Advertisements
Advertisements
प्रश्न
Find the unknown side in the following triangles
Advertisements
उत्तर
From ∆ABC, by Pythagoras theorem
BC2 = AB2 + AC2
Take AB2 + AC2
= 92 + 402
= 81 + 1600
= 1681
BC2 = AB2 + AC2
= 1681
= 412
BC2 = 412
⇒ BC = 41
∴ x = 41
APPEARS IN
संबंधित प्रश्न
ABC is an isosceles triangle right angled at C. Prove that AB2 = 2AC2
In the figure: ∠PSQ = 90o, PQ = 10 cm, QS = 6 cm and RQ = 9 cm. Calculate the length of PR.
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.
In the figure below, find the value of 'x'.

A ladder 25m long reaches a window of a building 20m above the ground. Determine the distance of the foot of the ladder from the building.
The foot of a ladder is 6m away from a wall and its top reaches a window 8m above the ground. If the ladder is shifted in such a way that its foot is 8m away from the wall to what height does its tip reach?
Each side of rhombus is 10cm. If one of its diagonals is 16cm, find the length of the other diagonals.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 = AD2 - BC x CE + `(1)/(4)"BC"^2`
PQR is an isosceles triangle with PQ = PR = 10 cm and QR = 12 cm. Find the length of the perpendicular from P to QR.
The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is ______.
