Exercise 3.5 — SASAS Construction

Construction when three sides and two included angles are given.

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Constructing a Quadrilateral Using the S.A.S.A.A. Method

Exercise 3.5 of Class 8 Mathematics — Chapter 3: Construction of Quadrilaterals (covered in CBSE, Telangana, and Andhra Pradesh syllabi) deals with a powerful and flexible construction case: building a quadrilateral when three consecutive sides and two included angles are known. This is the S.A.S.A.A. method — Side, Angle, Side, Angle, Angle (or equivalently, Side, Angle, Side, Side with an angle at each internal junction).

Unlike the S.S.S.S.S. method (five sides) or S.A.S.A.S. method (alternate sides and angles), the S.A.S.A.A. case combines measured arc-drawing with ray-angle construction to locate all four vertices step by step.

Why This Method Works

Once the first two sides and the angle between them are fixed, the third vertex is pinned by the compass arc. Placing the second angle at that vertex shoots a ray in a fixed direction; the third side then lands the fourth vertex precisely on that ray. The final side (joining the first and last vertices) closes the figure automatically.

Key Idea: In the S.A.S.A.A. method you do not need to calculate a missing angle first — the two given angles are "included" between known sides, so each angle is applied exactly where it naturally falls during construction. This is what makes this method different from the S.A.S.A.A. method in Exercise 3.4.

Comparison: Exercise 3.4 vs Exercise 3.5

Feature	Exercise 3.4	Exercise 3.5
Known sides	2 adjacent sides	3 consecutive sides
Known angles	3 angles (including a non-included one)	2 included angles
Missing angle?	Yes — must calculate it first	No — not needed
Last step	Intersection of two rays	Join first and last vertex
Example problems	HELP, GRAM, FLAG	PQRS, LAMP, Trapezium ABCD

Problem (a) — Quadrilateral PQRS

Question (a)

Construct quadrilateral PQRS with PQ = 3.6 cm, QR = 4.5 cm, RS = 5.6 cm, ∠RQP = 135° and ∠SRQ = 60°.

Understanding the Given Data

Notice that both given angles are included angles — ∠RQP sits between sides PQ and QR, while ∠SRQ sits between sides QR and RS. This means both angles are applied directly at the vertices where those sides meet, which is exactly the S.A.S.A.A. pattern.

Element	Value	Role in Construction
PQ	3.6 cm	First base side — drawn first
∠RQP	135°	Angle at Q — ray QX drawn here
QR	4.5 cm	Second side — arc from Q locates R on ray QX
∠SRQ	60°	Angle at R — ray RY drawn here
RS	5.6 cm	Third side — arc from R locates S on ray RY
PS	Not given	Closing side — join P to S at the end

Rough Sketch — Quadrilateral PQRS

Rough sketch of PQRS — rays QX and RY locate R and S respectively; PS closes the figure

Steps of Construction — PQRS

Draw line segment PQ = 3.6 cm as the base.
At Q, draw ray QX making an angle of 135° with PQ, using a protractor. The ray goes upward on the appropriate side.
With Q as centre and radius 4.5 cm, draw an arc that cuts ray QX at point R.
At R, draw ray RY making an angle of 60° with QR, on the side toward where S should lie.
With R as centre and radius 5.6 cm, draw an arc that cuts ray RY at point S.
Join P to S. Quadrilateral PQRS is complete.

Angle direction matters: At Q, the 135° angle opens away from P (making QR lean upward). At R, the 60° angle opens toward the interior of the figure, not outward. Sketching the rough figure first always helps you aim the rays in the correct direction.

Problem (b) — Quadrilateral LAMP

Question (b)

Construct quadrilateral LAMP with AM = MP = PL = 5 cm, ∠M = 90° and ∠P = 60°.

Understanding the Given Data

Three sides of the quadrilateral are equal (each 5 cm), while the fourth side AL is unknown — it will be determined by the construction itself. The two given angles, ∠M = 90° and ∠P = 60°, are both included angles between the known sides. This is a clean S.A.S.A.A. problem.

Note on the missing side LA: The side LA is not given and is not needed beforehand. Once A, M, P, and L are all located by the steps, you simply draw the segment AL to close the quadrilateral. Its length is determined by the geometry of the other four measurements.

