Can somewon help me i dont get this kind of math
2 answers:
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Answer:
See Below.
Step-by-step explanation:
We are given that ΔAPB and ΔAQC are equilateral triangles.
And we want to prove that PC = BQ.
Since ΔAPB and ΔAQC are equilateral triangles, this means that:
Likewise:
Since they all measure 60°.
Note that ∠PAC is the addition of the angles ∠PAB and ∠BAC. So:
Likewise:
Since ∠QAC ≅ ∠PAB:
And by substitution:
Thus:
Then by SAS Congruence:
And by CPCTC:
Difference between the area of the triangle and square is 25
Step-by-step explanation:
- Step 1: Find the area of the triangle given its 3 sides using the Heron's formula.
Area of the triangle = where s =
⇒ s = (6 + 8 + 10)/2 = 24/2 = 12
=
= =
= 24 sq. units
- Step 2: Find the area of the square with perimeter = 28 units.
Perimeter of the square = 4 × side = 28
⇒ Side of the square = 28/4 = 7 units
⇒ Area of the square = (side)² = 7² = 49 sq. units
- Step 3: Find the difference between the area of the square and triangle.
Difference = 49 - 24 = 25