Let O(0, 0), P(3, 4), Q(6,) be the vertices of the
triangle OPQ. The point Rinside the triangle OPQ
is such that the triangles OPR, PQR and OQR are
of equal area. The coordinates of Rare
Answers
Answer:
Co-ordinate of R = (3 , 4/3)
Step-by-step explanation:
O(0, 0), P(3, 4), Q(6,0) the vertices of the triangle OPQ
OP = 5
PQ = 5
OQ = 6
S = (5 + 5 + 6)/2 = 8
Area of Δ OPQ = √8(8-5)(8-5)(8-6)
= 12
Area of ΔOQR = (1/3)Area of Δ OPQ = (1/3)12 = 4
Let draw Perpendicular from R to OQ = h
(1/2)* 6 * h = 4
=> h = 4/3
Co-ordinate of R = (x , 4/3)
Area of Δ OPR = Area of Δ PQR
O(0, 0), P(3, 4) , R (x , 4/3) & P(3, 4), Q(6,0) R (x , 4/3)
=> (1/2)| 0(4 - 4/3) + 3(4/3 - 0) + x(0 - 4)| = (1/2)| 3(0-4/3) . + 6(4/3 - 4) + x(4-0)|
=> | 4 - 4x| = | -4 -16 + 4x |
=> | 4 - 4x| = | 4x - 20|
=> 4 - 4x = 4x - 20
=> x = 3
Co-ordinate of R = (3 , 4/3)
Co-ordinate of R = (3 , 4/3)
Step-by-step explanation:
O(0, 0), P(3, 4), Q(6,0) the vertices of the triangle OPQ
OP = 5
PQ = 5
OQ = 6
S = (5 + 5 + 6)/2 = 8
Area of Δ OPQ = √8(8-5)(8-5)(8-6)
= 12
Area of ΔOQR = (1/3)Area of Δ OPQ = (1/3)12 = 4
Let draw Perpendicular from R to OQ = h
(1/2)* 6 * h = 4
=> h = 4/3
Co-ordinate of R = (x , 4/3)
Area of Δ OPR = Area of Δ PQR
O(0, 0), P(3, 4) , R (x , 4/3) & P(3, 4), Q(6,0) R (x , 4/3)
=> (1/2)| 0(4 - 4/3) + 3(4/3 - 0) + x(0 - 4)| = (1/2)| 3(0-4/3) . + 6(4/3 - 4) + x(4-0)|
=> | 4 - 4x| = | -4 -16 + 4x |
=> | 4 - 4x| = | 4x - 20|
=> 4 - 4x = 4x-20
therfore x=3 answer is (3,4/3)