Question

Construct a quadrilateral ABCD in which AB = 4 cm, AD = 4.4 cm, BD = 5.4 cm, BC = 4.5 cm and AC = 5.2 cm.

Answer 1

Given:

AB = 4 cm

AD = 4.4 cm

BD = 5.4 cm

BC = 4.5 cm

AC = 5.2 cm

Step-wise calculation (Construction + Verification):

1. Draw base AC = 5.2 cm.

Place A at (0, 0) and C at (5.2, 0) for verification if desired.

This is the usual first step: construct triangle ABC from the three given lengths AB, BC and AC.

2. Construct B:

With centre A and radius 4 cm (AB), draw an arc.

With centre C and radius 4.5 cm (BC), draw an arc.

The intersection of these two arcs is point B   ...(Choose either intersection; pick the one on the same side of AC where you want the quadrilateral to lie).

Coordinate check Using A = (0, 0), C = (5.2, 0), the intersection gives

`x_B = (AB^2 - BC^2 + AC^2)/(2 xx AC)`

= `(16 - 20.25 + 27.04)/10.4`

= 2.192788... 

`y_B = +sqrt(AB^2 - x_B^2)`

= 3.346

So, B = (2.1928, 3.346).

This satisfies AB = 4 and BC = 4.5 by construction.

3. Locate D by the intersection of two arcs:

With centre A and radius AD = 4.4 cm, draw an arc.

With centre B and radius BD = 5.4 cm, draw an arc.

The intersection(s) of these two arcs give the possible positions for D.   ...(Choose the intersection that gives a non‑self‑intersecting quadrilateral in the order A–B–C–D).

Coordinate check Using the B above, the two intersection points

Solve x² + y²

= 4.4²

= 19.36 

And (x – x_B)² + (y – y_B)²

= 5.4²

= 29.16 

Subtracting yields the line 2(x_Bx + y_By) 

= x_B² + y_B² – 9.8

= 6.2

i.e., x_Bx + y_By = 3.1. 

Solving with the circle x² + y² = 19.36 gives two solutions;

Numerically, they are approximately D₁ = (–3.196, 3.022) and D₂ = (4.046, –1.726). 

Either is a valid D pick D₂ if you want the vertices in order A(0, 0) → B → C(5.2, 0) → D to form a simple quadrilateral. 

Both solutions satisfy AD = 4.4 and BD = 5.4 by construction.

4. Join the vertices in order: A – B, B – C, C – D, D – A to complete the quadrilateral ABCD.

This construction method build triangle ABC from three given sides then locate D by arcs from A and B is the standard approach for this data set.

Draw AC, draw arcs from A and C for AB and BC to get B; then draw arcs from A and B for AD and BD to get D; join.

A quadrilateral ABCD with the required lengths exists and can be constructed with straightedge and compass by:

Drawing (AC = 5.2 cm, 2) locating B as intersection of circles (A, 4 cm) and (C, 4.5 cm and 3) locating D as intersection of circles (A, 4.4 cm) and (B, 5.4 cm).