Dhaka Board SSC Math 2021 · Complete Solutions

Question	Chapter	Topic	Marks
Q1	Ch 3	Polynomial Factorization	10
Q2	Ch 4	Logarithm Equations	10
Q3	Ch 5	Quadratic Formula Application	10
Q4	Ch 8	Arithmetic Series	10
Q5	Ch 9	Pythagoras Theorem & Applications	10
Q6	Ch 14	Similarity Theorems	10
Q7	Ch 12	Sin/Cos Ratio Problems	10
Q8	Ch 16	Cylinder and Sphere Mensuration	10

Question 1 10 marks Ch 3 · Polynomial Factorization

STEM: Consider the polynomial p(x) = x³ − 6x² + 11x − 6.

(a) Show that x = 1 is a root of p(x). [2 marks]

(b) Factorize p(x) completely. [4 marks]

(c) Hence factorize: a³ − 6a²b + 11ab² − 6b³. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Substitute x=1

p(1) = 1 − 6 + 11 − 6 = 0 ✓

Answer

x = 1 is a root since p(1) = 0 ✓

(b) Solution [4 marks]

Divide by (x−1)

Using synthetic division or long division: x³−6x²+11x−6 ÷ (x−1)

Quotient

= x² − 5x + 6

Factor quotient

x²−5x+6 = (x−2)(x−3)

Complete factorization

p(x) = (x−1)(x−2)(x−3)

Verify

Roots: x=1,2,3. Check: p(2)=8−24+22−6=0 ✓, p(3)=27−54+33−6=0 ✓

Answer

p(x) = (x−1)(x−2)(x−3) ✓

(c) Solution [4 marks]

Replace x with a/b

a³−6a²b+11ab²−6b³ = b³·p(a/b)

Apply factor

= b³·(a/b−1)(a/b−2)(a/b−3)

Simplify

= (a−b)(a−2b)(a−3b)

Verify by expansion

(a−b)(a−2b) = a²−3ab+2b²; ×(a−3b) = a³−6a²b+11ab²−6b³ ✓

Answer

a³−6a²b+11ab²−6b³ = (a−b)(a−2b)(a−3b) ✓

Question 2 10 marks Ch 4 · Logarithm Equations

STEM: Solve equations and simplify using logarithm rules.

(a) Solve: log₂(x) = 5. [2 marks]

(b) Solve: log(x²−5) = log(2x+2). [4 marks]

(c) If log₄(x) = log₂(3), find x. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Convert

log₂(x) = 5 → x = 2⁵ = 32

Answer

x = 32 ✓

(b) Solution [4 marks]

Equal logs → equal arguments

x²−5 = 2x+2 → x²−2x−7 = 0

Quadratic formula

x = (2 ± √(4+28))/2 = (2 ± √32)/2 = 1 ± 2√2

Check domain

Need x²−5 > 0 and 2x+2 > 0. For x = 1+2√2 ≈ 3.83: both positive. For x = 1−2√2 ≈ −1.83: 2(−1.83)+2 = −1.66 < 0 (rejected)

Answer

x = 1 + 2√2 ✓

(c) Solution [4 marks]

Change of base

log₄(x) = log(x)/log(4) and log₂(3) = log(3)/log(2)

Also log(4) = 2log(2)

log₄(x) = log(x)/(2log2) = log₂(x)/2

Equation becomes

log₂(x)/2 = log₂(3) → log₂(x) = 2log₂(3) = log₂(9)

Solve

x = 9

Verify

log₄(9) = log₄(3²) = 2log₄(3) = 2×log₂(3)/log₂(4) = 2×log₂(3)/2 = log₂(3) ✓

Answer

x = 9 ✓

Question 3 10 marks Ch 5 · Quadratic Formula

STEM: A rectangular garden has perimeter 56 m and area 187 m².

