Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN
2 answers:
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
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Answer:
Step-by-step explanation:
Hi there!
Midpoint = where the two endpoints are
and
Plug in the given information:
Midpoint = (5,3), Endpoint = (5,5)
where
is the other endpoint
Solve for :
Solve for :
Therefore, the other endpoint is
.
I hope this helps!
For
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.
y = (x^2 -4)/((x +2)(x^2 -49))
the numerator factors to (x -2)(x +2), so the factor of (x +2) will cancel with that in the denominator, leaving
y = (x -2)/(x^2 -49)
There are points of discontinuity at the hole, x=-2, and at each of the vertical asymptotes, at x=-7, +7.
The horizontal asymptote is y=0.