CBSE Class 10 Answered
maths
Asked by krishpatel20108 | 01 Jan, 2023, 07:56: PM
Expert Answer
254 x + 309 y = -55 ...................................... (1)
309 x + 254 y = 55 .........................................(2)
By adding eqn.(1) and (2) , we get 563 x + 563 y = 0
Hence x + y = 0 or x = -y
if we substitute x = -y in eqn.(1) , we get y = -1, hence x = +1
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Question 23
(x2 - 3 ) = 0
( x + √3 ) ( x - √3 ) = 0
hence x = ± √3
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Question 24
If all the points are collinear , then a straight line ax +by + c = 0 is paasing through all the points
a x + b y + c = 0
Let us divide the above equation by a, then we get
x + (b/a ) y + ( c /a ) = 0 ..............................(1)
let ( b/a ) = p and ( c/a ) = q
Then using above substitution , we get
x + py = -q ................................. (2)
if the point (1, 2 ) is on the line , then 1 + 2p = -q
2p +q = -1 ........................................ (3)
if the point (7, 0 ) is on the line , then 7 = -q
By substituting q = -7 in eqn.(3) , we get p = 3
Hence the relation between x and y is obtained from eqn.(2)
x + 3 y = 7
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Question 25
Fig.(1) shows two concentric eircles. Radius of inner circle is OA = b and
radius of outer circle OB = a.
Cord CB of bigger circle is touching smaller circle at A , Hence CB is tangent
CB = 2 AB = 
-------------------------------------------------------
Fig.(2) shows that a tangent is drawn from P to a circle and this tangent meets the circle at Q.
QR is diamaeter of circle.
POR = 110o ( given )POQ = 180 - POR = 70o ΔPOQ is right angled triangle and Q is right angle .
Q is right angle . Hence OPQ = 20o
OPQ = 20o -------------------------------------------------------------------------------------------
Question 26
ΔABC is right angled triangle right angled at C.
AB = 29 units ( given )
BC = 21 units ( given )
AC = 
cosθ = BC / AB = 21/29
sinθ = AC/AB = 20/29
cos2θ + sin2θ = (21/29)2 + (20/29)2 = 1
cos2θ - sin2θ = (21/29)2 - (20/29)2 = 0.048
Answered by Thiyagarajan K | 01 Jan, 2023, 11:37: PM
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