Quant Boosters  Geometry  Raman Sharma  Set 1

Credits @swetabh_kumar
N is circumcenter of ABC (mid pt of hyp.)
so ADC is also a right triangle (line joining D to N is half of hyp.)
so ABCD is a rectangle.
so AD= BC = 12.
similar rectangle can be drawn below ML as well.
so for the overall rectangle, diagonal = 2R = 20.
AD= 12.
so (2AB)^2 + 144= 400....(2AB) = 16. AB = 8.

Q29) Shaded Area = ?
[OA: 97]

Q30)
[OA : 90]

Credits : @gaurav_sharma
Tan a = 1/1 = 1
Tan b = 1/2
Tan c = 1/3Tan (b + c) = (Tan b + Tan c)/ (1  Tan b. Tan c)
= (1/2+1/3) / (11/2.1/3)
=5/6 / 5/6 = 1 = Tan 45
b + c = 45
a = 45
So a + b + c = 90

@raman_sharma explanation?

credits : Claudia Samoila

[Credits  Deekonda Saikrishna]
y^2  x^2 = 1296 (Pythagoras theorem)
max value of x comes when y  x in min
so y  x =2
y + x = 648
y= 375 , x = 373 z = 36
min value of x > 36
smallest pyth triplets 3, 4, 5
36 = 3 * 12
so other sides = 4 * 12, 5 * 12
max  min areas = (18 * ( 323  48) = 4950

Credits : @swetabh_kumar
Let AM be angle bisector. let BAM=CAM=x. since AB=AC, so BM=BC.
so AM is median also. by congruency, AM is altitude also. so AMB=90
so sin x= (AB+AI)/2AB = 1/2+ 1/2 (AI/AB)..so 2sinx= 1+(AI/AB) cos X= AM/AB
AI/AM = (2AB)/(3AB+AI)=2/(3+AI/AB)=2/(3+2sinx 1)
AI/AM=1/(1+sin x) (1)
also, BC/2AM = tan x
(AB+AI)/AM= 2tan x
sec x + 1/((1+sinx) = 2tan x
1 + (cos x)/(1+sin x)= 2 sin x (multiplying throughout by cos x)
1+sin x+ cos x= 2sIn^2x+ 2sin x
12sin^2x = sinx  cos x
so cos 2x =sinxcos x
cos^2 2x=1sin 2x
so 1sin^2 2x= 1sin 2x
so sin^2 2x= sin 2x
sin 2x= 1, 2x= 90. = BAC.

LET the remaining areas in cyclic order from left be e,a, d, b.and the shaded area be c.
a+b+c =1/2(a+b+c+d+e+97)
a+b+c = d+e+97
or
d+e+97 = a+b+c (1)similarly,
e+c+d = 1/2(a+b+c+d+e+97)
=>e+c+d= a+b+97 (2)subtract (2) from (1)
c97 = 97c
=>c=97

BC =1 , CA =2 so AB =rt5
using similarity in triangle ABC and ADC
we get CD= 2/rt5 and diameter of circle as 4/rt5now using tangent secant theorem
CD^2 = CE * AC
4/5 = CE *2
CE =0.4
so AE=1.6
hence 1.6/0.4 = 4