Question

PLEASE HELP


Given: KLMN is a trapezoid, KL=MN, KM=15, m∠MKN=49° Find: The area of KLMN

Answer 1

Answer:

The area would be 111.41 unit² ( approx )

Step-by-step explanation:

Given,

KLMN is a trapezoid,

In which KL=MN, KM=15, m∠MKN=49°,

Suppose O and P are point in the segment LM, ( shown below )

Such that,

KN = OP,

In triangle MKO,

m∠KMO = 49° ( ∵ KM ║ LM, using alternative interior angle theorem ),

KM = 15 unit ( given )

$\sin 49^{\circ} = \frac{KO}{KM}$

$\implies KO = KM \sin 49^{\circ}=15 \sin 49^{\circ}$

i.e. height of the trapezoid KLMN is 15 sin 49°,

Again,

$\cos 49^{\circ} = \frac{OM}{KM}$

$\implies OM = KM \cos 49^{\circ}=15 \cos 49^{\circ}$

$OP + PM = 15 \cos 49^{\circ}$

$\implies KN + PM = 15 \cos 49^{\circ}----(1)$

Now, in right triangles KOL and NPM,

KL = MN,

OK = NP

By HL postulate of congruence,

Δ KOL ≅ Δ NPM

By CPCTC,

LO = PM,

⇒ LM = LO + OP + PM = PM + OP + PM = OP + 2PM = KN + 2PM-----(2),

Thus, the area of the trapezoid KLMN,

= 1/2 × height × sum of opposite parallel sides

$=\frac{1}{2}\times KO\times (KN+LM)$

$=\frac{1}{2}\times KO\times (KN+KN + 2PM)$

$=\frac{1}{2}\times KO\times (2KN+2PM)$

$=KO\times (KN+PM)$

$=15 \sin 49^{\circ}\times 15 \cos 49^{\circ}$

$=111.405157734$

$\approx 111.41\text{ square unit }$