All questions

• Radius of a sector of a circle is 3.5 cm and length of its arc is 2.2 cm. Find the area of the sector. Answer: Area of the sector = 3.85 cm² Area of a sector = 1/2 × length of arc × radius. A = 1/2 × 2.2 × 3.5 = 3.85 cm². Mensuration 1 answer
• Radius of a circle is 10 cm. Area of a sector of the circle is 100 cm². Find the area of its corresponding major sector. (π = 3.14) Answer: 214 cm² Area of circle =π r²=3.14×100=314 cm². Area of major sector = area of circle − area of minor sector =314-100=214 cm². Mensuration 1 answer
• Area of a sector of a circle of radius 15 cm is 30 cm². Find the length of the arc of the sector. Answer: 4 cm Area of sector =l× r/2 30=l×15/2⇒ l=60/15=4\ cm Mensuration 1 answer
• Radius of a circle with centre O is 7 cm and m(arc MBN) = 60°. Find (1) Area of the circle (2) A(O-MBN) (3) A(O-MCN). Answer: (1) 154 cm² (2) ≈ 25.67 cm² (3) ≈ 128.33 cm² (1) A=π r²=22/7×49=154 cm². (2) Minor sector: A(O-MBN)=60/360×154=77/3≈25.67 cm². (3) M... Mensuration 1 answer
• Radius of a circle with centre P is 3.4 cm and perimeter of sector P-ABC is 12.8 cm. Find A(P-ABC). Answer: A(P-ABC) = 10.2 cm² Perimeter of sector = 2r + length of arc ABC. 12.8=2(3.4)+l⇒ l=6\ cm A=l× r/2=6×3.4/2=10.2\ cm² Mensuration 1 answer
• O is the centre of the sector. ∠ROQ = ∠MON = 60°. OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. (π = 22/7) Answer: l(arc RXQ) = 22/3 ≈ 7.33 cm, l(arc MYN) = 22 cm Area of sector =θ/360×π r² ; length of arc =θ/360×2π r . l(RXQ)=60/360×2×22/7×7=22/3... Mensuration 1 answer
• If A(P-ABC) = 154 cm² and radius of the circle is 14 cm, find (1) ∠APC (2) l(arc ABC). Answer: (1) ∠APC = 90° (2) l(arc ABC) = 22 cm (1) θ/360×22/7×196=154⇒θ/360×616=154 θ=154×360/616=90^° (2) Area =l× r/2⇒154=l×14/2⇒ l=22 cm. Mensuration 1 answer
• Radius of a sector of a circle is 7 cm. If measure of arc of the sector is 30°, find the area of the sector. Answer: 154/12≈12.83 cm² Area of sector =θ/360×π r² ; length of arc =θ/360×2π r . Area of circle =22/7×49=154 cm². Area of sector =30/360×15... Mensuration 1 answer
• Radius of a sector of a circle is 7 cm. If measure of arc of the sector is 210°, find the area of the sector. Answer: 7×154/12≈89.83 cm² Area of sector =θ/360×π r² ; length of arc =θ/360×2π r . Area of circle =22/7×49=154 cm². Area of sector =210/360... Mensuration 1 answer
• Radius of a sector of a circle is 7 cm. If measure of arc of the sector is three right angles (270°), find the area of the sector. Answer: 3/4×154=115.5 cm² Area of sector =θ/360×π r² ; length of arc =θ/360×2π r . Area of circle =22/7×49=154 cm². Area of sector =270/360×... Mensuration 1 answer
• The area of a minor sector of a circle is 3.85 cm² and the measure of its central angle is 36°. Find the radius of the circle. Answer: r = 3.5 cm 36/360×22/7× r²=3.85 22/70r²=3.85⇒ r²=3.85×70/22=12.25 r=3.5\ cm Mensuration 1 answer
• □PQRS is a rectangle with PQ = 14 cm, QR = 21 cm. Find the areas of the parts x, y and z. Answer: x = 154 cm², y = 38.5 cm², z = 101.5 cm² x is a quarter circle with radius 14 cm: x=1/4×22/7×196=154 cm². y is a quarter circle with... Mensuration 1 answer
