Question

If two poles of height 10m and 15m are 8m apart then the height of the point of intersection of the lines joining the top of each pole to the foot of the opposite pole is

Answer 1

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,

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

#Learn more:

Two poles of height a meter and b meter are p meter apart . prove that height of point of intersection of the lines joining the top of each pole to the foot of opposite pole is given by ab/a+b

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