[urgent]help me solve class X physics questions

[Niilesh]

Hey Can someone solve these question and explain it?
I want the answer before 25th(till 24th)


I didn't even understand this-


A maths one(this one seems impossible)

81. The area of quadrilateral ABCD whose sides are 9 m, 40 m, 28 m and 15 m is
(A) 300 m2
(B) 306 m2
(C) 296 m2
(D) none of these

Update: solved the first and the third
can someone explain the second?
at least just what it means?

[avichandana20000]

YOUR QN REFRMAED:

I have a parcel with four sides with lengths of 9 on the north, 40 on the south, 28 on the west and 15 on the east. I do not know any of the angles and presume that none of them are right angles. Also presume that none of the sides are parallel to each other.

I can't divide it into two triangles and calculate those areas using the formula 1/2 b X h because neither of the triangles would have a right angle.

How do I go about calculating the area?



use Brahmagupta Formula:

A = SqRt[(s-a)(s-b)(s-c)(s-d)]

a, b, c and d are the lengths of the four sides of the quadrilateral, and s = (a + b + c + d)/2.

s=(9+40+28+15)/2=46

Area=Sq rt(46-9)(46-40)(46-28)(46-15)

=Sq rt(37)(6)(18)(31)

=Sqrt(123876)

=351.96

So Answer is D

hope this helps.

[Niilesh]

^
the formula only applies to cyclic quadrilateral
Answer is B (i have answer key)
actually the question is wrong but it has a special case in it a pythagoras triplet

Area of quad = 126 + 180 (by herons formula) = 306

Also i solved the first question

[dashing.sujay]

^How did you solved 3rd ?

[Niilesh]

^ you mean first?
I didn't solve the second one

[dashing.sujay]

No 3rd.

[Niilesh]

This is the fig is:


Now area of quadrilateral = area of two triangles
Area of triangle with sides 40,41 and 9 is 180(1/2*40*9)
area of other triangle is 126(herons formula)
total = 306 units

PS: i didn't label the image properly before

[dashing.sujay]

How can you consider the figure to be that way along with right angles?

[Niilesh]

Actually the question is wrong
But i found that it had a Pythagoras triplet so just put a right angle there
BTW in the question the order is 9,40,28,15
switch the 28 and 15 sides in my fig and there you have it - the required quadrilateral

also if you consider it to be a trapezium with height 9 and with parallel sides 40 and 28 the area is same

[vaibhav23]

^In that maths question if you do it the other way round that is 28^2+15^2 then the hypotenuse should be equal to 41 but it is not coming to 41.
In a quardilateral if one angle is 90 all other angles should be also 90 all this leading to that it is either a rectangle or a square

[Niilesh]

^In that maths question if you do it the other way round that is 28^2+15^2 then the hypotenuse should be equal to 41 but it is not coming to 41.
In a quardilateral if one angle is 90 all other angles should be also 90 all this leading to that it is either a rectangle or a square

hey man who said that other triangle is also right angled?

"In a quardilateral if one angle is 90 all other angles should be also 90 all this leading to that it is either a rectangle or a square" - this line is wrong. Go revise the basics
PS: no offence

[dashing.sujay]

Answer of second is?

[Niilesh]

^(c)

char

[vaibhav23]

hey man who said that other triangle is also right angled?

"In a quardilateral if one angle is 90 all other angles should be also 90 all this leading to that it is either a rectangle or a square" - this line is wrong. Go revise the basics
PS: no offence

^Sorry I mixed it up a little.
BTW what was the answer you got for the first one

[Niilesh]

Its again C
you will not understand it without knowing trigno and of course snells law

[vaibhav23]

^I have studied that chapter in class 9 and have took out the answer.
My one is also C but wanted to cross check with how you got the answer

[Niilesh]

^
I learned it in class X(since NCERT of class IX didn't have those topics and i was not preparing for competitive exams)

You want to know the solution?

[Gollum]

I give up, OMG I hate studies *runs away screaming*

[Niilesh]

^
BTW Test was held today
It went better than my expectation

[dashing.sujay]

^Share the results too

[Niilesh]

Ok ,they will come on 28th

finally i can enjoy the holidays

[vaibhav23]

^
I learned it in class X(since NCERT of class IX didn't have those topics and i was not preparing for competitive exams)

You want to know the solution?

Yup
You are taking pinnacle course from which city?
If you want you can take the pinnacle course from South Delhi as it is the best fiitjee institute

[Niilesh]

^
I live in Jaipur (you can see that with the post count on the left)

[kuki_295]

In second ques, magnetic force only acts on moving charged particles so it'll just sit there.
Is there magnetism chapter is 10th class.
Explaining the question: a charged particle is on a table tied to a string and there is a magnetic field perpendicular to table surface.

[Niilesh]

In second ques, magnetic force only acts on moving charged particles so it'll just sit there.
Is there magnetism chapter is 10th class.
Explaining the question: a charged particle is on a table tied to a string and there is a magnetic field perpendicular to table surface.

I didn't read that
BTW magnetism was in 9th

[Niilesh]

Got 185/270
90% scholarship (highest)

[dashing.sujay]

Can you share the questions ?

[Niilesh]

^ The question paper was taken with the OMR sheet but if you want i can post few questions that i can remember

[dashing.sujay]

Yes please.

[Niilesh]

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