Question

The angle between a diagonal and the longer base of the isosceles trapezoid MNFD is 45°. NK is an altitude to the longer base. If MD = 9 and NF = 5, find:
- NK
- MK
- MF
- $A_{AMNFD}$

Answer 1

Answer:

$NK \approx 7.001$, $MK = 2$, $MF \approx 7.001$, $A_{AMNFD} = 49.007$

Step-by-step explanation:

According to the statement, we find the following inputs:

$\angle MDK = \angle DMF = 45^{\circ}$ (Due to the condition of isosceles trapezoid)

$MD = 9$

$NF = 5$

Given than longer base and shorter base are parallel to each other, we conclude that:

$\angle KND = 180^{\circ} -\angle NKD - \angle KDN$

$\angle KND = 180^{\circ}-90^{\circ}-45^{\circ}$

$\angle KND = 45^{\circ}$

$\angle FND = 90^{\circ}-\angle KND$

$\angle FND = 45^{\circ}$ (By definition of complementary angles)

$\angle FND = \angle NFM = 45^{\circ}$ (Due to the condition of isosceles trapezoid)

$\angle MDK = \angle DMF = \angle FND = \angle NFM = 45^{\circ}$

$\angle NOF = \angle MOD = \angle MOF = \angle FOD = 90^{\circ}$ (By definitions of complementary and vertical angles and the theorem that states that sum of internal angles within a triangle equals 180º)

$MO = DO = \frac{\sqrt{2}}{2}\cdot MD$ (By theorem for 45-45-90 Right Triangle)

$NO = OF = \frac{\sqrt{2}}{2}\cdot NF$ (By theorem for 45-45-90 Right Triangle)

If we know that $MD = 9$ and $NF = 5$, then we find that:

$DO = MO \approx 6.364$

$NO = OF \approx 3.536$

The value of MK is obtained from the following relationship:

$MK = \frac{MD-NF}{2}$

$MK = \frac{9-5}{2}$

$MK = 2$

And the value of KD is calculated from this expression:

$KD = MD-MK$

$KD = 9-2$

$KD = 7$

Now by the Pythagorean Theorem we find that:

$NK = \sqrt{(NO+DO)^{2}-KD^{2}}$

$NK = \sqrt{9.9^{2}-7^{2}}$

$NK \approx 7.001$

And considering the symmetry characteristics of an isosceles trapezoid, we determine MF:

$MF = NO + DO \approx 7.001$

Lastly, the area of the isosceles trapezoid is determined by the following formula:

$A_{AMNFD} = NF\cdot NK + MK\cdot NK$

$A_{AMNFD} = NK\cdot (NF+MK)$

If we know that $NK \approx 7.001$, $NF = 5$ and $MK = 2$, then the area of the figure is:

$A_{AMNFD} = (7.001)\cdot (5+2)$

$A_{AMNFD} = 49.007$