A pilot in a helicopter sports a landing pad below. If the angle of elevation is73 degree and the horizontal distance to the pad
1 answer:
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Q1)Find KM
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
= 64 + 9x5
KM = √109
KM = 10.44
Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units
If we take the figure ∠K and ∠N are equal.
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units
KN = LM + 2x
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24
Q3)Find KN
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides
same right angle when the perpendicular lines cut KN
∠K = ∠N
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN
the distance ED = LM
Therefore KN = KE + ED + DN
since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
KN = 5 * 8 = 40
Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM)
= 1/2 * 3√5 * (40 + 24)
= 1/2 * 3√5 * 64
= 3 x 2.23 * 32
= 214.66 units²
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
= 64 + 9x5
KM = √109
KM = 10.44
Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units
If we take the figure ∠K and ∠N are equal.
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units
KN = LM + 2x
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24
Q3)Find KN
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides
same right angle when the perpendicular lines cut KN
∠K = ∠N
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN
the distance ED = LM
Therefore KN = KE + ED + DN
since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
KN = 5 * 8 = 40
Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM)
= 1/2 * 3√5 * (40 + 24)
= 1/2 * 3√5 * 64
= 3 x 2.23 * 32
= 214.66 units²
Answer:
You could use the polygon area formula OR here's an easier solution
We draw the lines DC and DE
Angle ADC = 120°
Triangle ADC is a 30 60 90 triangle
In such a triangle Line AE = hypotenuse * sq root (3) /2
Line AE = 3 * 0.8660254038 = 2.5980762114
Line DE = 1.5
Area of Triangle ADE = .5 * 1.5 * 2.5980762114 =
1.9485571586 square meters and entire area of Triangle
ABC = 6 * 1.9485571586
which equals 11.6913429513 square meters
Step-by-step explanation:
