ cm}\). (b) Arc AB is one twelfth of the circumference. Segment perimeter \(6+2\pi r/12=12.1\text{ cm}\).

(c) Apothem \(r\cos15^\circ\). Twelve triangle areas give prism volume \(12\left(\frac12(6)r\cos15^\circ\right)(2)=806\text{ cm}^3\) using unrounded r.

Arc length \(26-2(8)=10\) cm. \(\theta=\frac{10}{2\pi(8)}360=71.6^\circ\).

(a) Upper bounds 11.45, 14.85 and 15.75 give \(11.45+14.85+15.75=42.05\) cm. (b) \(A=\frac{150}{360}\pi(15.6)^2=318.557\ldots=319\text{ cm}^2\).

(c) Include both radii in each perimeter: \(\frac{360-x}{360}2\pi r+2r=3\left(\frac{x}{360}2\pi r+2r\right)\). Collecting terms gives \(\frac{4x}{360}2\pi r=2\pi r-4r\). Divide by positive r and solve: \(x=\frac{360(2\pi-4)}{8\pi}=\frac{90(\pi-2)}\pi\).

\(A=\frac14\pi5^2+\frac12(5)(6)=34.63495\ldots=34.6\text{ cm}^2\).

Each horizontal extension is \(d=18.2/\tan62^\circ\). Lower base \(15.4+2d\). Area \(A=\tfrac12(15.4+15.4+2d)(18.2)=456.4\ldots=456\text{ cm}^2\).

(a) The boundary is the 300° major arc and chord BC. \(P=\frac{300}{360}2\pi(12.4)+12.4=77.3\text{ cm}\). (b) Major angle = 319°. \(\frac{319}{360}\pi r^2=74.5\), so \(r=\sqrt{\frac{74.5(360)}{319\pi}}=5.17\text{ cm}\).

Perimeter = minor arc + chord. Arc \(\frac{80}{360}2\pi(12.6)=17.5929\ldots\) cm. Chord \(2(12.6)\sin40^\circ=16.1982\ldots\) cm. Total \(33.8\) cm to 3 significant figures.

(a) \(2(x+x+3)=20\) gives \(x=3.5\) cm. (b) Each rhombus side is 5 cm and the half-side at M is 2.5 cm. In the triangle containing the 20° angle, sine rule gives \(\sin\angle M=5\sin20^\circ/2.5\). Taking the acute angle at M (since y is obtuse), \(\angle M=43.1602\ldots^\circ\). Thus \(y=180-20-43.1602\ldots=116.8^\circ\). (c) \(2r+\frac{40}{360}2\pi r=20\) gives \(r=\frac{20}{2+2\pi/9}=7.4125366\ldots\) cm. Bisect the central angle: \(z=2r\sin20^\circ=5.07047\ldots\approx5.07\) cm.

Split the hexagon into six equilateral triangles. \(A=6\left(\frac12(7.4)^2\sin60^\circ\right)=142.2716\ldots=142\text{ cm}^2\).

Perimeter includes the diameter: \(19.02=\pi r+2r\). Thus \(r=\frac{19.02}{\pi+2}=3.70\text{ cm}\).

Radius = 6.5 cm. \(A=\frac{42}{360}\pi(6.5)^2=15.5\text{ cm}^2\).

Separate perimeters total \(5\times2(3+8)=110\) cm. Four joins each hide a 3 cm edge of each of two rectangles, so perimeter \(110-2(4)(3)=86\) cm.

Arc length \(s=r\theta\) and area \(A=r^2\theta/2\), so \(A=\frac12rs=\frac12(8)(6.4)=25.6\text{ cm}^2\).

\(A=\frac12(11+7)(5)=45\text{ cm}^2\).

\(A=\frac{60}{360}\pi(7.5)^2=29.5\text{ cm}^2\).

\(27=.5(6)h=3h\), so h=9.

