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Eukaryotes mare more related to bacteria or archae?
full form of ATP
adenosine triphosphate
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Test
6. In a square ABCD, diagonals meet at O. P is a point on BC such that OB = BP.
Show that:
(i) ∠POC = 22 ½o
(ii) ∠BDC = 2 ∠POC
(iii) ∠BOP = 3 ∠CPO
In the triangle BOC and triangle DOC
BD = BD [common side]
BO = CO
OD = OC [since diagonals cuts at O]
? BOC ? ? DOC [By SSS postulate]
Therefore,
?BOC = 90o
Now ?POC = 22.5
?BOP = 67.5
Again, in triangle BDC
?BDC = 45o [since ?B = 45o and ?C = 90o]
Therefore
?BDC = 2?POC
?BOP = 67.5o
?BOP = 2 ?POC
Hence the proof.
5. The difference between an exterior angle of (n – 1) sided regular polygon and an exterior angle of (n + 2) sided regular polygon is 6o find the value of n.
For (n-1) sided regular polygon:
Let measure of each angle is x.
Therefore
(n – 1) x = (2 (n -1) – 4) 90o
x = (n-3/ n – 1) 180o
For (n+1) sided regular polygon:
Let measure of each angle is y.
Therefore
(n + 2) y = (2 (n + 2) – 4) 90o
y = (n/ n + 2) 180o
now we have
y – x = 6o
(n/ n + 2) 180o – (n-3/ n – 1) 180o = 6o
(n /n + 2) – (n – 3/ n – 1) = 1/30
30 n(n – 1) – 30 (n – 3) (n + 2) = (n + 2) (n -1)
n2 + n – 182 = 0
(n – 13) (n + 14) = 0
n = 13, -14
Thus, the value of n is 13.
4. The ratio between an exterior angle and an interior angle of a regular polygon is 2 : 3. Find the number of sides in the polygon.
5