Home
Class 12
MATHS
If the circumcentre of the triangle whos...

If the circumcentre of the triangle whose vertices are `(0,2),(3,5)and (5,8)`is (h,k) then `4(h^(2)+k^(2))` equals`"_____".`

Text Solution

AI Generated Solution

The correct Answer is:
To find the value of \( 4(h^2 + k^2) \) where \( (h, k) \) is the circumcenter of the triangle with vertices \( (0, 2) \), \( (3, 5) \), and \( (5, 8) \), we can follow these steps: ### Step 1: Set up the equations based on the circumcenter property The circumcenter is equidistant from all three vertices of the triangle. Therefore, we can set up the following equations based on the distances from the circumcenter \( (h, k) \) to each vertex: 1. Distance from \( (h, k) \) to \( (0, 2) \): \[ (h - 0)^2 + (k - 2)^2 = d^2 \] which simplifies to: \[ h^2 + (k - 2)^2 = d^2 \tag{1} \] 2. Distance from \( (h, k) \) to \( (3, 5) \): \[ (h - 3)^2 + (k - 5)^2 = d^2 \] which simplifies to: \[ (h - 3)^2 + (k - 5)^2 = d^2 \tag{2} \] 3. Distance from \( (h, k) \) to \( (5, 8) \): \[ (h - 5)^2 + (k - 8)^2 = d^2 \] which simplifies to: \[ (h - 5)^2 + (k - 8)^2 = d^2 \tag{3} \] ### Step 2: Equate the distances From equations (1) and (2): \[ h^2 + (k - 2)^2 = (h - 3)^2 + (k - 5)^2 \] Expanding both sides: \[ h^2 + (k^2 - 4k + 4) = (h^2 - 6h + 9) + (k^2 - 10k + 25) \] Simplifying gives: \[ h^2 + k^2 - 4k + 4 = h^2 - 6h + 9 + k^2 - 10k + 25 \] \[ -4k + 4 = -6h + 34 - 10k \] \[ 6h + 6k = 30 \quad \Rightarrow \quad h + k = 5 \tag{4} \] ### Step 3: Use the second pair of distances Now, equate equations (2) and (3): \[ (h - 3)^2 + (k - 5)^2 = (h - 5)^2 + (k - 8)^2 \] Expanding both sides: \[ (h^2 - 6h + 9) + (k^2 - 10k + 25) = (h^2 - 10h + 25) + (k^2 - 16k + 64) \] Simplifying gives: \[ -6h - 10k + 34 = -10h - 16k + 89 \] \[ 4h + 6k = 55 \quad \Rightarrow \quad 2h + 3k = 27.5 \tag{5} \] ### Step 4: Solve the system of equations Now we have two equations: 1. \( h + k = 5 \) (from equation 4) 2. \( 2h + 3k = 27.5 \) (from equation 5) From equation (4), we can express \( h \) in terms of \( k \): \[ h = 5 - k \] Substituting into equation (5): \[ 2(5 - k) + 3k = 27.5 \] \[ 10 - 2k + 3k = 27.5 \] \[ k = 27.5 - 10 \] \[ k = 17.5 \] Substituting \( k \) back into equation (4): \[ h + 17.5 = 5 \quad \Rightarrow \quad h = 5 - 17.5 = -12.5 \] ### Step 5: Calculate \( 4(h^2 + k^2) \) Now we have \( h = -12.5 \) and \( k = 17.5 \): \[ h^2 = (-12.5)^2 = 156.25 \] \[ k^2 = (17.5)^2 = 306.25 \] \[ h^2 + k^2 = 156.25 + 306.25 = 462.5 \] \[ 4(h^2 + k^2) = 4 \times 462.5 = 1850 \] Thus, the final answer is: \[ \boxed{1850} \]
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-J (AAKASH CHALLENGERS QUESTIONS)|5 Videos
  • STRAIGHT LINES

    AAKASH INSTITUTE ENGLISH|Exercise SECTION-H (MULTIPLE TRUE-FALSE TYPE QUESTIONS)|2 Videos
  • STATISTICS

    AAKASH INSTITUTE ENGLISH|Exercise Section-C Assertion-Reason|15 Videos
  • THREE DIMENSIONAL GEOMETRY

    AAKASH INSTITUTE ENGLISH|Exercise ASSIGNMENT SECTION - J|10 Videos

Similar Questions

Explore conceptually related problems

Find the circumcentre of the triangle whose vertices are (2, 2), (4, 2) and (0, 4).

find the area of triangle whose vertices are (3,8),(-4,2)and (5,1)

Find the area of the triangle whose vertices are (3,8), (-4,2) and (5, -1).

Find the area of the triangle whose vertices are: (2,8),(-4,2),(5,1)

Find the area of that triangle whose vertices are (-4,3),(-2,1)and(5,2).

Find the circumcentre of the triangle whose vertices are (-2,\ -3),\ (-1,\ 0),\ (7,\ -6) .

Find the area of the triangle whose vertices are (3, 8) , (-4, 2) and (5, 1) .

The coordinates of circumcentre of the triangle whose vertices are (0,0), (4,0) and (0,-6) is

Find the centroid of the triangle whose vertices are A(-1,0), B(5, -2) and C(8, 2).

The coordinates of the circumcentre of the triangle whose vertices are at (0,0),(6,0) and (0,8) is