(a) \(OP=8\cos30^\circ\) and OA=8, so \(AP=8-8\cos30^\circ=1.07\text{ cm}\).

(b) BP=8 sin30°=4 cm. Subtract right triangle OPB from the sector: \(A=\frac{30}{360}\pi(8^2)-\frac12(8\cos30^\circ)(4)=2.90\text{ cm}^2\).

(a) \(P<2(8.55+10.75)=38.6\) cm, so the upper bound is 38.6 cm.

(b)(i) FD=AB=9. \(h=9\sin80^\circ=8.86\) cm. (ii) \(\angle CDF=180^\circ-80^\circ=100^\circ\); then \(\angle DCF=180^\circ-100^\circ-40^\circ=40^\circ\). Angles DCF and CFD are equal, so DF=CD. (iii) CD=9 and BD=12, hence BC=3. \(A=\tfrac12(12+3)(9\sin80^\circ)=66.5\text{ cm}^2\).

(c) Angles ABD and ACD subtend the same chord AD, so angle ACD=21°. Angle ADC=90° because AC is a diameter. Thus \(AC=12/\cos21^\circ\) and \(A=\pi\left(\frac{6}{\cos21^\circ}\right)^2=130\text{ cm}^2\) to 3 significant figures.

(d) The arc is \(4(8)-2(9.5)=13\) cm. \(\frac{x}{360}\,2\pi(9.5)=13\Rightarrow x=\frac{13(360)}{19\pi}=78.4^\circ\).

(a) \(\cos\angle EOD=OE/OD=.25/.5=1/2\), so angle EOD=60°. Minor angle AOD=90°+60°=150°, hence major sector210°. Outer straight edges total \(1.5+.5+1.5+.5=4\) cm. Perimeter \(4+\frac{210}{360}(2\pi(.5))=5.83\text{ cm}\).

(b) \(A=.5(1.5)+\frac{210}{360}\pi(.5)^2=1.2081489\ldots\approx1.21\text{ cm}^2\). (c)(i) Linear scale factor \(\sqrt{77.44/A}\), using unrounded A. Radius \(.5\sqrt{77.44/A}=4.00\text{ cm}\) to3 significant figures. (ii)15 mm=1.5 cm, volume \(77.44(1.5)=116.16\) cm³, mass \(116.16(19)/1000=2.20704\) kg.

\(.5(8.4)(3.5)=14.7\) cm².

\(A=\frac{217}{360}\pi(6.2)^2=72.7949\ldots\approx72.8\) cm².

\(A=\frac{45}{360}\pi(5^2-3^2)=2\pi=6.28\) cm².

(a) The square diagonal is the diameter, 16 cm. Area=\(16^2/2=128\) cm². (b)(i) The central angle for one side is 90°. Segment area=quarter-circle area minus the centre triangle: \(\frac14\pi8^2-\frac12(8)(8)=16\pi-32=18.3\) cm². (ii) Arc length=\(\frac14(2\pi8)=4\pi\) cm, chord=\(8\sqrt2\) cm. Total \(4\pi+8\sqrt2=23.9\) cm.

\(\pi(5.1)^2=81.7128\ldots\approx81.7\) cm².

Triangle area=\(\frac{\sqrt3}4(10)^2=25\sqrt3\). Sector radius 5 cm and angle 60°; area=\(\frac{60}{360}\pi(5)^2=\frac{25\pi}6\). Shaded area=\(25\sqrt3-\frac{25\pi}6=30.211\ldots\approx30.2\) cm².

Two radii total 12 cm. Arc length=\(\frac{72}{360}\times2\pi(6)=\frac{12\pi}5\). Hence p=12 and q=12/5.

(a) OB is perpendicular to ABC. \(\angle AOB=\tan^{-1}(15/8),\quad\angle BOC=\tan^{-1}(22.4/8)\). Their sum is 132.2737…°≈132.3°. (b) \(\frac{132.2737}{360}\times2\pi(8)\approx18.5\) cm. (c) Area of two right triangles=\(\frac12(15+22.4)(8)=149.6\) cm². Subtract sector XBY: \(149.6-\frac{132.2737}{360}\pi(8)^2\approx75.7\) cm².

