in the adjoining figure area is equal to the area of parallelogram abcd if altitude ef is 16cm long find the length of the altitude of the parallelogram to base ab of length 10 cm what is the area of triangle amd where m is mid point of side dc?
Answers
Answer:
The length of the altitude of the parallelogram = 12 cm
area of triangle AMD = 30 cm²
Step-by-step explanation:
Theorem: If 2 triangles are similar, sides are in the ratio a: b then the ratio of its areas can be written as a² : b²
Fro the figure we can see that,
ΔMCE and ΔABE are similar.
To prove ΔMCE and ΔABE
<E = <E (common)
<M = <A (corresponding angles)
<C = <B (corresponding angles)
we have also MC/AB = 5/10) = 1/2 (since M is the mid point of DC)
Therefore its areas are in the ratio 1²/2² = 1/4
To find the area of ΔABE
Area = bh/2 = (AB * FE)/2 = 10*16/2= 80 cm²
To find the area of ΔMCE
arΔMCE/arΔABE = 1/4
arΔMCE/80 = 1/4
arΔMCE = 80/4 = 20 cm²
To find height of ΔMCE
arΔMCE = MC*height = 20cm²
5*height = 20cm²
height = 20/5 = 4 cm
To find altitude of the parallelogram
altitude of the parallelogram = FE - height of ΔMCE = 16 - 4 = 12 cm
Area of triangle AMD
Area = bh/2
b = DM = 5cm and height = height of paralleogram = 12 cm
Area = (5*12)/2 = 60/2 = 30 cm²