Math, asked by shubhamcoolgupta2005, 4 months ago

Drills are designed to prepare students, teachers, administrators, and other people in the
school to leave the building quickly and in a pre-planned and organised fashion in the
event of danger such as a bomb threat, when conditions outside the building are safer

than the conditions inside the building.

Students of a school are standing in rows and columns in their playground for a drill

practice. A, B, C and D are the positions of four students as shown in figure.

(A) What are the Co-ordinate positions of A and B?

(i)(3,5)and(7,9)
(ii)(3,5)and(7,1)
(iii)(7,9)and(3,5)
(iv) (7,1)and(9,4)

(B) Distance between C and D is

(i) 4

(ii)4√2

(iii) 8

(iv) 2

(C) Jaspal in the drill is assigned a place in such a way that he is equidistant from each of

the four students A, B, C and D.
What should be his position?

(i) (7, 5)

(ii) (14, 10)

(iii) (6, 3)

(iv) Can be placed anywhere inside the square, formed after joining A, B, C and D

together.

(D) What is the mid point of AD?

(i) (5, 4)

(ii) (7, 5)

(iii) (3, 5)

(iv) (5, 3)




PLZZ .. ANSWER FAST ...........

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Answers

Answered by ashabariya719
1

Answer:

A bomb threat against this school may be received by phone, mail, ... students outside the building until further instructions are received. ... of the building rather than the interior. V. Type of

Step-by-step explanation:

A bomb threat against this school may be received by phone, mail, ... students outside the building until further instructions are received. ... of the building rather than the interior. V. Type of

Answered by pratik1332
1

✌️❣️_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_,_

Given P is equidistant from points A and B

PA=PB .....(1)

and Q is equidistant from points A and B

QA=QB .....(2)

In △PAQ and △PBQ

AP=BP from (1)

AQ=BQ from (2)

PQ=PQ (common)

So, △PAQ≅△PBQ (SSS congruence)

Hence ∠APQ=∠BPQ by CPCT

In △PAC and △PBC

AP=BP from (1)

∠APC=∠BPC from (3)

PC=PC (common)

△PAC≅△PBC (SAS congruence)

∴AC=BC by CPCT

and ∠ACP=∠BCP by CPCT ....(4)

Since, AB is a line segment,

∠ACP+∠BCP=180

Thus, AC=BC and ∠ACP=∠BCP=90

∴,PQ is perpendicular bisector of AB.

Hence proved.

Tusen Takk ✌️❣️

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