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
PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM . MR
Identify, with reason, if the following is a Pythagorean triplet.
(24, 70, 74)
In ∆PQR, point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.
If the sides of the triangle are in the ratio 1: `sqrt2`: 1, show that is a right-angled triangle.
In the following figure, OP, OQ, and OR are drawn perpendiculars to the sides BC, CA and AB respectively of triangle ABC.
Prove that: AR2 + BP2 + CQ2 = AQ2 + CP2 + BR2

Prove that `(sin θ + cosec θ)^2 + (cos θ + sec θ)^2 = 7 + tan^2 θ + cot^2 θ`.
Show that the triangle ABC is a right-angled triangle; if: AB = 9 cm, BC = 40 cm and AC = 41 cm
In a triangle ABC, AC > AB, D is the midpoint BC, and AE ⊥ BC. Prove that: AB2 + AC2 = 2(AD2 + CD2)
A man goes 18 m due east and then 24 m due north. Find the distance of his current position from the starting point?
A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.