The lens is the two sectors together minus the rhombus: \(2\left(\frac{40}{360}\pi10^2\right)-10^2\sin40^\circ=5.5344\ldots\approx5.53\) cm². Equivalently subtract the isosceles triangle area from each sector and double.

\(\frac12(33)h=528\Rightarrow h=1056/33=32\) cm.

Sector area=\((30/360)\pi6^2=3\pi\). Triangle area=\(\frac12(6)^2\sin30^\circ=9\). Segment=\(3\pi-9\) cm², so k=3,c=9.

Square area=10²=100 cm²; semicircle radius=5 cm. Total=\(100+\frac12\pi5^2=139.2699\ldots\approx139\) cm².

(a) The right isosceles triangle has legs x, so area=\(x^2/2\) and hypotenuse=\(x\sqrt2\), the rectangle’s width. The rectangle area is \(2\sqrt2x^2\). Sector area=\(3\pi x^2/4\). Thus \(k=3\pi/4+1/2+2\sqrt2=5.68462\ldots\approx5.68\). (b) \(x=\sqrt{110/k}\approx4.40\) cm.

\(AB=\frac{38}{360}(2\pi\times25)=16.58062789\ldots\) cm, which is 16.58 cm to 4 significant figures.

The fraction of the full circumference is \(\frac{6\pi}{2\pi(9)}=\frac13\). Hence the sector area is \(\frac13\pi(9^2)=27\pi\) cm², so \(k=27\).

The horizontal difference of the parallel sides is \(12-8=4\) cm. By Pythagoras, the perpendicular height is \(h=\sqrt{8^2-4^2}=\sqrt{48}=4\sqrt3\) cm. The area is \(\frac12(12+8)(4\sqrt3)=40\sqrt3=69.2820\ldots\approx69.3\) cm².

\(A=\text{base}\times\text{height}\), so \(5x=51.5\) and \(x=10.3\) cm.

\(A=32(8)+\frac{125}{360}\pi(48^2-40^2)=256+\frac{2200\pi}{9}=1023.9448\ldots\), giving 1024 cm².

\(A=\frac12(13+16)(4)=58\) cm². The perpendicular height is 4 cm.

(a) Six outside edges, each 5 cm, give \(6(5)=30\) cm. (b) Square area \(5^2=25\) cm². Rhombus area \(5(4.5)=22.5\) cm². Total \(25+22.5=47.5\) cm².

(a) \(\frac{w}{360}\pi r^2=r^2\Rightarrow w=360/\pi\approx114.6^\circ\). (b) Subtract the two radii to obtain arc length \(7\pi r/10\). Hence \(\frac{x}{360}(2\pi r)=\frac{7\pi r}{10}\Rightarrow x=126^\circ\). (c) The base is \(q\sqrt3\). The cosine rule gives \(3q^2=q^2+q^2-2q^2\cos y\Rightarrow\cos y=-1/2\). Therefore \(y=120^\circ\).

(a) \(2(3x+x)=60\Rightarrow x=7.5\) cm. (b) The hypotenuse is \(\sqrt{(3x)^2+(4x)^2}=5x\). Perimeter \(3x+4x+5x=12x=60\), so x = 5 cm. (c) The sector angle is 90°. Perimeter \(2x+\frac14(2\pi x)=60\). Hence \(x=\frac{60}{2+\pi/2}\approx16.8\) cm.

