Question
In ∆ABC, AD = DB , DE is parallel to BC, and the area
of ∆ABC is 24 cm2. What is the area of ∆ADE in cm2 ?Solution
As DE is parallel to BC & AD = BC. By mid point theorem, The line segment joining the mid-points of two sides of a triangle is parallel to the third side & vice-versa, Hence E is mid point of AC so area of ∆ADE = 1/4 of ∆ABC = 1/4 of 24 = 6 cm 2
I. 3p² - 14p + 15 = 0
II. 15q² - 34q + 15 = 0
Solve the quadratic equations and determine the relation between x and y:
Equation 1: x² - 30x + 221 = 0
Equation 2: y² - 28y + 189 = 0
I. 3x2 = 2x2 + 9x – 20
II. 3y2 = 75
In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between 'p' and 'q' and choose...
I. 6x2 - 47x + 77 =0
II. 6y2 - 35y + 49 = 0
I. 2x2 + 13x + 21 = 0
II. 3y2 + 34y + 63 = 0
I. 3x2 - 14x + 15 = 0
II. 15y2 - 34 y + 15 = 0
I. 2x² - 7x + 3 = 0
II. 8y² - 14y + 5 = 0
LCM of 'x' and 'y' is 30 and their HCF is 1 such that {10 > x > y > 1}.
I. 2p²- (x + y) p + 3y = 0
II. 2q² + (9x + 2) = (3x + y) q
I. 2x² - 15x + 27 = 0
II. 2y² - 13y + 20 = 0
