# The sides KL and MN of the KLMN trapezoid are 15 and 12, respectively, and the base LM = 3.

**The sides KL and MN of the KLMN trapezoid are 15 and 12, respectively, and the base LM = 3. The bisector of angle NKL passes through the midpoint of side MN. Find the area of the trapezoid.**

Let the bisector of angle NKL intersect the side МN at point E.

Line KE intersects the continuation of the smaller base LM at point C.

Line LC is parallel to KN.

∠ LCK = ∠ CKN as intersecting at the intersection of parallel lines of the secant CK.

But ∠ CKN = ∠ CKL by the condition (CK is the bisector of the angle NKL).

∠ LKC = ∠ LCK;

KLC triangle – isosceles;

KL = LC = 15;

MS = LC – LM = 15 – 3 = 12;

∠ CME = ∠ ENK as intersecting for parallel LC and KN and secant MN.

ME = EN by condition;

The angles at E are equal as vertical;

Triangles MCE and KNE are equal in side and adjacent angles KN = MC = 12;

From the vertex L draw LH parallel to MN;

NH = LM = 3 as sides of the parallelogram LMNH;

LH = MN = 12 as sides of a parallelogram (by construction);

KH = KN – NH;

KH = 12 – 3 = 9;

In the KLH triangle, the aspect ratio is KH: LH: KL = 3: 4: 5.

This is the ratio of a right-angled (Egyptian) triangle. (can be checked by T. Pythagoras)

⇒⊿ KLH is rectangular, LH is perpendicular to KN and is the height of the KLMN trapezoid.

The area of the trapezoid is equal to the product of the height and the half-sum of the bases.

S = LH * (LM + KN): 2;

S (KLMN) = 12 * (3 + 12): 2 = 90 (area units).

Answer: The area of the trapezoid is 90 area units.