Prove the diagonals of a square are perpendicular
1 answer:
Answer:
Given :- ABCD is a square.
To proof :- AC = BD and AC ⊥ BD
Proof :- In △ ADB and △ BCA
AD = BC [ Sides of a square are equal ]
∠BAD = ∠ABC [ 90° each ]
AB = BA [ Common side ]
△ADB ≅ △BCA [ SAS congruency rule ]
⇒ AC = BD [ Corresponding parts of congruent triangles are equal ]
In △AOB and △AOD
OB = OD [ Square is also a parallelogram therefore, diagonal of parallelogram bisect each other ]
AB = AD [ Sides of a square are equal ]
AO = AO [ Common side ]
△AOB ≅ △ AOD [ SSS congruency rule ]
⇒ ∠AOB = ∠AOD [ Corresponding parts of congruent triangles are equal]
∠AOB + ∠AOD = 180° [ Linear pair ]
∠ AOB = ∠AOD = 90°
⇒ AO ⊥ BD
⇒ AC ⊥ BD
Hence proved, AC = BD and AC ⊥BD
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Answer: -1066.65
-200 = (1/4)(-800)
-50 = (1/4)(-200)
-12.5 = (1/4)(-50)
Since we have a common ratio, r, of 1/4, we can use the properties of a geometric series to find the 8th term of the series.
Recall that to find the sum of the nth term of a geometric series, we have
where a is the first term of the series and r is the ratio.
So, for the first eight terms, we have
Therefore, the sum of the 8th series is approximately -1066.65.
Answer: -1066.65