Advertisements
Advertisements
प्रश्न
An isosceles triangle has equal sides each 13 cm and a base 24 cm in length. Find its height
Advertisements
उत्तर
In an isosceles triangle the altitude dives its base into two equal parts.
Now in the figure, ∆ABC is an isosceles triangle with AD as its height
In the figure, AD is the altitude and ∆ABD is a right triangle.
By Pythagoras theorem,
AB2 = AD2 + BD2
⇒ AD2 = AB2 – BD2
= 132 – 122 = 169 – 144 = 25
AD2 = 25 = 52
Height: AD = 5 cm
APPEARS IN
संबंधित प्रश्न
In a right triangle ABC right-angled at C, P and Q are the points on the sides CA and CB respectively, which divide these sides in the ratio 2 : 1. Prove that
`(i) 9 AQ^2 = 9 AC^2 + 4 BC^2`
`(ii) 9 BP^2 = 9 BC^2 + 4 AC^2`
`(iii) 9 (AQ^2 + BP^2 ) = 13 AB^2`
ABC is an equilateral triangle of side 2a. Find each of its altitudes.
PQR is a triangle right angled at P. If PQ = 10 cm and PR = 24 cm, find QR.
ABC is a triangle right angled at C. If AB = 25 cm and AC = 7 cm, find BC.
The given figure shows a quadrilateral ABCD in which AD = 13 cm, DC = 12 cm, BC = 3 cm and ∠ABD = ∠BCD = 90o. Calculate the length of AB.
In the following Figure ∠ACB= 90° and CD ⊥ AB, prove that CD2 = BD × AD
The sides of the triangle are given below. Find out which one is the right-angled triangle?
1.5, 1.6, 1.7
Two poles of height 9m and 14m stand on a plane ground. If the distance between their 12m, find the distance between their tops.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 = AD2 + BC x DE + `(1)/(4)"BC"^2`
In triangle ABC, line I, is a perpendicular bisector of BC.
If BC = 12 cm, SM = 8 cm, find CS
