Class 10 math all important theorems
Answers
Answer:
CIRCLES:
THEOREM 10.1 ( image - 1)
→the tangent at any point of a circle is perpendicular through the radius through the point of contact.
given: in a circle with Centre O, xy is a tangent and point P is point of contact.
To prove: OP⊥xy
proof: take any point Q and tangent x y join OQ consider OQ Insider
NOW, OQ> OR
OQ>OP {OR= OP , RADUS}
OP is the shortest of all distance of the point O to the tangent xy
∴OP is perpendicular to xy
THEOREM 10.2 ( image - 2)
→The lengths of tangents drawn from an external point to a circle are equal
Given : AP and AQ are two tangents from point A to a circle O
To Prove : AP=AQ
Construction : Join OA, OP, and OQ
Proof :AP is a tangent at P and OP is the radius through P.
∴ OP⊥AP
Similarly, AQ is a tangent at Q and OQ is the radius through Q
∴ OQ⊥AQ
In the right ∆OPA and ∆OQA, we have
OP=OQ (equal radii of the same circle)
AO=AO(common)
∠OPA=∠OQA (Each 90°)
∴∆OPA=∆OQA (By RHS congruence)
⟹ AP=AQ (By CPCT)
Hence AP=AQ proved { NOTE THAT YOU CAN PROVE THIS BY PYTHOGOARAS THEOREM ALSO}
TRIANGLES:
THEOREM 6.1 { IMAGE - 3}
if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides of the triangle in proportion.
Construction: ABC is a triangle, DE is a line parallel to BC and intersecting AB at D and AC at E, i.e. DE || BC.
Join C to D and B to E. Draw EM ⊥ AB and DN ⊥ AC.
To prove :
Area of a triangle, ADE = ½ × AD × EM
Similarly,
Ar(BDE) = ½ × DB × EM
Ar(ADE) = ½ × AE × DN
Ar(DEC) = ½ × EC × DN
Hence,
Ar(ADE)/Ar(BDE) = ½ × AD × EM / ½ × DB × EM = AD/DB
Similarly,
Ar(ADE)/Ar(DEC) = AE/EC
Triangles DEC and BDE are on the same base, i.e. DE and between same parallels DE and BC.
Hence,
Ar(BDE) = Ar(DEC)
From the above equations, we can say that
AD/DB = AE/EC.
Hence, proved.
THEOREM 6.2 { IMAGE - 4 }
Converse of Basic Proportionality Theorem
If a line divides any of the two sides of a triangle in the same ratio, then that line is parallel to the third side.
Suppose a line DE, intersects the two sides of a triangle AB and AC at D and E, such that;
AD/DB = AE/EC ……(1)
Assume DE is not parallel to BC. Now, draw a line DE’ parallel to BC.
Hence, by similar triangles,
AD/DB = AE’/E’C ……(2)
From eq. 1 and 2, we get;
AE/EC = AE’/E’C
Adding 1 on both the sides;
AE/EC + 1 = AE’/E’C +1
(AE +EC)/EC = (AE’+E’C)/E’C
AC/EC = AC/E’C
So, EC = E’C
This is possible only when E and E’ coincides.
But, DE’//BC
Therefore, DE//BC.
Hence, proved.
REAL NUMBERS :
Fundamental Theorem of Arithmetic:
The statement of the fundamental theorem of arithmetic is: "Every composite number can be factorized as a product of primes, and this factorization is unique, apart from the order in which the prime factors occur."
240 = 2 × 2 × 2 × 2 × 3 × 5.
This theorem further tells us that this factorization must be unique.
That is, there is no other way to express 240 as a product of primes
Of course, we can change the order in which the prime factors occur. For example, the prime factorization can be written as:
240 = 31 × 24 × 51 or 31× 22 × 51 × 22 etc.
But the set of prime factors (and the number of times each factor occurs) is unique.
That is, 240 can have only one possible prime factorization, with four factors of 2 that is 24, one factor of 3 that is 31, and one factor of 5 that is 51.
HENCE PROVED
ALL THEROEAM 2023 SYLLABUS
