M and N are midpoint of sides AB and BC of parallelogram ABCD. if the area parallelogram ABCD is 24cm2. what is the area of trapezium AMNC ?
Answers
Step-by-step explanation:
Join the diagonal AC From the figure we know that the diagonal AC divides the parallelogram ABCD into two triangles having the same area It can be written as Area of △ ADC = Area of △ ABC ……. (1) We know that △ ADC and parallelogram ABCD are on the same base CD and between the same parallel lines DC and AM. It can be written as Area of △ ADC = Area of △ ABC = ½ (Area of parallelogram ABCD) From the figure we know that M is the midpoint of AB So we get Area of △ AMC = Area of △ BMC = ½ (Area of △ ABC) = ½ (Area of △ ADC) It can be written as Area of AMCD = Area of △ ADC + Area of △ AMC By substituting the values 24 = Area of △ ADC + ½ (Area of △ ADC) It can be written as 24 = 3/2 (Area of △ ADC) By cross multiplication 24 × 2 = 3 × (Area of △ ADC) On further calculation 48 = 3 × (Area of △ ADC) So we get Area of △ ADC = 48/3 By division Area of △ ADC = 16 cm2 From equation (1) Area of △ ADC = Area of △ ABC = 16 cm2 Therefore, Area of △ ABC = 16 cm2.
Answer:
the diagonals of parallelogram divide it into 2 congruent triangles so the area of triangle ABC and triangle ADC are half of the area of parallelogram ABCD , MN is equal to the half of AC and is parallel to AC by mid point theorem you may also verify that the altitude of triangle MB and is half of the altitude of triangle ABC ,
since the base and height of triangle MBN is half of the triangle ABC its area will be 1/4 of triangle ABC so the area of trapezium AMNC is 3/4 of triangle ABC that is 9 CM square
area of parallelogram ABCD=1/2∆ ADC = 24/2=12
=3/4(12)
=9