Exercise 3.1 — SSSSA Construction

Constructing a quadrilateral when four sides and one angle are given.

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Exercise 3.1 – Constructing Quadrilaterals with Four Sides and One Angle (S.S.S.S.A)

In Exercise 3.1 of Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus), you learn to construct quadrilaterals when four sides and one included angle are given — a method abbreviated as S.S.S.S.A. The core idea is to split the quadrilateral into two triangles using an imaginary diagonal, construct each triangle using a compass, and then join the vertices to complete the shape.

All six problems in this exercise use the same six-step construction framework. Once you understand the logic behind even one problem, the others become straightforward — they differ only in the shape type (general quadrilateral, parallelogram, rhombus, rectangle, or square) and the given measurements.

Five Ways to Construct a Quadrilateral — Quick Reference

A quadrilateral has five independent measurements (sides + angles + diagonals) needed to fix its shape uniquely. Depending on which five are provided, the construction method changes:

#	Given Information	Short Form	Covered In
1	Four sides + one angle	S.S.S.S.A	Exercise 3.1
2	Four sides + one diagonal	S.S.S.S.D	Exercise 3.2
3	Three sides + two diagonals	S.S.S.D.D	Exercise 3.3
4	Two adjacent sides + three angles	S.A.S.A.A	Exercise 3.4
5	Three sides + two included angles	S.A.S.A.S	Exercise 3.5

General Steps for S.S.S.S.A Construction

All Exercise 3.1 problems follow this same logical sequence — only the numbers and vertex names change:

Draw the base side — the side whose angle is given (this is your starting line segment).
Construct the known angle at one vertex using a protractor; draw a ray in that direction.
Mark the adjacent vertex along the ray by drawing an arc with the known adjacent side as radius.
Draw intersecting arcs from the two free vertices using the two remaining sides as radii.
Mark the fourth vertex where the two arcs cross.
Join all vertices in order to complete the quadrilateral.

💡 Why does this work? The known angle pins one vertex precisely. The two remaining sides create arcs that can only intersect at one point — this uniquely fixes the fourth vertex, completing the quadrilateral.

Problem (a)

Quadrilateral ABCD — AB = 5.5 cm, BC = 3.5 cm, CD = 4 cm, AD = 5 cm, ∠A = 45°

S.S.S.S.A — General Quadrilateral

Here all four sides and the angle at vertex A are given. We start by drawing side AB (the side adjacent to the given angle ∠A = 45°), then use the angle to locate D along the ray from A, and finally use intersecting arcs to find C.

Measurement	Value	Role in Construction
AB	5.5 cm	Base side — draw first
AD	5 cm	Arc from A on the 45° ray to find D
CD	4 cm	Arc from D to locate C
BC	3.5 cm	Arc from B; intersects CD arc at C
∠A	45°	Direction of ray AX from AB

Rough sketch of ABCD

Completed quadrilateral ABCD

Steps of Construction

Draw a line segment AB = 5.5 cm.
At vertex A, draw ray AX making an angle of 45° with AB using a protractor.
With A as centre and radius 5 cm, draw an arc cutting ray AX at point D.
With D as centre and radius 4 cm, draw an arc (this will help locate C).
With B as centre and radius 3.5 cm, draw another arc cutting the previous arc at point C.
Join B–C and D–C. The quadrilateral ABCD is complete.

Problem (b)

Quadrilateral BEST — BE = 2.9 cm, ES = 3.2 cm, ST = 2.7 cm, BT = 3.4 cm, ∠B = 75°

S.S.S.S.A — General Quadrilateral

Same method as (a), now applied to quadrilateral BEST. The angle ∠B = 75° is at vertex B, so we start by drawing BE and constructing the 75° ray from B. The arc of radius 3.4 cm (= BT) from B along this ray gives us T, and intersecting arcs from E and T give us S.

Measurement	Value	Role in Construction
BE	2.9 cm	Base side — draw first
BT	3.4 cm	Arc from B on 75° ray → T
ST	2.7 cm	Arc from T to locate S
ES	3.2 cm	Arc from E; intersects ST arc at S
∠B	75°	Direction of ray BX from BE

Completed quadrilateral BEST

Steps of Construction

Draw a line segment BE = 2.9 cm.
At vertex B, draw ray BX making an angle of 75° with BE.
With B as centre and radius 3.4 cm, mark point T on ray BX.
With T as centre and radius 2.7 cm, draw an arc.
With E as centre and radius 3.2 cm, draw an arc cutting the previous arc at S.
Join E–S and T–S to complete quadrilateral BEST.

Problem (c)

Parallelogram PQRS — PQ = 4.5 cm, QR = 3 cm, ∠RQP = 60°

S.S.S.S.A — Parallelogram

This problem involves a parallelogram, which has an important property: opposite sides are equal. So even though only two sides are given, you can work out all four sides immediately before constructing.

PQ = RS = 4.5 cm | QR = PS = 3 cm

The angle given is ∠RQP = 60° at vertex Q. Start from P, draw PQ, then construct the 60° ray from Q to place R. Since PS = 3 cm and RS = 4.5 cm, intersecting arcs from P and R will locate S.

