Let circumcentre of a triangle ABC be O (0,0) and orthocentre of triangles OBC , OCA and OAB are A , B and C respectively. If OA =(3/2) and area of triangle ABC =( a √3/ b ) where a , b are respectively prime numbers, then find the value of (2 a-3 b).

Question

Let circumcentre of a triangle ABC be O (0,0) and orthocentre of triangles OBC , OCA and OAB are A , B and C respectively. If OA =(3/2) and area of triangle ABC =( a √3/ b ) where a , b are respectively prime numbers, then find the value of (2 a-3 b). (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

O is the circumcentre as well as orthocentre of the triangle.

∴ △ A BC

is equilateral triangle.

O A = \frac{3}{2} \Rightarrow A D = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}

Δ = \frac{( A D ) ^{2}}{3} = \frac{81}{16 3} = \frac{27 3}{16} \equiv \frac{a 3}{b}

∴ 2 a - 3 b = 54 - 48 = 6

Q. Let circumcentre of a triangle ABC be O(0,0) and orthocentre of triangles OBC,OCA and OAB are A,B and C respectively. If OA=23​ and area of △ABC=ba3​​ where a,b are respectively prime numbers, then find the value of (2a−3b).

O is the circumcentre as well as orthocentre of the triangle. ∴△ABC is equilateral triangle. OA=23​⇒AD=23​+43​=49​ Δ=3​(AD)2​=163​81​=16273​​≡ba3​​ ∴2a−3b=54−48=6

Q. Let circumcentre of a triangle $A BC$ be $O (0, 0)$ and orthocentre of triangles $OBC, OC A$ and $O A B$ are $A, B$ and $C$ respectively. If $O A = \frac{3}{2}$ and area of $△ A BC = \frac{a 3}{b}$ where $a, b$ are respectively prime numbers, then find the value of $(2 a - 3 b)$ .

O is the circumcentre as well as orthocentre of the triangle.

$∴ △ A BC$ is equilateral triangle.
$O A = \frac{3}{2} \Rightarrow A D = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$
$Δ = \frac{( A D ) ^{2}}{3} = \frac{81}{16 3} = \frac{27 3}{16} \equiv \frac{a 3}{b}$
$∴ 2 a - 3 b = 54 - 48 = 6$