Question

PLEASE HELP

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

Answer 1

The area of a triangle can be found using the formula 1/2 x base x height. For a base of 166 mm and height of 930 mm, the area is 77.19 cm². However, for trapezoid KLMN, additional information is needed to calculate its area.

Step-by-step explanation:

To find the area of a trapezoid, especially when it has special properties like equal non-parallel sides (as trapezoid KLMN does with KL equal to MN), we often need to break it down into simpler shapes such as triangles and rectangles. However, with the information given, it's not straightforward to calculate the area of the trapezoid without additional height or angle information. The area of a triangle can be calculated using the formula ½ × base × height. For example, if you have a triangle with a base of 166 mm and height of 930.0 mm, the area is calculated as ½ × 166 mm × 930.0 mm which equals 77190 mm² or 77.19 cm² when converted to square centimeters, maintaining the proper number of significant figures.

Answer 2

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) = (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) = (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

`=(1)/(2)* KO* (KN+LM)`

`=(1)/(2)* KO* (KN+KN + 2PM)`

`=(1)/(2)* KO* (2KN+2PM)`

`=KO* (KN+PM)`

`=15 \sin 49^(\circ)* 15 \cos 49^(\circ)`

`=111.405157734`

`\approx 111.41\text{ square unit }`