Question

plz solve these 2 questions plz it is important ​

Answer 1

Step-by-step explanation:

18

i)

ang.TQX=ang.SQX

QX=QX(common)

ang.QSX=ang.QTX(90°)

triangleXTQ is congruent to triangleXSQ {By aas}

19)congruent triangleABC and EDC by aas then find the value

similarly sss se dono triangle similar hoga

Answer 2

Answer:

PLS MARK ME AS THE BRAINLIEST FOR THIS......

Step-by-step explanation:

19) part one)

IN TRIANGLE ABC & TRIANGLE EDC,

BC = DC (GIVEN)

ANGLE ACB = ANGLE DEC (VERTICALLY OPPOSITE ANGLES)

ANGLE ABC = ANGLE EDC (ALT. INT. ANGLE)

BY ASA CONGRUENCE RULE

TRIANGLE ABC ≅ TRIANGLE EDC

SO,

AB = ED (BY CPCT)

AC = EC (BY CPCT)

2X-4 = 14

2X = 14 + 4

2X = 18

X = 18/2

X = 9

.

3Y+5 = 20

3Y = 20-5

3Y = 15

Y = 15/3

Y = 5

PART TWO)

PROVE BY CONGRUENCY.. THEN

Y - 6 = 30 (ANGLES OPPOSITE TO EQUAL SIDES ARE EQUAL)

Y = 30 + 6 = 36

.

2X + 7 = 63

2X = 63-7 = 56

X = 56/2

X = 28