Hello Kinshuk !
PQR is an Isosceles triangle so PE bisects QR
QE = ER
In ∆PQE => QE² = PQ² - PE² = 7² - PE²
In ∆PDE , DE² = PD² - PE² = 5² - PE²
QD × DR = (QE + DE) (QE - DE)
=> QD × DR = ( QE)²- ( DE)²
=> QD × DR = 7² -5² = 24
So they can only be 3 × 8 or 4 × 6.
Hence 3+8 = 11,
Option ( B )
In triangle ABC, BC = 4 units, AC = 2 units and sin A = 1. What is angle ABC?
In triangle ABC ,Produce a line from B to meet AC,meeting at D and From C to AB,meeting at E.let BD and CE meeet at X.let triangle BXE have area a,triangle BXC have area b and triangle CXD have area c.find the area of quadrilateral AEXD in terms of a,b,and c
ake a special case. Assume ABC is an equilateral triangle.
Let's assume D is the mid point on AC and E is the mid point on AB.
Now Area(BXE) = Area(CXD) = a = c
Then for this case Area(BXC) = b = 2a
And area of AEXD = 2a (it's very simple to calculate now)
Now, check all options:
- a) 2a*a(2a+2a+a)/4a^2-a^2 = 10a/3
- b) a*a(6a)/4a^2-a^2 = 2a
c)a*2a*a/4a^2-a^2 = 2a/3
- d) a*a(a+2a+2a)/4a^2-a^2 = 5a/3
Only b) matches. So, it is the answer.
The perimeter of a right triangle is 12+8√3. The sum of the squares of the three sides is 294. Find the length of the hypotenuse.
Given that P2+ B2 + H2 = 294
and P2 + B2 = H2
So, H2 + H2 = 294
2H2 = 294
H2 = 147 = 49 x 3
H = 7 sqrt(3)
The sides of a right triangle are 3 4 5
A point is taken on the hypotenuse at a distance of 2 cm from the vertex adjacent to the 4 side. Find the distance from this point to the vertex of the right triangle.
Let's say MN intersect AX at O ,
MO = 1/2 BX
∆AMO and ∆ABX are similar triangles so
Area∆ AMO/ Area∆ABX = MO²/BX² = 1/4
So if Area of ∆AMO = x ,then Area ABX = 4x and Area MOXB = 4x - x = 3x ( shaded area )
Area of ∆ABC = 2 × Area ∆ABX = 8x
Hence required ratio = 3x/ 8x = 3 : 8