Measurement	Value	How we know it
PQ	4.5 cm	Given directly
QR	3 cm	Given directly
RS	4.5 cm	Opposite to PQ (parallelogram property)
PS	3 cm	Opposite to QR (parallelogram property)
∠RQP	60°	Given directly

Rough sketch of PQRS

Completed parallelogram PQRS

Steps of Construction

Draw a line segment PQ = 4.5 cm.
At Q, draw ray QX making an angle of 60° with PQ.
With Q as centre and radius 3 cm, mark point R on ray QX.
With P as centre and radius 3 cm, draw an arc (this arc will pass through S).
With R as centre and radius 4.5 cm, draw an arc cutting the previous arc at S.
Join P–S and R–S to complete parallelogram PQRS.

Problem (d)

Rhombus MATH — AT = 4 cm, ∠MAT = 120°

S.S.S.S.A — Rhombus

In a rhombus, all four sides are equal. So knowing just one side gives you all four. Only the side AT = 4 cm and angle ∠MAT = 120° are needed to uniquely construct this rhombus.

MA = AT = TH = HM = 4 cm (all sides equal in a rhombus)

💡 In rhombus MATH, adjacent angles are supplementary: since ∠MAT = 120°, then ∠ATH = 60°. This is a useful check — the four angles must sum to 360°.

Completed rhombus MATH

Steps of Construction

Draw a line segment MA = 4 cm.
At A, draw ray AX making an angle of 120° with MA.
With A as centre and radius 4 cm, mark point T on ray AX.
With M as centre and radius 4 cm, draw an arc.
With T as centre and radius 4 cm, draw another arc cutting the previous arc at H.
Join M–H and T–H to complete rhombus MATH.

Problem (e)

Rectangle FLAT — FL = 5 cm, LA = 3 cm

S.S.S.S.A — Rectangle (∠ = 90°)

In a rectangle, opposite sides are equal and every angle is exactly 90°. So even though only two sides are given, you immediately know all four sides and all four angles.

FL = AT = 5 cm | LA = FT = 3 cm | ∠F = ∠L = ∠A = ∠T = 90°

The angle at every vertex is 90°, so you use a protractor to draw a perpendicular ray at L, then set LA = 3 cm along it. The fourth vertex T is found where arcs from F and A intersect.

Completed rectangle FLAT

Steps of Construction

Draw a line segment FL = 5 cm.
At L, draw ray LX making an angle of 90° with FL (perpendicular to FL).
With L as centre and radius 3 cm, mark point A on ray LX.
With F as centre and radius 3 cm, draw an arc.
With A as centre and radius 5 cm, draw an arc cutting the previous arc at T.
Join F–T and A–T to complete rectangle FLAT.

Problem (f)

Square LUDO — LU = 4.5 cm

S.S.S.S.A — Square (all sides equal, all angles 90°)

A square is the simplest quadrilateral to construct because just one piece of information — the side length — determines everything. All four sides are equal and all four angles are 90°.

LU = UD = DO = OL = 4.5 cm | ∠L = ∠U = ∠D = ∠O = 90°

Completed square LUDO

Steps of Construction

Draw a line segment LU = 4.5 cm.
At U, draw ray UX making an angle of 90° with LU (perpendicular at U).
With U as centre and radius 4.5 cm, mark point D on ray UX.
With L as centre and radius 4.5 cm, draw an arc.
With D as centre and radius 4.5 cm, draw another arc cutting the previous arc at O.
Join L–O and D–O to complete square LUDO.

Exercise 3.1 at a Glance — All Six Problems Compared

Problem	Shape	Given Angle	Key Trick
(a) ABCD	General quadrilateral	∠A = 45°	No shortcuts — all four sides given independently
(b) BEST	General quadrilateral	∠B = 75°	Same method, different vertex orientation
(c) PQRS	Parallelogram	∠RQP = 60°	Use opposite-sides property to derive RS & PS
(d) MATH	Rhombus	∠MAT = 120°	All four sides = given side (4 cm)
(e) FLAT	Rectangle	∠L = 90°	Opposite sides equal; angle always 90°
(f) LUDO	Square	∠U = 90°	All sides equal; angle always 90° — only 1 measurement needed

Common Mistakes to Avoid in Exercise 3.1

Starting from the wrong vertex: Always start drawing the base side from the vertex where the given angle sits (or adjacent to it), so the angle can be constructed immediately.
Not using parallelogram/rhombus/square properties first: For problems (c), (d), (e), and (f), always derive the missing sides using shape properties before picking up the compass. Skipping this step leads to wrong constructions.
Arcs not intersecting: If two arcs don't meet, it usually means a measurement was entered incorrectly or the compass slipped. Double-check your radii against the given data.
Drawing the angle on the wrong side: The angle opens inward toward the quadrilateral. A rough sketch drawn first prevents this error every time.
Forgetting to join all sides: Construction is incomplete until every pair of adjacent vertices is connected and the shape is closed.

📐 Board exam tip: In Telangana and AP board exams, you are expected to write the steps of construction alongside the figure. Practise writing all six steps clearly — examiners award marks for each step, not just the final figure.

What Exercise 3.1 Prepares You For

Mastering the S.S.S.S.A method gives you the core compass-and-ruler fluency needed for every other exercise in this chapter. In Exercise 3.2, you will construct quadrilaterals when a diagonal replaces the angle — a slightly different strategy but built on the same arc-intersection logic.

The parallelogram, rhombus, rectangle, and square constructions you practised here will also reappear when you study congruence and similarity of triangles in Class 9, where knowing how to uniquely fix a polygon from minimum data becomes essential.