Circles m and k are congruent, qr, is congruent to ln and op is congruent to vw. find x and y

Given, arc QR is congruent to arc LN and arc OP is congruent to arc VW.

And the expressions for each arc in the diagram also given as:

Arc QR = 2x - y, arc LN = 11 , arc OP= 10 and arc VW=5x+y.

Hence, we will get the system of equations as following:

Arc QR = Arc LN

2x - y = 11 ...(1)

Arc OP = Arc VW

5x + y = 10 ...(2)

Next step is to add the two equation to eliminate y so that we can solve the equations for x. Therefore,

2x+5x = 11 + 10

7x = 21

Divide each sides by 7.

So, x= 3

Now plug in x=3 in equation (2) to get the value of y.

5(3) + y = 10

15 + y =10

15 + y - 15 = 10 - 15 Subtracting 15 from each sides.

y = -5

So, x=3 and y =-5

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The value of x and y is 2 and 1 respectively.

Step-by-step explanation:

Given that circles M and K are congruent, QR is congruent to LN and OP is congruent to VW.

QR=y , WV=x-1

PO=y,  LN=2x-3

we have to find the value of x and y

As QR is congruent to LN

 →    (1)

OP is congruent to VW

   →   (2)

From equation (1) and (2), we get

(2) ⇒

Hence, the value of x and y is 2 and 1 respectively.

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step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.