• ΔLMN is an equilateral triangle. LM = 14 cm. Three sectors are drawn with vertices as centres and radius 7 cm. Find (1) A(ΔLMN) (2) Area of any one of the sectors (3) Total area of all the three sectors (4) Area of the shaded region. Answer: (1) ≈ 84.87 cm² (2) ≈ 25.67 cm² (3) 77 cm² (4) ≈ 7.87 cm² (1) A(Δ LMN)=sqrt3/4×14²=49sqrt3≈49×1.732=84.87 cm². (2) Each angle of the... Mensuration 1 answer
• A is the centre of the circle. ∠ABC = 45° and AC = 7√2 cm. Find the area of segment BXC. Answer: ≈ 27.93 cm² (π = 3.14), or 28 cm² with π = 22/7 AB = AC (radii), so ∠ ACB=∠ ABC=45^° and ∠ BAC=90^° . Radius r=7sqrt2 , r²=98 . Area... Mensuration 1 answer
• O is the centre of the circle. m(arc PQR) = 60°, OP = 10 cm. Find the area of the shaded region (minor segment PQR). (π = 3.14, √3 = 1.73) Answer: ≈ 9.08 cm² ∠ POR=60^° and OP = OR, so ΔPOR is equilateral with side 10 cm. Area of segment = area of sector − area of triangle forme... Mensuration 1 answer
• A is the centre of the circle. ∠PAR = 30°, AP = 7.5, find the area of the segment PQR. (π = 3.14) Answer: 0.65625 sq. unit Area of segment = area of sector − area of triangle formed by the two radii and the chord. A(sector)=30/360×3.14×7.... Mensuration 1 answer
• O is the centre of the circle, PQ is a chord. ∠POQ = 90°, area of shaded region (minor segment PRQ) is 114 cm², find the radius of the circle. (π = 3.14) Answer: r = 20 cm Area of segment = area of sector − area of triangle formed by the two radii and the chord. 90/360×3.14 r²-1/2r²=114 0.785r... Mensuration 1 answer
• A chord PQ of a circle with radius 15 cm subtends an angle of 60° with the centre of the circle. Find the area of the minor as well as the major segment. (π = 3.14, √3 = 1.73) Answer: Minor segment ≈ 20.44 cm², major segment ≈ 686.06 cm² The triangle formed by the radii and chord PQ is equilateral (central angle 60... Mensuration 1 answer
• If tanθ = 2, find the values of other trigonometric ratios. Answer: sinθ = 2/√5, cosθ = 1/√5, cosecθ = √5/2, secθ = √5, cotθ = 1/2 cotθ=dfrac1tanθ=1/2 . sec²θ=1+tan²θ=1+4=5⇒secθ=sqrt5 cosθ=dfrac1sqrt5... Trigonometry 1 answer
• If secθ = 13/12, find the values of other trigonometric ratios. Answer: cosθ = 12/13, sinθ = 5/13, tanθ = 5/12, cosecθ = 13/5, cotθ = 12/5 cosθ=dfrac1secθ=12/13 . tan²θ=sec²θ-1=169-144/144=25/144 So tanθ=... Trigonometry 1 answer
• Prove that: secθ (1 – sinθ) (secθ + tanθ) = 1 Answer: LHS simplifies to RHS, as shown below. secθ(1-sinθ)=secθ-secθsinθ=secθ-tanθ , since sinθ/cosθ=tanθ . LHS=(secθ-tanθ)(secθ+tanθ) =sec... Trigonometry 1 answer
• Prove that: (secθ + tanθ) (1 – sinθ) = cosθ Answer: LHS simplifies to RHS, as shown below. secθ+tanθ=1+sinθ/cosθ LHS=(1+sinθ)(1-sinθ)/cosθ=1-sin²θ/cosθ =cos²θ/cosθ=cosθ=RHS Trigonometry 1 answer
• Prove that: cot²θ – tan²θ = cosec²θ – sec²θ Answer: LHS simplifies to RHS, as shown below. Using cot²θ=csc²θ-1 and tan²θ=sec²θ-1 : LHS=(csc²θ-1)-(sec²θ-1) =csc²θ-sec²θ=RHS Trigonometry 1 answer
• Prove that: tan⁴θ + tan²θ = sec⁴θ – sec²θ Answer: LHS simplifies to RHS, as shown below. LHS=tan²θ(tan²θ+1)=tan²θsec²θ =(sec²θ-1)sec²θ =sec⁴θ-sec²θ=RHS Trigonometry 1 answer