Question

Construct a quadrilateral ABCD in which AB = 5.2 cm, BC = 6 cm, ∠B = 70°, CD = 4.8 cm and DA = 4.0 cm.

Answer 1

Given:

AB = 5.2 cm

BC = 6.0 cm

∠B = 70°

CD = 4.8 cm

DA = 4.0 cm

Step wise calculation (Construction + Numeric check):

1. Draw BC = 6.0 cm.

Mark B and C on your paper, put B at the origin for a coordinate check: B(0, 0), C(6, 0).

2. At B construct ∠ABC = 70°.

On the ray that makes 70° with BC, mark A so that BA = 5.2 cm.

Compass–straightedge: Open the compass to 5.2 cm, with centre B cut the 70° ray at A.

Coordinate check (Optional): With BC on the x-axis and the 70° ray above it.

A = (5.2 cos 70°, 5.2 sin 70°)

= 1.779, 4.886

3. We must locate D so that DA = 4.0 cm and DC = 4.8 cm.

Using compass: Draw the circle with centre A radius 4.0 cm and the circle with centre C radius 4.8 cm; their intersection(s) give the possible position(s) of D.

This is the standard method of locating D.

4. Optional numeric verification / find coordinates of D.

Vector AC

= C – A

= 6 – 1.779, 0 – 4.886

= 4.221, –4.886

Distance d = |AC|

= `sqrt(4.221^2 + (-4.886)^2)`

= 6.459

Using circle–intersection formulas:

`A = (rA^2 - rC^2 + d^2)/(2d)` where rA = 4.0, rC = 4.8.

Compute `a = (16 - 23.04 + 41.694)/(2 xx 6.459)` 

= 2.684

`h = sqrt(rA^2 - a^2)`

= `sqrt(16 - 2.684^2)`

= 2.967

Point P on AC at distance a from A:

`P = A + (a/d)(C - A)`

= 3.534, 2.855

Two intersection points D = P ± (h) × (Unit vector perpendicular to AC).

Numerically:

D₁ = 5.780, 4.797

D₂ = 1.288, 0.913

Check distances (Approx):

AD₁ = 4.00, CD₁ = 4.80

AD₂ = 4.00, CD₂ = 4.80   ...(Both intersections satisfy the required lengths; choose either to get a quadrilateral)

5. Join A to D and C to D to finish the quadrilateral ABCD.

Construction: Draw BC = 6 cm, at B make ∠B = 70° and mark A with BA = 5.2 cm.

Draw circle centre A radius 4.0 cm and circle centre C radius 4.8 cm; their intersection(s) are the possible D points; join AD and CD.

There are two possible quadrilaterals two circle intersections. 

In coordinates, if you placed B at (0, 0) and C at (6, 0), the two possible D positions are approximately D₁(5.780, 4.797) and D₂(1.288, 0.913); both give AD = 4.0 cm and CD = 4.8 cm.