Advertisements
Advertisements
प्रश्न
The top of a ladder of length 15 m reaches a window 9 m above the ground. What is the distance between the base of the wall and that of the ladder?
Advertisements
उत्तर

Let LN be the ladder of length 15 m that is resting against a wall. Let M be the base of the wall and L be the position of the window.
The window is 9 m above the ground. Now, MN is the distance between the base of the wall and that of the ladder.
In the right-angled triangle LMN, ∠M = 90°. Hence, side LN is the hypotenuse.
According to Pythagoras' theorem,
l(LN)2 = l(MN)2 + l(LM)2
⇒ (15)2 = l(MN)2 + (9)2
⇒ 225 = l(MN)2 + 81
⇒ l(MN)2 = 225 − 81
⇒ l(MN)2 = 144
⇒ l(MN)2 = (12)2
⇒ l(MN) = 12
∴ Length of seg MN = 12 m.
Hence, the distance between the base of the wall and that of the ladder is 12 m.
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`
In the given figure, ∠B = 90°, XY || BC, AB = 12 cm, AY = 8cm and AX : XB = 1 : 2 = AY : YC.
Find the lengths of AC and BC.

In the figure, given below, AD ⊥ BC.
Prove that: c2 = a2 + b2 - 2ax.
O is any point inside a rectangle ABCD.
Prove that: OB2 + OD2 = OC2 + OA2.
In the figure below, find the value of 'x'.

Find the length of the perpendicular of a triangle whose base is 5cm and the hypotenuse is 13cm. Also, find its area.
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AC2 - AB2 = 2BC x ED
In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that: 9AQ2 = 9AC2 + 4BC2
Find the unknown side in the following triangles
Jiya walks 6 km due east and then 8 km due north. How far is she from her starting place?
