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
Accepted Solution
A:
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