Q7 of 19 Page 54

Find the altitude of an equilateral triangle of side 10 cm.

Let us consider ΔABC an equilateral triangle with side 10 cm.


Construction: Draw a perpendicular bisector at C to AB such that AD = DB = 5 cm and the triangle is cut into two halves.


We have to find the altitude i.e. CD.



Consider ΔBCD,


We know that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.


⇒ BC2 = BD2 + CD2


⇒ 102 = 52 + CD2


⇒ 100 = 25 + CD2


⇒ CD2 = 100 – 25 = 75


∴ CD = √75 = = 5√3 cm


∴ Altitude = 5√3 cm


More from this chapter

All 19 →
5

Q and R of a triangle PQR are 25° and 65°. Is PQR a right-angled triangle? Moreover, PQ is 4cm and PR is 3 cm. Find QR.

 6

A 15 m long ladder reached a window 12m high from the ground. On placing it against a wall at a distance x m. Find x.

 8

Are the numbers 12, 5 and 13 form a Pythagorean Triplet?

 9

A painter sets a ladder up to reach the bottom of a second storey window 16 feet above the ground. The base of the ladder is 12 feet from the house. While the painter mixes the paint a neighbour’s dog bumps the ladder which moves the base 2 feet farther away from the house. How far up side of the house does the ladder reach?