(a) Form two equations using length l and width w. [2 marks]

(b) Find l and w by forming and solving a quadratic equation. [4 marks]

(c) Using the quadratic formula, solve: 3x² − 7x + 2 = 0 and verify using Vieta's formulas. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Perimeter

2(l+w) = 56 → l+w = 28

Area

lw = 187

Answer

l+w = 28 and lw = 187 ✓

(b) Solution [4 marks]

Quadratic in l

From l+w=28: w = 28−l. Substituting: l(28−l) = 187 → 28l−l² = 187 → l²−28l+187 = 0

Factorize

(l−11)(l−17) = 0 → l = 11 or l = 17

Both solutions valid

l=17,w=11 or l=11,w=17 (rectangle)

Answer

Length = 17 m, Width = 11 m ✓

(c) Solution [4 marks]

Quadratic formula

x = (7 ± √(49−24))/6 = (7 ± √25)/6 = (7 ± 5)/6

Roots

x = 12/6 = 2 or x = 2/6 = 1/3

Vieta's check

Sum of roots = 7/3. Check: 2 + 1/3 = 7/3 ✓. Product = 2/3. Check: 2 × 1/3 = 2/3 ✓

Answer

x = 2 or x = 1/3 ✓

Question 4 10 marks Ch 8 · Arithmetic Series

STEM: The sum of first n terms of an AP is Sₙ = 3n² + 5n.

(a) Find the first term and common difference. [2 marks]

(b) Find the 12th term and the sum of the first 12 terms. [4 marks]

(c) Which term of the AP is 119? [4 marks]

▶ Show Solution

(a) Solution [2 marks]

a₁ = S₁

S₁ = 3(1)² + 5(1) = 8 = a₁

a₂ = S₂ − S₁

S₂ = 3(4)+10 = 22, a₂ = 22−8 = 14

Common difference

d = a₂ − a₁ = 14 − 8 = 6

Answer

a₁ = 8, d = 6 ✓

(b) Solution [4 marks]

12th term via formula

aₙ = Sₙ − Sₙ₋₁ = (3n²+5n) − (3(n−1)²+5(n−1)) = 6n+2

a₁₂

a₁₂ = 6(12)+2 = 74

S₁₂

S₁₂ = 3(144)+5(12) = 432+60 = 492

Answer

a₁₂ = 74, S₁₂ = 492 ✓

(c) Solution [4 marks]

General term

aₙ = 6n + 2 = 119

Solve

6n = 117 → n = 19.5

Not an integer!

119 is not a term. Check: a₁₉ = 6(19)+2 = 116; a₂₀ = 6(20)+2 = 122

Conclusion

119 is NOT a term of this AP. The nearest terms are 116 (n=19) and 122 (n=20) ✓

Question 5 10 marks Ch 9 · Pythagoras Theorem

STEM: In right triangle ABC, ∠C = 90°, BC = 8 cm, AB = 17 cm.

(a) Find AC using Pythagoras theorem. [2 marks]

(b) An altitude CD is drawn from C to hypotenuse AB. Find CD, AD, and DB. [4 marks]

(c) Prove that AB² = AD × AB + DB × AB (i.e., verify Pythagoras via the altitude). [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Pythagoras

AC² = AB² − BC² = 289 − 64 = 225

Answer

AC = 15 cm ✓

(b) Solution [4 marks]

Altitude rule: CD² = AD × DB

Also: AD = AC²/AB = 225/17 and DB = BC²/AB = 64/17

Check AD+DB

225/17 + 64/17 = 289/17 = 17 = AB ✓

CD

CD = AC×BC/AB = 15×8/17 = 120/17 ≈ 7.06 cm

Verify

CD² = (120/17)² = 14400/289 = AD×DB = (225/17)×(64/17) = 14400/289 ✓

Answer

AD = 225/17 cm, DB = 64/17 cm, CD = 120/17 cm ✓

(c) Solution [4 marks]

In △ACD

AC² = AD × AB (geometric mean relation)

In △BCD

BC² = DB × AB (geometric mean relation)

Add

AC² + BC² = AD×AB + DB×AB = AB(AD+DB) = AB×AB = AB²

This is Pythagoras

AB² = AC² + BC² verified via altitude theorem ✓

Question 6 10 marks Ch 14 · Similarity

STEM: In △ABC, DE ∥ BC where D is on AB and E is on AC. AD = 3, DB = 5, AE = 4.5.

