Tamil Nadu Board of Secondary EducationSSLC (English Medium) Class 10th
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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 - Mathematics

Sum

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.

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Solution

“C” is the position of the foot of the ladder “A” is the position of the top of the ladder.

In the right ∆ABC,

BC2 = AC2 – AB2 

= 52 – 42

= 25 – 16

= 9

BC = `sqrt(9)` = 3 m

When the foot of the ladder moved 1.6 m toward the wall.

The distance between the foot of the ladder to the ground is

BE = 3 – 1.6 m = 1.4 m

Let the distance moved upward on the wall be “h” m

The ladder touch the wall at (4 + h) M

In the right ∆BED,

ED2 = AB2 + BE2

52 = (4 + h)2 + (1.4)2

25 – 1.96 = (4 + h)

∴ 4 + h = `sqrt(23.04)`

4 + h = 4.8 m

h = 4.8 – 4

= 0.8 m

Distance moved upward on the wall = 0.8 m

Concept: Converse of Pythagoras Theorem
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APPEARS IN

Samacheer Kalvi Mathematics Class 10 SSLC Tamil Nadu State Board
Chapter 4 Geometry
Exercise 4.3 | Q 6 | Page 187
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