Math, asked by joelbijuv, 4 months ago

Two poles of height 50m and 10m stand upright on a plane ground. If the distance
between their tops is 41m, find the distance between their feet?

Answers

Answered by chauhanvikrantsingh
3

Answer:

sorry for bad handwriting but if you liked the solution please mark it as beanliest

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Answered by zakibinlatheef
1

Answer:      9m.  ( Figure attached at the end.)

Step-by-step explanation:

From the given figure we know that,

triangle ABP is a right-angled triangle (PB is drawn perpendicular to AC)

in quad. BPQC,

∠PBC = ∠BCQ = ∠PQC = 90°

Hence it is a rectangle. (All angles are 90°)

So, BC = PQ = 10m (Opposite sides of a rectangle are equal)

We know, AB + BC = AC

                         AB = AC - AB

                               = 50 - 10

                               = 40m

In Δ ABP,

AB² + BP² = AP²

         BP² = AP² - AB²

                = (41)² - (40)²

                = 1681 - 1600

                = 81

         BP² = 81

          BP =√81

   So, BP = 9m

We know that,

BP = CQ= 9m  ( opposite sides of a recyangl are equal)

∴The distance between the feet of the two poles is 9m.

   

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