Latest Elitmus Aptitude Question SOLUTION: trapezium is a quadrilater with at least two sides parallel to each other. four triangles are forme using its diagonals. Area of triangles adjacent to parallel sides is A and B. fi

• Placement Papers
• /
• Elitmus
• /
• Numerical Ability
• /
• Geometry

Elitmus Question Papers:

(#M40155964) ELITMUS Numerical Ability question area of trapezium Keep an EYE

trapezium is a quadrilater with at least two sides parallel to each other. four triangles are forme using its diagonals. Area of triangles adjacent to parallel sides is A and B. find area of trapezium in terms of A and B.

Asked In Elitmus Mukesh Kumar Singh (8 years ago)

Unsolved Read Solution (11)

Is this Puzzle helpful?   (6)   (1) Submit Your Solution Geometry

Read Solution (11) :

mamta kumari  (7 years ago)

let sides of trap is l1 &l2,and heights of traingle is h1&h2
area of trap=1/2(l1+l2)(h1+h2)=A+B+1/2*l1*h2+1/2*l2*h1
also h1/h2=l1/l2 i.e l1*h2=l2*h1
so area of trap=A+B+l1*h2
now A=1/2*l1h1 i.e l1h1=2A
similarly l2h2 =2B
so l1h1*l2h2=4AB i.e l1h2^2=4AB i.e l1h2=2sqrtAB
so area of trap = A+B+2sqrtAB

Is this solution Helpfull? Yes (14) | No (1)  

Sneha Singh  (8 years ago)

where are the options??

Is this solution Helpfull? Yes (2) | No   

Mukesh Kumar Singh  (8 years ago)

I don't remember the options guys, i could not solve it. First option was (√A√B)^2. Other options i don't recall. But they were like 2(A+B) + √AB, 2AB + A+B.

It was also given (not sure) that the other two triangles have same area.

I started solving it by taking a square of side 2 (since square is also a trapezium), taking two opposite triangles as having area A and B. But, I didn't had time so i just moved on.

Is this solution Helpfull? Yes (2) | No   

santosh choudhary  (8 years ago)

C is the answer, so pls go with it ...

Is this solution Helpfull? Yes | No   

Jasminder Singh  (8 years ago)

u didn't give the options!!

Is this solution Helpfull? Yes | No (1)  

juhi  (8 years ago)

2sqrt of A and sqrt of B

Is this solution Helpfull? Yes | No   

richa gargi  (8 years ago)

product of areas of two opposite triangles is equall to other two opposite triangles
AB=CD (1)
Let K be the area of trepezium
thus K=A+B+C+D
or K=A+B+2SQRT(AB)-2SQRT(CD) +C+D -( FRM (1))
or K=(SQRT A+SQRT B)^2 +(SQRT C-SQRT D)
NOW in a trepezium
area of triangles adjacent two parallel sides and formed by diagonal is
(a^2 h)/2(a+b) and (b^2 h)/(a+b)
and other two is
abh/(2(a+b)
thus C AND D have equall areas thus C-D=0

and K=(SQRT A+SQRT B)^1/2

Is this solution Helpfull? Yes | No (5)  

RISHABH RAJ SRIVASTAVA  (7 years ago)

ans.-(root A+root B)^2

Is this solution Helpfull? Yes | No   

suvasree chatterjee  (7 years ago)

sm1 plz explain me in detail

Is this solution Helpfull? Yes | No   

pooja kumari  (7 years ago)

ar trapezium=1/2*(add of parallel lines)*height
Area of triangles adjacent to parallel sides is A and B
now, ar trapezium= 1/2*(l1+l2)*h
=(1/2*l1*h)+(1/2*l2*h)
=A+B

Is this solution Helpfull? Yes | No (3)  

intakhab  (6 years ago)

any one explain the equation plz
area of trap=1/2(l1+l2)(h1+h2)=A+B+1/2*l1*h2+1/2*l2*h1

when i m calculating it is coming like this 1/2(l1+l2)(h1+h2)=A+B+1/2*l1*h1+1/2*l2*h2

Is this solution Helpfull? Yes | No   

Please Login to Read Solution.