Question

A girl of height 90 cm is walking away from the base of the lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6 m above the ground, find the length of her shadow after 4 seconds.​

Answer 1

✰ Qᴜᴇsᴛɪᴏɴ :-

A girl of height 90 cm is walking away from the base of the lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6 m above the ground, find the length of her shadow after 4 seconds.

☢ Gɪᴠᴇɴ :-

• Height of the lamp post = 3.6 m
• Speed of the girl = 1.2 m/sec

✦ Tᴏ Fɪɴᴅ :-

• Length of her shadow after 4 seconds

✪ Sᴏʟᴜᴛɪᴏɴ :-

❝ ʀᴇғᴇʀ ᴛʜᴇ ᴀᴛᴛᴀᴄʜᴍᴇɴᴛ ғᴏʀ ᴛʜᴇ ᴅɪᴀɢʀᴀᴍ ❞

Distance travelled in 4 sec = Speed × Time

⇒ 1.2 × 4

⇒ 4.8 m

⇒ 480 cm

• Height of the girl (CD) = 90 cm

Let the length of the shadow at a distance of 4.8m from the lamp post = x cm

From the diagram,

∆ABE ∼ ∆DCE

∵ ∠B = ∠C = 90°

∠E = ∠C common

(A.A. similarity)

Hence, AB/DC = BE/CE = AE/DE

∴ 360/90 = 480 + x / x

➸ 4 = 480 + x / x

➸ 4x = 480 + x

➸ 4x - x = 480

➸ 3x = 480

➸ x = 480/3

➸ x = 160cm

➸ x = 1.6m

⛬ Length of the shadow = 1.6 m

Answer 2

Qᴜᴇsᴛɪᴏɴ ✔

✯.A girl of height 90 cm is walking away from the base of the lamp post at a speed of 1.2 m/sec. If the lamp post is 3.6 m above the ground, find the length of her shadow after 4 seconds.

$\large\underline\bold{GIVEN,}$

$\sf\dashrightarrow height\:of\:the\:lamp=3.6m(360CM)$

$\sf\dashrightarrow height\:of\:the\:girl\:90cm$

$\sf\dashrightarrow speed\:of\:the\:girl\:1.2 m/s$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow LENGTH\:OF\:HER\:SHADOW\:AFTER\:4\:SECONDS$

FORMULA IN USED,

$\sf\therefore DISTANCE=SPEED \times TIME$

$\large\underline\bold{SOLUTION,}$

$\large{\boxed{\bf{please\:refer\:the\:given\:diagram }}}$

$\sf\therefore time=4\:seconds$

$\sf\implies d=1.2 \times 4$

$\sf\implies d=4.8cm$

✯conversion,

$\sf\therefore centimetre \:into\:metre$

$\sf\implies 4.8cm$

$\sf\implies 1m=100cm$

$\sf\implies 480cm$

$\large{\boxed{\bf{480\:CM}}}$

✯ACCORDING TO THE QUESTION,

$\sf\therefore LET \:US\:ASSUME\:THE\:LENGTH\:OF\:THE\:SHADOW\:AS\:"X"$

$\sf\therefore height\:of\:the\:girl\:90cm$

$\sf\therefore NOW,\:IN\:TRIANGLE\: \triangle ABE \:and \: \triangle DCE$

$\sf\implies \angle B= \angle C.....90\degree \:each$

$\sf\implies \angle E= \angle C ......common$

$\sf\implies \triangle ABE \sim \triangle ...... A.A. RULE CRITERIA$

THEREFORE,

$\sf\implies \dfrac{AB}{DC}= \dfrac{BE}{CE}= \dfrac{AE}{DE}$

NOW,

$\sf\implies \dfrac{360}{90}= \dfrac{480+x}{x}$

$\sf\implies \dfrac{\cancel{360}}{\cancel{90}}= \dfrac{480+x}{x}$

$\sf\implies 4x=480+x$

$\sf\implies 4x-x=480$

$\sf\implies 3x=480$

$\sf\implies \dfrac{480}{3}$

$\sf\implies \cancel \dfrac{480}{3}$

$\sf\implies 160cm$

✯FROM CM TO M,

$\sf\implies 1.6M$

$\sf{\boxed{\bf{ LENGTH\:OF \:SHADOW=1.6M}}}$

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