Latest Elitmus Aptitude Question SOLUTION: A Quadrilateral PQRS circumscribes a circle with center O and PQ||RS and PQ thrice of RS.length of QR,SP are equal but not parallel.Perimeter of quadrilateral iss 24...then area of

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(#M40033075) ELITMUS Numerical Ability question Quadrilateral Keep an EYE

A Quadrilateral PQRS circumscribes a circle with center O and PQ||RS and PQ thrice of RS.length of QR,SP are equal but not parallel.Perimeter of quadrilateral iss 24...then area of quadrilateral=?

Asked In Elitmus Vivek Jain (10 years ago)

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RAKESH  (10 years ago)

from diagram,
PQ=3RS => if RS=x, then PQ = 3x
PS=QR = y(let)

perimeter = 24
=> x+3x+y+y=24
=> 4x+2y=24

now, PQ+RS=PS+QR
=> X+3X=Y+Y
=> 4x=2y

solving eqns we get y=6,x=3
sides of quadilateral are 3,9,6,6

s=(a+b+c+d)/2 = (3+9+6+6)/2 = 12
A=[(s-a)(s-b)(s-c)(s-d)]^1/2 = [(12-3)(12-9)(12-6)(12-6)]^1/2 = 18√3

Is this solution Helpfull? Yes (47) | No (4)  

shlagha sharma  (10 years ago)

@rakesh- how PQ+RS=PS+QR ??

Is this solution Helpfull? Yes (14) | No   

RAKESH  (10 years ago)

@SHLAGHA SHARMA
we know that,two tangents from a point outside a circle to the circle are equal,
using this property it can be proved that PQ+RS=PS+QR
If the lengths of the tangents from vertices P, Q, R, S of a quadrilateral
circumscribed around a circle are denoted by p, q, r, s then

PQ=p+q
RS=r+s
PS=p+s
QR=q+r

then PQ+RS=PS+QR=(p+q+r+s)

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

aditi  (10 years ago)

@rakesh
area of quadrilateral??
From where have you applied this formulae??

Is this solution Helpfull? Yes (1) | No   

Madhu  (6 years ago)

xplanation :


As the area of square is 36cm2 its side will be 6cm

So the perimeter of quadrilateral = perimeter of square = 6x4cm = 24cm

Therefore (2a + 2b + 2c + 2d) = 24cm or (a + b + c + 2) = 12cm

As RS = 3PQ , therefore (c + d) = 3(a + b)

Therefore: (a +b + 3a + 3b) = 12cm or (a +b ) = 3cm

therefore PQ = 3cm , RS = 9cm . Also PS = QR = 6cm

now the height of quadrilateral can be calculated as: h2 = 62 - 32 = 27

Therefore height of quadrilateral = √27 = 3√3

As it is a trapezium ,so its area = 1/2 [PQ + RS] x height

= 1/2 x (3 + 9) x 3√3

= 1/2 x 12 x 3√3 = 18√3 cm2

Is this solution Helpfull? Yes (1) | No   

Anuj Gupta  (7 years ago)

@RAKESH
can u please tell me where u have got the area of quadrilateral = ((s-a)(s-b)(s-c)(s-d))^1/2

i know that
area of quadrilateral = ((s-a)(s-b)(s-c)(s-d) - abcd (cos^2(theta)/2))^1/2
where theta = sum of two opposite angles

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Shivam  (6 years ago)

@sir pls upload full question atleast......

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Ayush Barnawal  (6 years ago)

check out thi solution:-
http://www.logicguns.com/q/a-quadrilateral-pqrs-circumscribes-a-circle-with-centre-oit-is-given-that-pq-is-parallel-to-rsalso-length-of-pq-is-thrice-that-of-rs-while-length-of-q/5603a2d82b78726726a80909/

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