Given that : AB² = 2AC² ⇒ AB² = AC² + AC² ⇒ AB² = AC² + AB² [Because AC = BC] These sides satisfy the pythagoras theorem. Hence, the triangle ABC is a right angled triangle.
Given that : AB² = 2AC²
⇒ AB² = AC² + AC²
⇒ AB² = AC² + AB² [Because AC = BC]
These sides satisfy the pythagoras theorem.
Hence, the triangle ABC is a right angled triangle.
Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC. We know that the altitude in equilateral triangle, bisects the opposite sides. Therefore, ∴ BD = DC = a In triangleADB, by Pythagoras theorem AB² = AD² + BD² ⇒ (2a)² = AD² + a² [Because AB = 2aRead more
Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC.
We know that the altitude in equilateral triangle, bisects the opposite sides.
Therefore, ∴ BD = DC = a
In triangleADB, by Pythagoras theorem
AB² = AD² + BD²
⇒ (2a)² = AD² + a² [Because AB = 2a]
⇒ 4a² = AD² + a²
⇒ AD² = 3a²
⇒ AD = √3a
Hence, the lenght of each altitude is √3a.
PQR is a triangle right angled at P and M is a point on QR such that PM perpendicular QR. Show that PM² = QM . MR.
ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².
ABC is an isosceles triangle with AC = BC. If AB² = 2AC² , prove that ABC is a right triangle.
Given that : AB² = 2AC² ⇒ AB² = AC² + AC² ⇒ AB² = AC² + AB² [Because AC = BC] These sides satisfy the pythagoras theorem. Hence, the triangle ABC is a right angled triangle.
Given that : AB² = 2AC²
See less⇒ AB² = AC² + AC²
⇒ AB² = AC² + AB² [Because AC = BC]
These sides satisfy the pythagoras theorem.
Hence, the triangle ABC is a right angled triangle.
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC. We know that the altitude in equilateral triangle, bisects the opposite sides. Therefore, ∴ BD = DC = a In triangleADB, by Pythagoras theorem AB² = AD² + BD² ⇒ (2a)² = AD² + a² [Because AB = 2aRead more
Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC.
See lessWe know that the altitude in equilateral triangle, bisects the opposite sides.
Therefore, ∴ BD = DC = a
In triangleADB, by Pythagoras theorem
AB² = AD² + BD²
⇒ (2a)² = AD² + a² [Because AB = 2a]
⇒ 4a² = AD² + a²
⇒ AD² = 3a²
⇒ AD = √3a
Hence, the lenght of each altitude is √3a.
Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.