Question

Given: KLMN is a trapezoid m∠N = m∠KML

Answer 1

Answer:

The length of sides KM, LM and KN are $\sqrt{109}, \frac{48}{5}$ and 16 respectively. The area of KLMN is $\frac{192\sqrt{5}}{5}$ square unit.

Step-by-step explanation:

According to given information:  KLMN is a trapezoid, ∠N= ∠KML, $\frac{LM}{KN}=\frac{3}{5}$, ME ⊥ KN, KE=8 $ME=3\sqrt{5}$.

Use pythagoras theorem is triangle EKM

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Let angle N and angle KML be θ.

Since angle KML and angle MKE are alternate interior angles, therefore angle MKE is θ.

In triangle KME,

$\tan\theta=\frac{3\sqrt{5} }{8}$     .... (1)

In triangle KNE,

$\tan\theta=\frac{3\sqrt{5} }{EN}$ .... (2)

Equate (1) and (2),

$\frac{3\sqrt{5} }{8}=\frac{3\sqrt{5} }{EN}$

$EN=8$

The length of side KN is

$KN=KE+EN=8+8=16$

Sides LM:KN are in the ratio 3:5. Let the side lengths are 3x and 5x respectively.

$5x=16$

$x=\frac{16}{5}$

$3x=3\times \frac{16}{5}=\frac{48}{5}$

The area of KLMN is

$Area=\frac{1}{2}(b_1+b_2)h$

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

$A=\frac{1}{2}\times (\frac{48}{5}+16)\times 3\sqrt{5}$

$A=\frac{192\sqrt{5}}{5}$

Therefore length of sides KM, LM and KN are $\sqrt{109}, \frac{48}{5}$ and 16 respectively. The area of KLMN is $\frac{192\sqrt{5}}{5}$ square unit.