(a) Half AB is 12 cm and the perpendicular distance from O to AB is 18 − r. Pythagoras gives \(12^2+(18-r)^2=r^2\Rightarrow144+324-36r=0\Rightarrow r=13\) cm. (b) \(\theta=2\sin^{-1}(12/13)=134.760\ldots^\circ\approx134.8^\circ\). (c)(i) Use the unrounded angle: \(A_{\rm major}=\frac{360-\theta}{360}\pi(13)^2\approx332.2\) cm². (ii) Add triangle OAB, whose height is 5 cm: \(A_{\rm water}=A_{\rm major}+\tfrac12(24)(5)\approx392.2\) cm². (iii) \(V=40A_{\rm water}\approx15700\) cm³ (3 significant figures). (d) Water volume is unchanged: \(h=\frac{40A_{\rm water}}{\pi(13)^2}\approx29.5\) cm.

(a) Equal square sides give \(x-4=7\Rightarrow x=11\). (b) Square perimeter \(4(7)=28\) cm. The triangle’s equal sides are 10 cm, so \(FG=28-2(10)=8\) cm.

(a) \(2\pi r=30\Rightarrow r=15/\pi\approx4.77\) cm. (b) \(A=\tfrac12\pi(15/\pi)^2=225/(2\pi)\approx35.8\) cm².

Include the arc and diameter: \(P=\pi(6.1)+2(6.1)=31.3637\ldots\approx31.4\) cm.

Perimeter includes two radii and the arc: \(P=30+(26/360)(2\pi\cdot15)=36.8068\ldots\approx36.8\) cm.

Subtract triangle OBC from sector OAB. \(BC=8\sin30^\circ=4\) and \(OC=8\cos30^\circ=4\sqrt3\). Hence \(A=(30/360)\pi(8)^2-\tfrac12(4)(4\sqrt3)=16\pi/3-8\sqrt3=2.8987\ldots\approx2.90\) cm².

Rectangle height \(AE=8\sin30^\circ=4\) cm and \(ED=8\cos30^\circ=4\sqrt3\) cm. Rectangle area = 32 cm². Triangle ADE area \(\tfrac12(4)(4\sqrt3)=8\sqrt3\) cm²; sector DAB area \((30/360)\pi(8)^2=16\pi/3\) cm². Shaded area \(32-8\sqrt3-16\pi/3=1.3886\ldots\approx1.39\) cm².

Each sector is 120°. The shaded outer arcs total 240°: \((2/3)(2\pi6)=8\pi\) cm. The exposed inner arcs from the shaded ring and inner sector together form a full small circumference: \(2\pi4=8\pi\) cm. Straight edges: two lengths 6 − 4 and two radii of 4, giving 12 cm. Total \(16\pi+12=62.265\ldots\approx62.3\) cm.

First: equal base segments give equal triangular areas, so ⅓. Second: \(72/360=1/5\). Third: linear factor ½ gives area factor ¼. Fourth: the shaded triangle is one sixth of a regular hexagon’s area, so ⅙. Fifth: for radius r, sector area \(\pi r^2/4\) and triangle area \(r^2/2\). Thus \(k=\frac{\pi r^2/4-r^2/2}{\pi r^2/4}=1-2/\pi\approx0.363\).

(a) ABC and ABD are equilateral; each contributes 60° at B. (b) \(S=(120/360)\pi(8)^2=64\pi/3\approx67.0\) cm². (c)(i) Subtract triangle BCD: \(S-\tfrac12(8)^2\sin120^\circ=39.308\ldots\approx39.3\) cm². (ii) Twice this segment area = 78.6 cm².

(a) \(A=(56/360)\pi(2.25)^2=2.474004\ldots\approx2.47\) m². (b) \(V=0.3A=0.742201\ldots\approx0.742\) m³.

(a) \(A=\tfrac12(8)^2\sin56^\circ=26.5292\ldots\approx26.5\) cm². (b)(i) \((x/360)\pi(6.5)^2=A\Rightarrow x=71.953\ldots^\circ\approx72.0^\circ\). (ii) \(P=13+(x/360)2\pi(6.5)=13+2A/6.5=21.1628\ldots\approx21.2\) cm. (c)(i) Sector minus triangle: \(\frac{30}{360}\pi r^2-\frac12r^2\sin30^\circ=\frac{\pi r^2}{12}-\frac{r^2}4=\frac{r^2}4(\pi/3-1)\). (ii) \(r=\sqrt{20/(\pi/3-1)}=20.585\ldots\approx20.6\) cm.

