Quant Boosters - Geometry - Raman Sharma - Set 1
Ratio of sides
1/12 = 1/20 = 1/h
5h : 3h : 60
Sum of two sides > third
5h +3h > 60
Diff of two sides < 60
5h -3h < 60
8h > 60 > 2h
60/8 < h < 60/2
credits : @gaurav_sharma
Q11) Semi perimeter of a right angled triangle is 28 units. If the sum of the product of the sides a , b , c taken two at a time = 943 units. Find the circumradius of the triangle.
(a+b+c)^2 =a^2+b^2+c^2+ 2 * 943
56^2= 2c^2 + 2 * 943
c= 25..so c/2=25/2
Q12) A triangle with area 'x' units have sides of lengths 2, 5 and x. Find 'x'.
[OA : √21]
Area = 1/2 × base x height
So for base => x , height 2. But 2 is one side length of the triangle.
That means triangle is right angled triangle with 2 and x as two perpendicular sides and 5 as hypotenuse,
so x = √(5² - 2²) = √21
Q13) Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP = 6 cms, PB = 4 units and DP = 3 units. What is the area of the circle?
Q14) There is a circumscribed circle around hexagon ABCDEF , such that AE is diameter of circumscribed circle. then find the sum of angles ABC, CDE & EFA.
Raman_Sharma last edited by Raman_Sharma
Q15) In triangle ABC, AC = 12. If one of the trisectors of angle B is the median to AC and the other trisector of angle B is the altitude to AC, find the length of the altitude.
Q16) The lengths of the sides of a triangle with positive area are log 12 , log 75 and log n (all base 10). Find the number of possible value of n. ( Where n is a positive integer )
credits : shiv kishore
log12+log75 > logn
=> 900 > n
Also log12+logn > log75
=> 12n > 75
Since n is a integer => n ≥ 7
But 900 > n
=> n has 893 integer values.
Q17) A paper of dimension 30 cm by 40 cm is folded such that opposite vertices meet. Find the length of the crease.
shortcut: a * root(a^2 + b^2)/b where a < b
Q18) The length indicated on the rectangle shown are in centimeters. What is the number of square centimeters in the area of the shaded region?