Question

.A circle is inscribed in quadrilateral ABCD. If AB = 4, BC = 5, and CD = 8,
what is DA?

Answer 1

Explanation:

I just answered this. how often was that posted ?

the distance of a vertex to its 2 neighboring circle touching points is the same, because tangents emanating from the same point to the same circle must have the same distance to the touching points.

so, AB splits into a (from A to the circle touching point on AB) and b (from the circle touching point on AB to B).

b = AB - a

BC splits into b (from B to the circle touching point on BC) = AB - a, and c (from the circle touching point on BC to C).

c = BC - (AB - a) = BC - AB + a

CD splits into c (from C to the circle touching point on CD) = BC - AB + a, and d (from the circle touching point on CD to D).

d = CD - (BC - AB + a) = CD - BC + AB - a

DA now splits into d (from D to the circle touching point on DA) and ... a (from the circle touching point on DA to A).

a = DA - (CD - BC + AB - a) = DA - CD + BC - AB + a

0 = DA - CD + BC - AB

DA = CD - BC + AB = 8 - 5 + 4 = 7

Answer 2

DA = 7

Explanation:

You have quadrilateral ABCD with an inscribed circle and side lengths AB=4, BC=6, CD=8, and you want to know the length of DA.

Inscribed circle

The tangents from each vertex of the quadrilateral to the inscribed circle are the same length, so we can work our way around the quadrilateral to find the missing length. The attached figure shows the tangent lengths working counterclockwise from vertex A.

The sum of the tangents from A and D is ...

x + (7 -x) = 7

The length of DA is 7 units.

__

Additional comment

The sum of the lengths of opposite sides is the same in any quadrilateral with an inscribed circle. Here, that sum is 4+8 = 12, so the unknown side is 12-5 = 7.