#ssc 10th
- Q.1 (A)(3) Distance of point (3, 4) from the origin is .................... . Answer: C. 5 Answer: (C) 5 Distance from origin = √(x² + y²) = √(9 + 16) = √(25) = 5 . A distance can never be negative. 1
- Q.1 (B)(3) Angle made by a line with the positive direction of X-axis is 30º. Find slope of that line. Answer: Slope = 1/sqrt3 Slope m = tanθ = tan 30^° = 1/√(3) . 1
- Q.1 (B)(4) In cyclic quadrilateral ABCD m∠A = 100º, then find m∠C. Answer: m∠C = 80º Opposite angles of a cyclic quadrilateral are supplementary: m∠C = 180º – 100º = 80º. 1
- Q.2 (B)(4) Verify, whether points, A(1, –3), B(2, –5) and C(–4, 7) are collinear or not. Answer: Yes, A, B and C are collinear. Slope of AB = -5 - (-3)/2 - 1 = -2 . Slope of BC = 7 - (-5)/-4 - 2 = 12/-6 = -2 . Slopes of AB and BC... 1
- Q.2 (B)(5) If \(\sin\theta = \dfrac{11}{61}\), find the values of cos θ using trigonometric identity. Answer: cosθ = 60/61 sin²θ + cos²θ = 1 cos²θ = 1 - 121/3721 = 3600/3721 cosθ = 60/61 (taking θ acute). 1
- Q.1 (A)(2) ∠ACB is inscribed angle in a circle with centre O. If m∠ACB = 65º, then what is measure of its intercepted arc AXB ? Answer: D. 130º Answer: (D) 130º The measure of an inscribed angle is half the measure of its intercepted arc, so m(arc AXB) = 2 × 65º = 130... 1
- Q.1 (A)(4) If radius of cone is 5 cm and its perpendicular height is 12 cm, then the slant height is ..................... . Answer: C. 13 cm Answer: (C) 13 cm l = √(r² + h²) = √(25 + 144) = √(169) = 13 cm. 1
- Q.1 (B)(1) In the following figure Δ ABC, B – D – C and BD = 7, BC = 20, then find \(\dfrac{A(\Delta ABD)}{A(\Delta ABC)}\).[Figure: Δ ABC with point D on side BC; segment AD drawn.] Answer: 7/20 Δ ABD and Δ ABC have the same height (the perpendicular from A to BC), so the ratio of their areas equals the ratio of their ba... 1
- Q.1 (B)(2) In the following figure ∠MNP = 90º, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.[Figure: right-angled Δ MNP with the right angle at N; Q on hypotenuse MP with NQ ⊥ MP.] Answer: NQ = 6 By the theorem of geometric mean, NQ² = MQ × QP = 9 × 4 = 36 , so NQ = 6. 1
- Q.2 (A)(1) The radius of a circle with centre ‘P’ is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (π = 3.14)[Figure: circle with centre P, chord AB, ∠APB = 90º, X on minor ar… Answer: Area of minor sector = 78.5 sq. cm A(P–AXB) = θ/360 × π r² = 90/360 × 3.14 × 10² = 1/4 × 314 = 78.5 sq. cm. 1
- Q.2 (A)(2) In the following figure chord MN and chord RS intersect at point D. If RD = 15, DS = 4, MD = 8, find DN by completing the following activity :[Figure: circle with chords MN and RS intersecting at D inside the circle.]Activity :∴ MD × DN =… Answer: DN = 7.5 MD × DN = RD × DS (theorem of internal division of chords). 8 × DN = 15 × 4 = 60, so DN = 60/8 = 7.5. 1
- Q.2 (A)(3) An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60º. To find the height of tree complete the activity. (\(\sqrt3\) = 1.73)[Figure: right-angled Δ ABC, right angle at B; AB is the tree, BC… Answer: Height of the tree = 17.3 m tan θ = AB/BC ... (I); tan 60º = sqrt3 ... (II). So AB = BC × sqrt3 = 10sqrt3 = 10 × 1.73 = 17.3 m. 1
- Q.2 (B)(1) In Δ ABC, DE || BC. If DB = 5.4 cm, AD = 1.8 cm, EC = 7.2 cm, then find AE.[Figure: Δ ABC with D on AB and E on AC, DE parallel to BC.] Answer: AE = 2.4 cm By the basic proportionality theorem, AD/DB = AE/EC . 1.8/5.4 = AE/7.2 , so AE = 1.8 × 7.2/5.4 = 7.2/3 = 2.4 cm. 1
