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.
Let OA is well and AB is ladder in the figure. In triangle AOB, by pythagoras theorem AB² = AO² + OB² ⇒ 10² = 8² + BO² ⇒ 100 = 64 + BO² ⇒ BO² = 36 ⇒ BO = 6m Hence, the distance of the root of the ladder from the base of the wall is 6 m.
Let OA is well and AB is ladder in the figure.
In triangle AOB, by pythagoras theorem
AB² = AO² + OB²
⇒ 10² = 8² + BO²
⇒ 100 = 64 + BO²
⇒ BO² = 36
⇒ BO = 6m
Hence, the distance of the root of the ladder from the base of the wall is 6 m.
Let OB is vertical pole in the figure. In triangle AOB, by pythagoras theorem AB² = OB² + OA² ⇒ 24² = 18² + OA² ⇒ 576 = 324 + OA² ⇒ OA² = 252 ⇒ OA = 6 √7m Hence, the distance of stake from the base of the pole is 6 √7m.
Let OB is vertical pole in the figure.
In triangle AOB, by pythagoras theorem
AB² = OB² + OA²
⇒ 24² = 18² + OA²
⇒ 576 = 324 + OA²
⇒ OA² = 252
⇒ OA = 6 √7m
Hence, the distance of stake from the base of the pole is 6 √7m.
Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.
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.
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Let OA is well and AB is ladder in the figure. In triangle AOB, by pythagoras theorem AB² = AO² + OB² ⇒ 10² = 8² + BO² ⇒ 100 = 64 + BO² ⇒ BO² = 36 ⇒ BO = 6m Hence, the distance of the root of the ladder from the base of the wall is 6 m.
Let OA is well and AB is ladder in the figure.
See lessIn triangle AOB, by pythagoras theorem
AB² = AO² + OB²
⇒ 10² = 8² + BO²
⇒ 100 = 64 + BO²
⇒ BO² = 36
⇒ BO = 6m
Hence, the distance of the root of the ladder from the base of the wall is 6 m.
A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Let OB is vertical pole in the figure. In triangle AOB, by pythagoras theorem AB² = OB² + OA² ⇒ 24² = 18² + OA² ⇒ 576 = 324 + OA² ⇒ OA² = 252 ⇒ OA = 6 √7m Hence, the distance of stake from the base of the pole is 6 √7m.
Let OB is vertical pole in the figure.
See lessIn triangle AOB, by pythagoras theorem
AB² = OB² + OA²
⇒ 24² = 18² + OA²
⇒ 576 = 324 + OA²
⇒ OA² = 252
⇒ OA = 6 √7m
Hence, the distance of stake from the base of the pole is 6 √7m.
An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 3/2 hours?
E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that Triangles ABE ~ Triangles CFB.