Element	Value	Role
AM	5 cm	First side — base of construction
∠M (∠AMP)	90°	Right angle at M — ray MX drawn upward
MP	5 cm	Arc from M marks P on ray MX
∠P (∠MPL)	60°	Angle at P — ray PY drawn toward L
PL	5 cm	Arc from P marks L on ray PY
LA	To be found	Closing side — join L to A

Rough Sketch — Quadrilateral LAMP

Rough sketch of LAMP — ∠M = 90° places P directly above M; ∠P = 60° shoots ray PY toward L

Steps of Construction — LAMP

Draw line segment AM = 5 cm as the base.
At M, draw ray MX making an angle of 90° with AM. Since 90° is a right angle, a set-square gives the most accurate result.
With M as centre and radius 5 cm, draw an arc that cuts ray MX at point P.
At P, draw ray PY making an angle of 60° with MP, directed toward the interior (same side as A).
With P as centre and radius 5 cm, draw an arc that cuts ray PY at point L.
Join A to L. Quadrilateral LAMP is complete.

Using a set-square for 90°: While a protractor works, a set-square or the corner of a piece of paper gives a more accurate right angle at M. Always use the sharpest available tool for right-angle constructions to avoid accumulated error.

Problem (c) — Trapezium ABCD

Question (c)

Construct trapezium ABCD in which AB ∥ CD, AB = 8 cm, BC = 6 cm, CD = 4 cm and ∠B = 60°.

Step 0 — Find ∠C Using the Parallel-Lines Property

A trapezium is a quadrilateral with exactly one pair of parallel sides. Here, AB ∥ CD. The sides BC acts as a transversal cutting these two parallel lines. Therefore, angles B and C are co-interior angles (also called same-side interior or consecutive interior angles), and they must add up to 180°.


AB ∥ CD  ⟹  ∠B + ∠C = 180°  (co-interior angles)

∠C = 180° − 60°

∠C = 120°

Why co-interior angles sum to 180°: When a transversal crosses two parallel lines, the two angles on the same side of the transversal and between the lines always add up to 180°. This property is the key to solving and constructing trapeziums — you only need one base angle to find the other.

Element	Value	Source / Role
AB	8 cm	Given — longer parallel base, drawn first
∠B	60°	Given — ray BX drawn here
BC	6 cm	Given — arc from B locates C on ray BX
∠C	120° (calculated)	Co-interior angle — ray CY drawn here
CD	4 cm	Given — arc from C locates D on ray CY
AD	Not given	Closing side — join A to D

Rough Sketch — Trapezium ABCD

Rough sketch of trapezium ABCD — AB ∥ CD confirmed by equal tick marks; ∠C = 120° is calculated

Steps of Construction — Trapezium ABCD

Draw line segment AB = 8 cm as the longer parallel base.
At B, draw ray BX making an angle of 60° with AB, on the upper side.
With B as centre and radius 6 cm, draw an arc that cuts ray BX at point C.
At C, draw ray CY making an angle of 120° with BC, directed toward A's side. (This ensures CD will be parallel to AB.)
With C as centre and radius 4 cm, draw an arc that cuts ray CY at point D.
Join A to D. Trapezium ABCD is complete. Verify that AB ∥ CD by checking that AD is a transversal cutting both at supplementary co-interior angles.

Verify the parallel condition: After construction, measure ∠A and ∠D. If AB ∥ CD, then ∠A + ∠D should equal 180°. This is a quick and reliable accuracy check that examiners often expect students to mention.

Common Mistakes to Avoid in Exercise 3.5

Forgetting to calculate ∠C in the trapezium — In problem (c), ∠C = 120° must be derived from the parallel-line property before construction begins. Drawing a 60° angle at C instead is the most common error.
Wrong direction for rays — The ray at each angle vertex must open toward the interior of the quadrilateral. Pointing a ray outward results in a figure that is open or crosses itself.
Compass slip while drawing arcs — If the compass shifts while you draw a large arc (like 5.6 cm), the radius changes mid-arc. Press the compass firmly and draw in a single smooth motion.
Not joining the closing side — In all three problems, the last step is to join the first and last vertex (PS, AL, or AD). Forgetting this leaves an incomplete quadrilateral.
Rough sketch skipped — Students who skip the rough sketch often apply the angle in the wrong direction. Always draw the rough sketch with all labels first — it takes 30 seconds and prevents major errors.

What This Exercise Prepares You For

The S.A.S.A.A. construction in Exercise 3.5 is one of the most exam-relevant topics in Class 8 Mathematics for CBSE, Telangana, and Andhra Pradesh boards. The ability to combine arc-and-compass work with protractor-based angle construction is tested both in the 2-mark and 4-mark sections of annual and quarterly exams.

Beyond Class 8, these skills carry directly into higher classes. Understanding how a trapezium's co-interior angles work reinforces the parallel lines chapter from Class 7 and leads naturally into the properties studied in Class 9 and 10.