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