Exercise 3.6 — Special Quadrilaterals
Construction of special quadrilaterals like rhombus and square.
Exercise 3.6 — Construction of a Rhombus and a Square Using Diagonals
Class 8 Mathematics · CBSE, Telangana & Andhra Pradesh Syllabus · Chapter 3: Construction of Quadrilaterals
Exercise 3.6 is the final exercise of Chapter 3, Construction of Quadrilaterals, in Class 8 Mathematics (CBSE, Telangana & Andhra Pradesh syllabus). Unlike the earlier exercises in this chapter — which built general quadrilaterals from sides, angles, and one diagonal — this exercise focuses on two special quadrilaterals, the rhombus and the square, and shows how either one can be constructed using only the lengths of its two diagonals.
Below is a full explanation of the key property that makes this possible, the general method using a ruler and compass, and complete step-by-step constructions for all three problems in this exercise — two rhombi and one square — each with a labelled rough sketch and the actual construction diagram.
Why Diagonals Are Enough: The Key Property
A rhombus is a parallelogram with all four sides equal, and a square is a rhombus whose angles are also all 90°. Both shapes share one powerful property that the earlier exercises' quadrilaterals (like a general parallelogram or trapezium) do not automatically have:
- The diagonals of a rhombus (and therefore also of a square) bisect each other — they cross exactly at their own midpoints.
- The diagonals intersect at 90° — they are always perpendicular to each other.
- In a square specifically, the two diagonals are also equal in length, unlike in a general rhombus where they can be different.
Rhombus Diagonals
- Bisect each other at the centre
- Cross at exactly 90°
- Usually of different lengths
- Need both diagonal lengths to construct it
Square Diagonals
- Bisect each other at the centre
- Cross at exactly 90°
- Always equal in length
- Only one diagonal length needs to be given
Half-diagonal radius = (length of the other diagonal) ÷ 2
General Method: Constructing From Two Diagonals
Every problem in this exercise — whatever the actual diagonal lengths happen to be — follows exactly the same four-step method:
- Draw the first diagonal as a straight line segment of the given length, using a ruler.
- Construct the perpendicular bisector of this diagonal using a compass — this locates the centre point and guarantees the 90° crossing angle.
- With the centre as the compass point, draw two arcs with radius equal to half of the second diagonal's length, cutting the perpendicular bisector on both sides of the centre — these two points are the other two vertices.
- Join the four vertices in order with straight lines to complete the rhombus or square.
Diagonal 1 (purple) is drawn first; its perpendicular bisector XY locates the centre; arcs of radius = half of Diagonal 2 mark Vertex 3 and Vertex 4; joining all four vertices completes the quadrilateral.
Problem (a) — Rhombus CART With Diagonals CR = 6 cm, AT = 4.8 cm
Question: Construct a rhombus CART in which the diagonal CR = 6 cm and the diagonal AT = 4.8 cm.
CR and AT are the diagonals of rhombus CART, bisecting each other at O. So ∠AOC = 90°, and OA = OT = AT ÷ 2 = 4.8 ÷ 2 = 2.4 cm (the half-diagonal radius needed for the arcs).
CR = 6 cm and AT = 4.8 cm are the diagonals, crossing at O with OC = OR = 3 cm and OA = OT = 2.4 cm.
- Draw a line segment CR of length 6 cm.
- Construct the perpendicular bisector XY of CR, which intersects CR at O.
- With O as centre and radius 2.4 cm, draw two arcs cutting XY on either side of CR at A and T.
- Join C–A, A–R, R–T, and T–C to complete the required rhombus.
CR is drawn first (blue); the perpendicular bisector XY (teal) locates O; orange arc-marks show the construction of XY itself, and the teal arcs of radius 2.4 cm cut XY at A and T; finally C–A–R–T (green) is joined.
Problem (b) — Rhombus SOAP With Diagonals SA = 4.3 cm, OP = 5 cm
Question: Construct a rhombus SOAP in which the diagonal SA = 4.3 cm and the diagonal OP = 5 cm.
