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

---------------------------------------------------------------------------------------------------

Question 23

(x²- 3 ) = 0

( x + √3 ) ( x - √3 ) = 0

hence x = ± √3

--------------------------------------

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

------------------------------------------------------------------------

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

begin mathsize 14px style A B space equals space square root of left parenthesis O B right parenthesis squared minus left parenthesis O A right parenthesis squared end root space equals space square root of a squared minus b squared end root end style

CB = 2 AB =

begin mathsize 14px style 2 space square root of a squared minus b squared end root end style

-------------------------------------------------------

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.

begin mathsize 14px style angle end style

POR = 110^o ( given )

begin mathsize 14px style angle end style

POQ = 180 -

begin mathsize 14px style angle end style

POR = 70^o

ΔPOQ is right angled triangle and

begin mathsize 14px style angle end style

Q is right angle .

Hence

begin mathsize 14px style angle end style

OPQ = 20^o

-------------------------------------------------------------------------------------------

Question 26

ΔABC is right angled triangle right angled at C.

AB = 29 units ( given )

BC = 21 units ( given )

AC =

begin mathsize 14px style square root of A B squared space minus space B C squared end root space equals space square root of left parenthesis 29 right parenthesis squared minus left parenthesis 21 right parenthesis squared end root space equals space 20 space u n i t s end style

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

CBSE Class 10 Answered

Application Videos

Concept Videos

Elimination Method

Equations Reducible to Linear Form

CALL US / WHATSAPP

CBSE Class 10 Answered

Application Videos

Concept Videos

Elimination Method

Equations Reducible to Linear Form

Choose Filters

Get in Touch

Call / Whats App