Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC. ∴ BE = EC = BC/2 = a/2 In ΔABE, by pythagoras theorem AB² = AE² + BE² ⇒ a² = AE² + (a/2)² = AE² + (a²/4) ⇒ AE² = a² - (a²/4) = 3a²/4 4AE² = 3a² ⇒ 4 × (Altitude) = 3 × (Side)
Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC.
∴ BE = EC = BC/2 = a/2
In ΔABE, by pythagoras theorem
AB² = AE² + BE²
⇒ a² = AE² + (a/2)² = AE² + (a²/4)
⇒ AE² = a² – (a²/4) = 3a²/4
4AE² = 3a²
⇒ 4 × (Altitude) = 3 × (Side)
Let AB and CD are the two pole with height 6 m and 11 m respectively. Therefore, CP = 11 - 6 = 5 m and AP = 12 m In triangle APC, by pythagoras theorem AP² + PC² = AC² ⇒ 12² = 5² = AC² ⇒ AC² = 144 + 25 = 169 ⇒ AC = 13m Hence, the distance between the tops of two poles is 13 m.
Let AB and CD are the two pole with height 6 m and 11 m respectively.
Therefore, CP = 11 – 6 = 5 m and AP = 12 m
In triangle APC, by pythagoras theorem
AP² + PC² = AC²
⇒ 12² = 5² = AC²
⇒ AC² = 144 + 25 = 169
⇒ AC = 13m
Hence, the distance between the tops of two poles is 13 m.
In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC. ∴ BE = EC = BC/2 = a/2 In ΔABE, by pythagoras theorem AB² = AE² + BE² ⇒ a² = AE² + (a/2)² = AE² + (a²/4) ⇒ AE² = a² - (a²/4) = 3a²/4 4AE² = 3a² ⇒ 4 × (Altitude) = 3 × (Side)
Let triangle ABC be an equilaterral triangle with side a. Altitude AE is Drown from A to BC.
See less∴ BE = EC = BC/2 = a/2
In ΔABE, by pythagoras theorem
AB² = AE² + BE²
⇒ a² = AE² + (a/2)² = AE² + (a²/4)
⇒ AE² = a² – (a²/4) = 3a²/4
4AE² = 3a²
⇒ 4 × (Altitude) = 3 × (Side)
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Let AB and CD are the two pole with height 6 m and 11 m respectively. Therefore, CP = 11 - 6 = 5 m and AP = 12 m In triangle APC, by pythagoras theorem AP² + PC² = AC² ⇒ 12² = 5² = AC² ⇒ AC² = 144 + 25 = 169 ⇒ AC = 13m Hence, the distance between the tops of two poles is 13 m.
Let AB and CD are the two pole with height 6 m and 11 m respectively.
See lessTherefore, CP = 11 – 6 = 5 m and AP = 12 m
In triangle APC, by pythagoras theorem
AP² + PC² = AC²
⇒ 12² = 5² = AC²
⇒ AC² = 144 + 25 = 169
⇒ AC = 13m
Hence, the distance between the tops of two poles is 13 m.
D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE² + BD² = AB²+ DE².
Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.
If the areas of two similar triangles are equal, prove that they are congruent.