Question

If two poles of height 10 metre and metre are 8 metre apart then the height of the point of intersection of the lines joining the top of each pole to root the Fort of the opposite Pole is .......A) 150m B)25m C)5m D) 6m​

Answer 1

Answer:

It would be 6 meters

Step-by-step explanation:

Let Pole AB has height 10 meters and pole CD has height 15 meters ( where A and C are top while B and D are bottoms of poles ),

Now, suppose AD and BC intersects at O such that ON ⊥ BD where N ∈ BD

Again BN = x, ND = y and ON = h,

In triangles ABD and OND,

∠ABD ≅ ∠OND ( right angles )

∠ADB ≅ ∠OND ( common angles )

By AA similarity postulate,

\triangle ABD\sim \triangle OND

Similarly, \triangle CBD\sim \triangle OBN

∵ Corresponding sides of similar triangles are in same proportion

\implies \frac{AB}{ON}=\frac{BD}{ND} and \frac{CD}{ON}=\frac{BD}{BN}

\implies \frac{10}{h}=\frac{8}{y} and \frac{15}{h}=\frac{8}{x}

\implies h =\frac{10}{8}y, h=\frac{15x}{8}

\implies \frac{10y}{8}x=\frac{15x}{8}

10y = 15x

2y = 3x

y =\frac{3}{2}x .....(1)

Since, BD = 8 m,

⇒ x + y = 8

⇒ x+\frac{3}{2}x = 8

⇒ \frac{2x+3x}{2}=8

⇒ \frac{5x}{2}=8

⇒ 5x = 16

⇒ x = 3.2

Hence, height of the intersection point,

h = \frac{15}{8}\times 3.2 = 15\times 0.4 = 6\text{ m}

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