# Question Bank - Geometry - Hemant Malhotra

• Q29) • Q30) Five equal squares are joined together to form a rectangle whose perimeter is 372. What is the area of one square?

• Q31) 3 identical right angle cones with base radius r are placed on their bases so that each is touching the other two.The radius of circle drawn through their vertices is:-
a. smaller than r
b. Equal to r
c. larger than r
d. Depends on height

• It's like three circles touching each other & when u meet center it will form equilateral triangle with side 2r .. and Radius of circle going through it will be equal to circumradius = 2r/sqrt3 so greater then r

• Q32) • Q33) • Q34) • Q35) In the figure given below, seg EQ is the bisector of angle FEG. Seg EQ is perpendicular to side FG, seg QP is perpendicular to side FE and seg QR is perpendicular to side EG. Find EP x PF, if ER = 23 cm and RG = 13 cm. • Q36) • Q37) In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 17. What is the greatest possible perimeter of the triangle?

• Q38) • Q39) In a triangle with integer side lengths, one side is three times as long as a second side, and the length of the third side is 17. What is the greatest possible perimeter of the triangle?

• Q40) In the figure given below, ABCD is a rectangle and GB = DH = 2HC. What is the area (in sq. cm) of the shaded region if AD = 1 cm and AB = 3 cm? • Q41) • Q42) A (-1, 2) and B(1, 4) are two points and C is a moving point on the x-axis. What will be the x-coordinate of C when angle ACB is maximum

• Method1- you will get 3 points
x = 1 , -7 , -3
so at -3 , angle is minimum i.e 0 degrees
again at x = -7 is a local maxima with angle less than 45 degrees
and at x = 1 global maximum as alpha angle is maximum with 45 degrees so at x=1 OA=1

Method2- If a circle passes through A,B andC, then angle ACB will be maximum when radius of the circle is as small as possible
C = (k, 0)
AB is a chord, center of the circle pass through the perpendicular bisector of AB
equation will be x + y = 3
Center will be O (k, 3 - k)
AO = CO
(k+ 1)^2 + (1 - k)^2 = (3 - k)^2
so k = 1, -7

89

22

30

107

207

45

68