[urgent]help me solve class X physics questions

Niilesh

Niilesh

Padawan
Hey Can someone solve these question and explain it?
I want the answer before 25th(till 24th)
*i.imgur.com/E6RFX.jpg

I didn't even understand this-
*i.imgur.com/OIeAe.jpg

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?
 
Last edited:
avichandana20000

avichandana20000

Cyborg Agent
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.
 
OP
Niilesh

Niilesh

Padawan
^
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
*i.imgur.com/WRmI0.jpg
Area of quad = 126 + 180 (by herons formula) = 306

Also i solved the first question
 
OP
Niilesh

Niilesh

Padawan
This is the fig is:
*i.imgur.com/wXkfF.jpg

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 :oops:
 
OP
Niilesh

Niilesh

Padawan
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

vaibhav23

In the zone
^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
 
OP
Niilesh

Niilesh

Padawan
vaibhav23 said:
^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
Click to expand...

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
 
vaibhav23

vaibhav23

In the zone
Niilesh said:
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
Click to expand...
^Sorry I mixed it up a little.:lol:
BTW what was the answer you got for the first one
 
vaibhav23

vaibhav23

In the zone
^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
 
OP
Niilesh

Niilesh

Padawan
^
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?
 
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