\(P=2(12)+\tfrac{135}{360}\,2\pi(12)=24+9\pi=52.2743\ldots\approx52.3\) cm.

\(42=\tfrac12(BC)(8)\Rightarrow BC=10.5\) cm.

\(A=\frac{120}{360}\pi(5)^2-\tfrac12(5)(5)\sin120^\circ=15.3546\ldots\approx15.4\) cm².

Split into two congruent halves. Each has a 4 cm by 10 cm vertical rectangle and an 18 cm by 5 cm horizontal rectangle. \(A=2[4(10)+18(5)]=260\) cm².

\(C=2\pi(2.5)=5\pi=15.7079\ldots\approx15.7\) cm.

Parallel sides are 12 cm and 22 cm, perpendicular height 10 cm. \(A=\tfrac12(12+22)(10)=170\) cm².

(a) Start with the lower semicircle and subtract the segment below AB. \(A=\tfrac12\pi(3)^2-\left[\tfrac{120}{360}\pi(3)^2-\tfrac12(3)(3)\sin120^\circ\right]=1.5\pi+\tfrac{9\sqrt3}4=8.6095\ldots\approx8.61\) m². (b) \(V=50A=430.4751\ldots\approx430\) m³.

OT is perpendicular to PT, so \(PT=5\tan78^\circ\). Tangents from P have equal length and OP bisects ROT; minor ROT = 156°, giving major arc angle 204°. \(L=2(5\tan78^\circ)+(204/360)(2\pi\cdot5)=64.8487\ldots\approx64.8\) cm.

\(2\pi r=2(8+6)=28\Rightarrow r=14/\pi=4.4563\ldots\approx4.46\) cm.

\(A=\pi(6^2-0.5^2)=35.75\pi\approx112\) cm².

\(8=\frac{x}{360}\,2\pi(4)\Rightarrow x=\frac{360}{\pi}=114.59\ldots\approx114.6^\circ\).

Sector area is half radius × arc length: \(A=\tfrac12(5r)(4r)=10r^2\).

(a) Radius is k/2, so \(A=k^2-\pi(k/2)^2=k^2-\pi k^2/4\). Multiply by 4. (b) \(4A=(4-\pi)k^2\Rightarrow k=\sqrt{4A/(4-\pi)}\), taking the positive root for the side length.

(a) \(A=\pi(50)^2=2500\pi\approx7854\) cm². (b) Divide by 10000: \(0.25\pi\approx0.7854\) m².

Subtract the triangle from the sector: \(A=(50/360)\pi(9)^2-\tfrac12(9)^2\sin50^\circ=4.317\ldots\approx4.32\) cm².

\(A=0.8(1.4)+\tfrac14\pi(0.8)^2=1.12+0.16\pi\approx1.62\) m².

(a) \(A=64\pi\approx201\) m². (b) \(L=4[(45/360)2\pi(12)]+2\pi(8)=12\pi+16\pi=28\pi\approx88.0\) m.

(a) \(A=(40/360)\pi(5.6)^2\approx10.9\) cm². (b) Include both radii and the arc: \(P=2(5.6)+(40/360)2\pi(5.6)\approx15.1\) cm.

The straight boundary totals \(8+6+6+(8-6)=22\) cm. The semicircle arc is \(\pi(3)\) cm. Total \(22+3\pi\approx31.4\) cm.

\(18^2-\pi(5^2)=324-25\pi\approx245.46\) cm², or 245 cm² to 3 significant figures.

Remove the two radii: arc length \(14\pi\). Thus \((x/360)(2\pi\cdot8)=14\pi\Rightarrow x=315^\circ\).