CAT 2007: Quantitative Questions a Day 134 till the end - The Discussions

Aarav

Quote:

Originally Posted by kekambus

Thanx 4 the approach won't violate the tread again, sorry

No worries

Use co-ordinates and you will easily see that PA^2 + PC^2 = PB^2 + PD^2.

ma013851

Answer to the quant q # 157

say AD=DC= x
in triangle AEC, ED is a median so ED=AD=DC so ED = x.

now in triangle ABC,
by apollonious theorem:
AB^2 + BC^2 = 2(BD^2 + AD^2)
=> 25 = 2((1+x^2) + x^2)
=> 4x^2 + 4x -17 = 0

solving for x and then AC = 2AD = 2x .
so the option 3 is correct.

krishp

Solution for QQAD : 157.

Since AD=DC & <AED=90, therefore AC is diameter of circle passing through A,C,E.
So DE=DA=DC=x

Area of ADB = Area of BDC

(x+5/2)(x-1/2)(5/2)(1/2) = (x+3)(x-1)(1)(3)

Let x=y-1

Equation bcoms:

(5/4)(y^2 - 9/4)=3(y^2 - 4)

So, y=(21^(1/2))/2

x = (21^(1/2))/2) - 1

AC = 2x = 21^(1/2) - 2

So Option (5)
None of these

varun nakra1

Quote:

Originally Posted by vineet.nitd

Let ABC be an acute angled triangle;AD be the bisector of <BAC with D on BC;and BE be the altitude from B on AC.Which of the following statements is definitely false?

(a)<CED is necessarily lesser than 45 degrees
(b)<CED is necessarily greater than 45 degrees
(c)<CED may be greater than 60 degrees
(d)<CED may be lesser than 30 degrees
(e) <CED may or may not be lesser than 45 degrees

Since the triangle is an acute one..so <BAC is strictly <90..Lets consider the limiting case in which <BAC approaches 90...In that case the perpendicular drawn from point B on AC is the same as BA (coz <BAC =90)..Hence as <BAC approaches 90,point E approaches point A..after this point D will be the point where <BAC is bisected..now since <BAC is approaching 90,E is approaching A and hence <CED or CAD is approaching 45..but it can never reach 45 unless <BAC =90...Hence <CED is necessarily <45 degrees..

Rohit .Sinha

ans

25/2 = a2 + 4 + a2 + 4a

2a2 + 4a -17/2->

a= -1+/- sqrt(4 + 17)

2a = 2(sqrr(21) -1)

answer non eof these!!!
(5)
----------------
Rohit

Rohit .Sinha

Let ABC be an acute angled triangle;AD be the bisector of <BAC with D on BC;and BE be the altitude from B on AC.Which of the following statements is definitely false?

(a)<CED is necessarily lesser than 45 degrees
(b)<CED is necessarily greater than 45 degrees
(c)<CED may be greater than 60 degrees
(d)<CED may be lesser than 30 degrees
(e) <CED may or may not be lesser than 45 degrees

My answer is 1.

krish_mec · 04:03 PM

by appolinous therom,

Let

3^2+4^2= 2(a^2+(2+x)^2)

But x=a. As D is midpoint of AC and AEC is a right angled triangle. Hence circumradius is x=a.

Solving for a we get a= 21^1/2-1

Hence 2a (which is AC) is 2a= 21^1/2-2

ans. 5.