Exercise 3.3 — SSSDD Construction

Construction when three sides and two diagonals are given.

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Exercise 3.3 – Construction of Quadrilaterals (Two Diagonals) | Class 8 Maths

Class 8 · Chapter 3 · Exercise 3.3

Construction of Quadrilaterals — When Three Sides and Two Diagonals Are Given (S.S.S.D.D.)

Learn to accurately construct any quadrilateral using compass and ruler when both diagonals and three sides are known. Fully solved for CBSE, Telangana, and Andhra Pradesh board students.

Class 8 Maths Chapter 3 CBSE Telangana Board AP Board Geometry S.S.S.D.D. Method

What Is the S.S.S.D.D. Method?

Every unique quadrilateral requires exactly five independent measurements to be drawn. Exercise 3.3 introduces a different combination: three sides and both diagonals — abbreviated as S.S.S.D.D. (Side-Side-Side-Diagonal-Diagonal).

This is slightly trickier than Exercise 3.2 because the two diagonals cross each other inside the quadrilateral. You must carefully identify which triangle to draw first using the diagonal that connects known vertices, then use the second diagonal to place the remaining vertex.

Core Idea: With two diagonals given, the quadrilateral can be split into triangles in two different ways. The trick is to pick the triangle formed by one complete side + both diagonals as your starting triangle (S.S.S. by two diagonals and a side), then use the remaining sides to fix the last vertex.
Example — Quadrilateral ABCD with sides AB, BC, CD and diagonals AC, BD: Step 1: Construct △ABD using AB, diagonal BD, and diagonal AC is not yet drawn — Actually pick the triangle whose 3 measurements are all known Step 2: From completed triangle, use remaining measurements to fix last vertex → Two SSS triangles together complete the quadrilateral
Given Data
5
3 sides + 2 diagonals
Method Name
S.S.S.D.D.
Side-Side-Side-Diagonal-Diagonal
Tools Needed
Ruler + Compass
No protractor required

How Exercise 3.3 Differs from Exercise 3.2

Students often confuse these two methods. The table below clarifies the key differences to help you pick the right approach in an exam.

Feature Exercise 3.2 — S.S.S.S.D. Exercise 3.3 — S.S.S.D.D.
Measurements given 4 sides + 1 diagonal 3 sides + 2 diagonals
Diagonal usage One diagonal splits figure into 2 triangles Both diagonals cross inside; used across triangles
Starting triangle Always the triangle with the base side + diagonal The triangle formed by a side and both diagonals
Difficulty level Moderate Slightly higher — careful triangle identification needed
Example problems ABCD, PQRS, Parallelogram, Rhombus GOLD (irregular), PQRS (irregular)

General 7-Step Framework for S.S.S.D.D.

Study this general procedure before tackling individual problems. The pattern is the same for both questions in this exercise.

StepActionPurpose
1Draw a base side (one of the given sides)Anchors the figure
2Draw arc from one endpoint using first diagonalBegins locating third vertex
3Draw arc from other endpoint using second diagonalIntersects to fix third vertex
4Join endpoints to complete Triangle 1 (S.S.S.)First triangle fully constructed
5Draw arc from a vertex using one remaining sideBegins locating fourth vertex
6Draw arc from another vertex using last sideIntersects to fix fourth vertex
7Join all four vertices to complete the quadrilateralFinal figure is complete
Problem (a)

Quadrilateral GOLD

Given: OL = 7.5 cm  |  GL = 6 cm  |  LD = 5 cm  |  DG = 5.5 cm  |  Diagonal OD = 10 cm

Notice what is given: sides OL, GL, LD, DG and diagonal OD. The side GO is not given — only its diagonal OD is. This means we must identify the triangle that is fully determined by the available measurements.

The diagonal OD connects O and D. Triangle OLD has sides OL = 7.5 cm, LD = 5 cm, and OD = 10 cm — all three are known. So △OLD is our starting triangle. Then vertex G is fixed using GL = 6 cm and DG = 5.5 cm.

TriangleSides UsedWhat It Fixes
△OLD (first)OL = 7.5 cm, LD = 5 cm, OD = 10 cmVertices O, L, D
△GLD (second)GL = 6 cm, DG = 5.5 cm, LD = 5 cm (shared)Vertex G
OL = 7.5 cm LD = 5 cm DG = 5.5 cm GO OD = 10 cm GL = 6 cm O L D G Diagonal OD GL (ref)

Figure: Rough sketch of Quadrilateral GOLD. The diagonal OD (blue dashed) is the given diagonal. Starting triangle △OLD is highlighted.

Why start with △OLD? Because all three of its sides are given — OL, LD, and diagonal OD. Once O, L, D are placed, vertex G is found using GL and DG from vertex L and D respectively.

Steps of Construction

  1. Draw a line segment OL = 7.5 cm as the base.
  2. With L as centre, draw an arc of radius 5 cm (= LD).
  3. With O as centre, draw an arc of radius 10 cm (= OD, the diagonal). These two arcs intersect at point D.
  4. Join LD and OD to complete triangle △OLD.
  5. With L as centre, draw an arc of radius 6 cm (= GL).
  6. With D as centre, draw an arc of radius 5.5 cm (= DG). These arcs intersect at point G.
  7. Join GL, GO (inferred), and GD to complete the required quadrilateral GOLD.
Watch out: In this problem the side GO is not explicitly given — it is the connecting side between G and O. After fixing G, join it to both L, O, and D to close the quadrilateral. Examiners expect all four sides to be drawn.
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Problem (b)

Quadrilateral PQRS

Given: PQ = 4.2 cm  |  QR = 3 cm  |  PS = 2.8 cm  |  Diagonal PR = 4.5 cm  |  Diagonal QS = 5 cm

Here both diagonals PR and QS are given, along with three sides PQ, QR, and PS. The fourth side SR is not given. We need to identify which triangle is fully solvable first.

