Advertisements
Advertisements
рдкреНрд░рд╢реНрди
Find the length of the altitude of an equilateral triangle of side 2a cm.
рдЙрддреНрддрд░
We know that the altitude of an equilateral triangle bisects the side on which it stands and forms right angled triangles with the remaining sides.
Suppose ABC is an equilateral triangle having AB = BC = CA = 2a.
Suppose AD is the altitude drawn from the vertex A to the side BC.
So, it will bisects the side BC
∴ DC = a
Now, In right triangle ADC
By using Pythagoras theorem, we have
`AC^2=CD^2+DA^2`
⇒` (2a)^2=a^2+DA^2`
⇒` DA^2=4a^2-a^2`
⇒` DA^2=3a^2`
⇒` DA=sqrt3a`
Hence, the length of the altitude of an equilateral triangle of side 2a cm is `sqrt3`ЁЭСО ЁЭСРЁЭСЪ
APPEARS IN
рд╕рдВрдмрдВрдзрд┐рдд рдкреНрд░рд╢реНтАНрди
Construct a triangle ABC with sides BC = 7 cm, ∠B = 45° and ∠A = 105°. Then construct a triangle whose sides are `3/4` times the corresponding sides of тИЖABC.
Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm find the length of the other diagonal.
In тИЖABC, ∠A is obtuse, PB ⊥ AC and QC ⊥ AB. Prove that:
(i) AB тЬХ AQ = AC тЬХ AP
(ii) BC2 = (AC тЬХ CP + AB тЬХ BQ)
In an equilateral ΔABC, AD ⊥ BC, prove that AD2 = 3BD2.
An aeroplane leaves an airport and flies due north at a speed of 1000km/hr. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km/hr. How far apart will be the two planes after 1 hours?
State the converse of Pythagoras theorem.
ΔABC~ΔDEF such that ar(ΔABC) = 64 cm2 and ar(ΔDEF) = `169cm^2`. If BC = 4cm, find EF.
Find the side and perimeter of a square whose diagonal is `13sqrt2` cm.
From given figure, In тИЖABC, AB ⊥ BC, AB = BC, AC = `5sqrt(2)` , then what is the height of тИЖABC?
In the given figure, ΔPQR is a right triangle right angled at Q. If PQ = 4 cm and PR = 8 cm, then P is ______.