SA and OP are the diagonals of rhombus SOAP, bisecting each other at X. So ∠OXS = 90°, and XO = XP = OP ÷ 2 = 5 ÷ 2 = 2.5 cm.
SA = 4.3 cm and OP = 5 cm are the diagonals, crossing at X with XS = XA = 2.15 cm and XO = XP = 2.5 cm.
- Draw a line segment SA of length 4.3 cm.
- Construct the perpendicular bisector MN of SA, which intersects SA at X.
- With X as centre and radius 2.5 cm, draw two arcs cutting MN on either side of SA at O and P.
- Join S–O, O–A, A–P, and P–S to complete the required rhombus.
SA is drawn first; its perpendicular bisector MN locates X; arcs of radius 2.5 cm cut MN at O and P; joining S–O–A–P completes the rhombus.
Problem (c) — Square JUMP With Diagonal 4.2 cm
Question: Construct a square JUMP whose diagonal is 4.2 cm.
JM and UP are the diagonals of square JUMP, bisecting each other at X. So ∠UXJ = 90°, and XJ = XM = JM ÷ 2 = 4.2 ÷ 2 = 2.1 cm — and since JM = UP, the same 2.1 cm radius is used for both halves of both diagonals.
JM = UP = 4.2 cm — both diagonals are equal, crossing at X with all four half-diagonal segments equal to 2.1 cm.
- Draw a line segment JM of length 4.2 cm.
- Construct the perpendicular bisector LN of JM, which intersects JM at X.
- With X as centre and radius 2.1 cm, draw two arcs cutting LN on either side of JM at U and P.
- Join J–U, U–M, M–P, and P–J to complete the required square.
JM is drawn first; its perpendicular bisector LN locates X; arcs of radius 2.1 cm cut LN at U and P; joining J–U–M–P completes the square.
Common Mistakes to Avoid
- Using the full diagonal length as the compass radius: The arcs in Step 3 must use half of the second diagonal, not the full length — this is the single most common error in this exercise.
- Drawing the perpendicular bisector freehand: Always use the compass-arc method (equal arcs from both endpoints of the diagonal) to guarantee a true 90° angle — an estimated "by eye" perpendicular will throw off the whole figure.
- Joining the vertices in the wrong order: The four vertices alternate between the two diagonals — always join a vertex from one diagonal to a vertex from the other diagonal (never the two ends of the same diagonal directly, since those are not sides of the quadrilateral).
- Forgetting that a square's diagonals are equal: In Problem (c), both arcs use the same 2.1 cm radius as the first diagonal's half-length — don't be tempted to look for a second given measurement that the question never provides.
- Not verifying the construction: After construction, measure all four sides — in a true rhombus or square they must all come out equal, which is a simple, reliable check that your arcs and bisector were accurate.
Quick Reference — All 3 Constructions at a Glance
| Part | Shape & Given Diagonals | Half-Diagonal Radius Used | Side Length (check) |
|---|---|---|---|
| (a) | Rhombus CART — CR = 6 cm, AT = 4.8 cm | 2.4 cm | ≈ 3.84 cm |
| (b) | Rhombus SOAP — SA = 4.3 cm, OP = 5 cm | 2.5 cm | ≈ 3.30 cm |
| (c) | Square JUMP — diagonal = 4.2 cm | 2.1 cm | ≈ 2.97 cm |
What This Exercise Prepares You For
Exercise 3.6 builds on the constructions practised throughout the rest of Chapter 3, particularly the use of a perpendicular bisector to locate a centre point, which is the same compass technique used whenever a figure has a point of symmetry. Revising the earlier exercise on constructing a parallelogram and a rectangle is also useful, since it shares the same step-by-step construction style.
The Pythagorean check used here — finding a rhombus's side length from its half-diagonals — is also a preview of how right triangles are used throughout geometry, a theme that returns in greater depth when the Triangles chapter is studied in Class 9.