Math, asked by guntikavitha833, 5 months ago

1. If x– coordinate of a point is zero, then this point always lies:

(a) I quadrant (b) II quadrant (c) x – axis (d) y – axis


2. The point (1, –1), (2, –2), (4, –5), ( –3, –4) lies in:

(a) II quadrant
(b) III quadrant

(c) IV quadrant
(d) do not lie in the same quadrant


3. Point (5, 0) lies on the:

(a) I quadrant
(b) II quadrant
(c) x – axis
(d) y –axis


4. Point ( –6, 4) lies in the quadrant:

(a) I
(b) II
(c) III
(d) IV


5. The point ( –4, –3) means: [ ]

(a) x = –4, y = –3
(b) x = –3, y = –4
(c) x = 4, y = 3
(d) None of these


6. Point (0, 4) lies on the:

(a) I quadrant
(b) II quadrant
(c) x – axis
(d) y – axis


7. Point (1, 7) lies in the quadrant: [ ]

(a) I
(b) II
(c) III
(d) IV


8. On joining points (0, 0), (0, 2), (2,2) and (2, 0) we obtain a [ ]

(a) Square
(b) Rectangle
(c) Rhombus
(d) Parallelogram


9. Signs of the abscissa and ordinate of a point in the first quadrant are respectively:

(a) +, +
(b) , +
(c) +, 

(d) , 

10. The point whose ordinate is 4 and which lies on y  axis is: [ ]

(a) (4, 0)
(b) (0, 4)
(c) (1, 4)
(d) (4, 2)

11. Abscissa of a point is positive in:

(a) I and II quadrant
(b) I and IV quadrant
(c) I quadrant only
(d) II quadrant only


12. The point where the two axes meet, is called

(a) x-coordinate
(b) y- coordinate
(c) quadrant
(d) origin


13. The point ( –5, 4) and (4, –5) are situated in [ ]

(a) same quadrant
(b) I and III quadrant, respectively

(c) Different quadrants
(d) IV and II quadrant, respectively​

Answers

Answered by citivhiyiyfifyfyfi
5

Answer:

hcgchiiucguggvuugvgufufyuyfifyyfiifyiyguiAnswer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.Answer:

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

ANSWER

AD=AC ( Given )

So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]

Now ∠ACD is exterior angle of △ABD

∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]

So, ∠ADC>∠ABD

∠ACD>∠ABD ( from (1))

AB>AC [ side opposite to greater angle is longer ]

∴AB>AD ( As AC=AD given )

Hence, proved.

Answered by rachanakrishna
3

Answer:

the above anwer us right

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