- Q.2 (B)(2) In the figure given below, find RS and PS using the information given in Δ PSR.[Figure: Δ PSR with ∠S = 90º, ∠P = 30º and hypotenuse RP = 12.] Answer: RS = 6, PS = 6sqrt3 In Δ PSR, ∠S = 90º, ∠P = 30º, so ∠R = 60º; it is a 30º–60º–90º triangle. Side opposite 30º = 1/2 × hypotenuse: R... 1
- Q.2 (B)(3) In the following figure, circle with centre D touches the sides of ∠ACB at A and B. If ∠ACB = 52º, find measure of ∠ADB.[Figure: circle with centre D; CA and CB are tangents touching the circle at A and B; radii DA and DB drawn.] Answer: m∠ADB = 128º A radius is perpendicular to the tangent at the point of contact, so ∠DAC = ∠DBC = 90º. The sum of the angles of quadri... 1
- Q.3 (A)(1) In the following figure, XY || seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.[Figure: Δ ABC with X on side AB and Y on side BC, XY parallel to AC.]Activity :2AX = 3BX (Given)∴ \(\dfrac{AX}{BX} = \dfrac{3}{\… Answer: AC = 22.5 AX/BX = 3/2 ; by componendo AX + BX/BX = 5/2 , i.e. AB/BX = 5/2 ... (I). Δ BCA ~ Δ BYX by the AA test (∠B common, ∠BXY ≅ ∠... 1
- Q.3 (A)(2) Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.[Figure: rhombus PQRS with diagonals PR and QS intersecting at T.]Given : PQRS is a rhombus. Diagona… Answer: the blanks are 2PT², 2RT², RT², PT², 2PT and QS². Diagonals of a rhombus bisect each other, so T is the midpoint of PR and of QS (PT... 1
- Q.3 (B)(1) Show that points P(1, –2), Q(5, 2), R(3, –1), S(–1, –5) are the vertices of a parallelogram. Answer: The diagonals PR and QS have the same midpoint (2, –3/2), so PQRS is a parallelogram. Midpoint of diagonal PR = (1 + 3/2, -2 - 1/2)... 1
- Q.3 (B)(2) Prove that tangent segments drawn from an external point to a circle are congruent. Answer: Proof by the hypotenuse-side test (Δ PAO ≅ Δ PBO). Given: a circle with centre O; P is an external point; PA and PB are tangent segm... 1
- Q.3 (B)(3) Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre. Answer: Construction steps (each tangent will measure about 6 cm). Draw a circle with centre O and radius 4.1 cm. Take a point P such that O... 1
- Q.3 (B)(4) How many solid cylinders of radius 10 cm and height 6 cm can be made by melting a solid sphere of radius 30 cm ? Answer: 60 cylinders Volume of sphere = 4/3π r³ = 4/3π(30)³ = 36000π cm³. Volume of one cylinder = π r² h = π(10)²(6) = 600π cm³. Number of... 1
- Q.4 (1) In the following figure DE || BC, then :[Figure: Δ ABC with D on side AB and E on side AC, DE parallel to BC.](i) If DE = 4 cm, BC = 8 cm, A(Δ ADE) = 25 cm2, find A(Δ ABC).(ii) If DE : BC = 3 : 5, then find A(Δ ADE) : A(□ DBCE). Answer: (i) A(Δ ABC) = 100 cm²; (ii) A(Δ ADE) : A(□ DBCE) = 9 : 16 Since DE || BC, Δ ADE ~ Δ ABC (AA test: ∠A common, ∠ADE ≅ ∠ABC as corresp... 1
- Q.4 (2) Δ ABC ~ Δ PQR. In Δ ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of Δ ABC and Δ PQR are in the ratio 2 : 3, construct Δ ABC and Δ PQR. Answer: PQ = 5.4 cm, QR = 6 cm, PR = 6.3 cm; construct both triangles from their sides. AB/PQ = BC/QR = AC/PR = 2/3 . PQ = 3/2 × 3.6 = 5.4 c... 1
- Q.4 (3) The radii of the circular ends of a frustum of a cone are 14 cm and 8 cm. If the height of the frustum is 8 cm, find : (π = 3.14)(i) Curved surface area of frustum.(ii) Total surface area of the frustum.(iii) Volume of the frustum. Answer: (i) 690.8 cm²; (ii) 1507.2 cm²; (iii) 3114.88 cm³ r₁ = 14, r₂ = 8, h = 8. Slant height l = √(h² + (r₁ - r₂)²) = √(64 + 36) = 10 cm.... 1