Diagonal QS connects Q and S. Triangle PQS has sides PQ = 4.2 cm, PS = 2.8 cm, and QS = 5 cm — all three are known! So △PQS is the starting triangle. Then vertex R is fixed using QR = 3 cm (from Q) and diagonal PR = 4.5 cm (from P).

TriangleSides UsedWhat It Fixes
△PQS (first)PQ = 4.2 cm, PS = 2.8 cm, QS = 5 cmVertices P, Q, S
△PQR (second)QR = 3 cm, PR = 4.5 cm, PQ = 4.2 cm (shared)Vertex R
PQ = 4.2 cm QR = 3 cm SR PS = 2.8 cm QS = 5 cm PR = 4.5 cm P Q R S Diagonal QS Diagonal PR

Figure: Rough sketch of Quadrilateral PQRS. Both diagonals QS (purple) and PR (blue) cross inside the figure.

Key observation: When both diagonals are given, they create four triangles inside the quadrilateral. Your job is to find the one outer triangle whose three sides are all known, construct it first, then add the remaining vertex using the other given measurements.

Steps of Construction

  1. Draw a line segment PQ = 4.2 cm as the base.
  2. With Q as centre, draw an arc of radius 5 cm (= QS diagonal).
  3. With P as centre, draw an arc of radius 2.8 cm (= PS). These arcs intersect at point S.
  4. Join PS and QS to complete triangle △PQS.
  5. With Q as centre, draw an arc of radius 3 cm (= QR).
  6. With P as centre, draw an arc of radius 4.5 cm (= PR, the other diagonal). These arcs intersect at point R.
  7. Join QR, PR (the diagonal), and SR to complete the required quadrilateral PQRS.

Reading the Construction Diagrams

The diagrams in Exercise 3.3 look more complex than those in 3.2 because both diagonals cross each other and create an X pattern inside the figure. Here is how to read them correctly:

  • Solid red/green lines — The four sides of the quadrilateral that form the outer boundary.
  • Dashed coloured lines — The diagonals. These are measurement lines used in construction but are not sides of the quadrilateral.
  • Arc intersections (✕ marks) — The points where two arcs cross, giving the location of a vertex. Always mark these clearly with a dot and label.
  • The X pattern in the middle — This is formed by the two diagonals crossing. This crossing point is inside the quadrilateral, not a vertex.

Common Mistakes to Avoid

  • Wrong starting triangle — Always check that all three sides of your first triangle are given before you begin. In GOLD, △OLD works; in PQRS, △PQS works. Choosing the wrong triangle leads to a construction that cannot be completed.
  • Confusing a side with a diagonal — Diagonals connect opposite vertices (like O to D in GOLD, skipping G and L). Sides connect adjacent vertices. Read the problem statement carefully and label the rough sketch before touching compass.
  • Forgetting the fourth side — In both problems, one side of the quadrilateral is not given (GO in GOLD; SR in PQRS). After fixing all four vertices, you must still draw this side to close the figure. Examiners deduct marks if the quadrilateral is not closed.
  • Erasing arcs — CBSE, Telangana, and AP board examiners award marks for construction arcs. Never erase them — they are proof of your method.
  • No rough sketch — The rough sketch helps you decide which diagonal belongs to which triangle. Skipping it is the single biggest cause of errors in construction questions.
  • Incorrect compass width — Always verify your compass opening against the ruler before drawing each arc. A 0.5 cm error compounds across the construction and misplaces vertices.
Board Exam Alert: Both CBSE and Telangana/AP board marking schemes award separate marks for the rough sketch, correct construction arcs, and the final labelled figure. A correct figure without visible arcs can lose 1–2 marks.

Summary: Both Problems at a Glance

Detail Problem (a) — GOLD Problem (b) — PQRS
Base side drawnOL = 7.5 cmPQ = 4.2 cm
First arc (Step 2)Radius 5 cm from L (LD)Radius 5 cm from Q (QS diagonal)
Second arc (Step 3)Radius 10 cm from O (OD diagonal)Radius 2.8 cm from P (PS)
Third vertex foundDS
Starting triangle△OLD△PQS
Fourth arc (Step 5)Radius 6 cm from L (GL)Radius 3 cm from Q (QR)
Fifth arc (Step 6)Radius 5.5 cm from D (DG)Radius 4.5 cm from P (PR diagonal)
Fourth vertex foundGR
Missing side (not given)GOSR

What This Exercise Prepares You For

Exercise 3.3 completes your understanding of the S.S.S.D.D. method. Together with Exercise 3.2 (S.S.S.S.D.), you now have two powerful tools for constructing any quadrilateral when sides and diagonals are given.

Syllabus alignment: Exercise 3.3 is part of Chapter 3 — Construction of Quadrilaterals in the Class 8 Mathematics textbooks of CBSE, Telangana State Board (SCERT), and Andhra Pradesh State Board (APSCERT).

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