Advertisements
Advertisements
प्रश्न
Show that the points (-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` are the vertices of an equilateral triangle.
उत्तर
Let the given points be(-3, -3),(3,3) and C (-3 `sqrt(3) , 3 sqrt(3))` Now
`AB = sqrt((-3-3)^2 +(-3-3)^2) = sqrt((-6)^2 +(-6)^2)`
`= sqrt(36+36) = sqrt(72) = 6sqrt(2)`
`BC = sqrt((3+3sqrt(3))^2 +(3-3sqrt(3))^2)`
`= sqrt(9+27+18sqrt(3) +9+27-18sqrt(3)) = sqrt(72)=6 sqrt(2)`
`AC = sqrt((-3+3sqrt(3))^2 +(-3-3sqrt(3))^2 )= sqrt((3-3sqrt(3))^2 +(3+3sqrt(3))^2)`
`= sqrt(9+27-18sqrt(3)+9+27+18sqrt(3)`
`= sqrt(72)=6sqrt(2)`
Hence, the given points are the vertices of an equilateral triangle.
APPEARS IN
संबंधित प्रश्न
Prove that the points (2, – 2), (–3, 8) and (–1, 4) are collinear
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the coordniates of any point P, if PA = PB and Area of ∆PAB = 10
The vertices of a ΔABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that `(AD)/(AB) = (AE)/(AC) = 1/4`Calculate the area of the ΔADE and compare it with the area of ΔABC. (Recall Converse of basic proportionality theorem and Theorem 6.6 related to ratio of areas of two similar triangles)
Find the area of a triangle with vertices at the point given in the following:
(−2, −3), (3, 2), (−1, −8)
Show that the following sets of points are collinear.
(2, 5), (4, 6) and (8, 8)
In a ΔABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC ?
Using integration, find the area of the triangle whose vertices are (2, 3), (3, 5) and (4, 4).
In ∆PQR, PR = 8 cm, QR = 4 cm and PL = 5 cm.
Find:
(i) the area of the ∆PQR
(ii) QM.
If the points (2, -3), (k, -1), and (0, 4) are collinear, then find the value of 4k.
Triangles having the same base have equal area.