Answer:
Proved
Step-by-step explanation:
Given: EC || AC, DB || AC, ∠A = ∠F
Prove: ΔMDF ∼ ΔNCA
Solution
See diagram attached to the solution to better understand the following workings.
Redrawing ΔMDF or rotating to be facing the same direction.
EC is parallel to AC
DB parallel to AC
Using similar triangle theorem:
If ΔMDF ∼ ΔNCA
Ratio of Corresponding sides would be equal
(adjacent of ΔMDF)/(adjacent of ΔNCA) = (Opposite of ΔMDF)/(opposite of ΔNCA) = (hypotenuse of ΔMDF)/(hypotenuse of ΔNCA)
DF/ CA = MD/NC = FM/AN
∠A = ∠F
∠M = ∠N
∠D = ∠C
Since the ratio of Corresponding sides and angle are equal, ΔMDF is similar to ΔNCA.
ΔMDF ∼ ΔNCA
Answer:5/10
Step-by-step explanation:
1/2 x 5 = 5/10
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Option C:
Area of the remaining paper = (3x – 4)(3x + 4) square centimeter
Solution:
Area of the square paper =
sq. cm
Area of the square corner removed = 16 sq. cm
Let us find the area of the remaining paper.
Area of the remaining paper = Area of the square paper – Area of the corner
Area of the remaining = 
= 
Using algebraic formula: 

Area of the remaining paper = (3x – 4)(3x + 4) square centimeter
Hence (3x – 4)(3x + 4) represents area of the remaining paper in square centimeters.