(a) Find EC using the Basic Proportionality Theorem. [2 marks]

(b) Prove △ADE ∼ △ABC and find the ratio of their areas. [4 marks]

(c) If BC = 12 cm, find DE. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

BPT

DE ∥ BC → AD/DB = AE/EC

Solve

3/5 = 4.5/EC → EC = 4.5×5/3 = 7.5 cm

Answer

EC = 7.5 cm ✓

(b) Solution [4 marks]

∠A = ∠A (common)

Shared angle at vertex A

DE ∥ BC → ∠ADE = ∠ABC

Corresponding angles

AA criterion

△ADE ∼ △ABC by AA

Ratio of sides

AD/AB = 3/(3+5) = 3/8

Ratio of areas

Area(△ADE)/Area(△ABC) = (AD/AB)² = (3/8)² = 9/64

Answer

△ADE ∼ △ABC, area ratio = 9:64 ✓

(c) Solution [4 marks]

Similar triangles ratio

DE/BC = AD/AB = 3/8

Find DE

DE = BC × 3/8 = 12 × 3/8 = 4.5 cm

Answer

DE = 4.5 cm ✓

Question 7 10 marks Ch 12 · Trigonometry

STEM: Angle A is acute and 3 sin A = 2 cos A.

(a) Find tan A, sin A and cos A. [2 marks]

(b) Find sin 2A and cos 2A using double angle formulas. [4 marks]

(c) Verify: sin²A + cos²A = 1 and calculate sec A + cosec A. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

From 3 sin A = 2 cos A

tan A = sin A/cos A = 2/3

Right triangle: opp=2, adj=3, hyp=√13

sin A = 2/√13, cos A = 3/√13

Answer

tan A = 2/3, sin A = 2/√13, cos A = 3/√13 ✓

(b) Solution [4 marks]

Double angle sin

sin 2A = 2 sin A cos A = 2×(2/√13)×(3/√13) = 12/13

Double angle cos

cos 2A = cos²A − sin²A = 9/13 − 4/13 = 5/13

Answer

sin 2A = 12/13, cos 2A = 5/13 ✓

(c) Solution [4 marks]

Verify identity

sin²A+cos²A = 4/13+9/13 = 13/13 = 1 ✓

sec A

sec A = 1/cos A = √13/3

cosec A

cosec A = 1/sin A = √13/2

Sum

sec A + cosec A = √13/3 + √13/2 = √13(1/3+1/2) = √13×5/6 = 5√13/6

Answer

sec A + cosec A = 5√13/6 ≈ 3.007 ✓

Question 8 10 marks Ch 16 · Cylinder and Sphere

STEM: A solid sphere has radius 6 cm. A cylinder has height 9 cm and the same base radius of 6 cm.

(a) Find the volume and surface area of the sphere. (π = 22/7) [2 marks]

(b) Find the volume and total surface area of the cylinder. [4 marks]

(c) A hollow cylinder with inner radius 4 cm and outer radius 6 cm has height 9 cm. Find its volume. [4 marks]

▶ Show Solution

(a) Solution [2 marks]

Sphere volume

V = (4/3)πr³ = (4/3)×(22/7)×216 = (4×22×216)/(3×7) = 19008/21 = 904.76 cm³

Sphere SA

SA = 4πr² = 4×(22/7)×36 = 3168/7 = 452.57 cm²

Answer

V ≈ 904.76 cm³, SA ≈ 452.57 cm² ✓

(b) Solution [4 marks]

Cylinder volume

V = πr²h = (22/7)×36×9 = 7128/7 = 1018.29 cm³

Total SA

TSA = 2πr(r+h) = 2×(22/7)×6×(6+9) = 2×(22/7)×6×15 = 3960/7 = 565.71 cm²

Answer

V ≈ 1018.29 cm³, TSA ≈ 565.71 cm² ✓

(c) Solution [4 marks]

Hollow cylinder volume

V = π(R²−r²)×h where R=6, r=4

Calculate

V = (22/7)×(36−16)×9 = (22/7)×20×9 = 3960/7 ≈ 565.71 cm³

Answer

Volume of hollow cylinder ≈ 565.71 cm³ ✓

Dhaka Board · SSC Mathematics · 2021

Chapter Coverage

Creative Questions

(a) Solution [2 marks]

(b) Solution [4 marks]

(c) Solution [4 marks]

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