THREE QUESTIONSS PLEASE HELP !!!
2 answers:
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Answer:
The perimeter of polygon ABCD, to the nearest thousandth units is
22.227 units
The perimeter of polygon ABCD', to the nearest thousandth, would be 20.980 units
The area of polygon ABCD' is 19.5 units²
Step-by-step explanation:
The coordinates forming the polygon are
A (-4,-1), B(-2,3), C(2,2), and D(4,-3)
The perimeter then is given by the sum of the length of the sides as follows;
Length of line between two X and Y points distance, xi and yj apart is
length XY =
Therefore, the length between points AB is
length AB =
= = 4.472 units
Similarly, length BC is given by
Length BC = = 4.123 units
Length CD = = 5.385 units
Length DA = = 8.246 units
The perimeter is equal to;
length AB + Length BC + Length CD + Length DA
= 4.472 + 4.123 + 5.385 + 8.246 = 22.227 units
If the point D is moved up 2 units and left 1 unit we have
D' = x-1, y+2 where x and y are the coordinates of point D
D(4, -3) → D'((4-1), (-3+2)) = D'(3, -1)
The length D'A = =7 units
The perimeter of polygon ABCD'
length AB + Length BC + Length CD + Length D'A
= 4.472 + 4.123 + 5.385 + 7 = 20.980 units.
The area is given by the determinant of the 3 by 3 matrix using Cramer's Rule as follows
=
=
Triangle ABC we have
A(-4,-1)
B(-2,3)
C(2,2)
=
=
Triangle ACD'
A(-4,-1)
C(2,2)
D'(3, -1)
=
= 10.5 units²
Area of polygon = =
+
= (9 + 10.5) units²
Area of polygon ABCD' = 19.5 units².
Given: NQ = NT , QS Bisect NT(∴ NS=ST ) , TV Bisects QN (∴ NV=VQ )
To Prove: QS=TV
Proof: In ΔNQT
NQ=NT
∴ VQ=ST
In a isosceles triangle, If two sides are equal then their opposites angles are equal.
∴ ∠NQT=∠NTQ ( ∵ NQ=NT)
In ΔQST and TVQ
ST=VQ (sides of isosceles triangle)
∠NQT=∠NTQ (Prove above)
QT=TQ (Common)
So, ΔQST ≅ TVQ by SAS congruence property
∴ QS=TV (CPCT)
CPCT: Congruent part of congruence triangles.
Hence Proved