Answer:
Solution given:
m∠ADB=(4x−12)°
m∠CDB=(3x+6)°
m∠ADC =?
Since diagonal BD bisect the angle <ADC
so
m∠ADB= m∠ADC
(4x-12)°=(3x+6)°
4x-3x=6-12
x=12+6
x=18°
again.
<ADB=m∠ADB+ m∠ADC=4×18-12+3×18+6=120°
So
<u>the m∠</u><u>ADC</u><u> </u><u>=</u><u>1</u><u>2</u><u>0</u><u>°</u>
Answer:30%
Step-by-step explanation:
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
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