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Ilia_Sergeevich [38]
3 years ago
10

HELP ME WITH QUESTION 5 AND 6 PLEASEEE I DONT UNDERSTAND IT???

Mathematics
1 answer:
Ilya [14]3 years ago
7 0
I would need to have the same book and go to page 73 to be able to answer it. By the way, do you have any friends that have the answer?
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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
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A) Find KM∠KEM is a right angle hence ΔKEM is a right angled triangle Using Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides we can answer the 
KM² = KE² + ME²KM² = 8² + (3√5)²       = 64 + 9x5KM = √109KM = 10.44
b)Find LMThe ratio of LM:KN is 3:5 hence if we take the length of one unit as xlength of LM is 3xand the length of KN is 5x ∠K and ∠N are equal making it a isosceles trapezoid. A line from L that cuts KN perpendicularly at D makes KE = DN
KN = LM + 2x 2x = KE + DN2x = 8+8x = 8LM = 3x = 3*8 = 24
c)Find KN Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have the same height ME = LD.
∠K = ∠N  Hence KE = DN the distance ED = LMhence KN = KE + ED + DN since ED = LM = 24and KE + DN = 16KN = 16 + 24 = 40
d)Find area KLMNArea of trapezium can be calculated using the  formula below Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)substituting values into the general equationArea = 1/2 * ME * (KN+ LM)          = 1/2 * 3√5 * (40 + 24)         = 1/2 * 3√5 * 64         = 3 x 2.23 * 32         = 214.66 units²
7 0
3 years ago
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