Question

a girl of height 180 cm is walking away from a lamp post of 6.6m at the speed of 2m/s. find the length of her shadow after 8s.​

Answer 1

Answer:

6m.

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Step-by-step explanation:

[Refer to the diagram]

Lamppost ➝ AB ➝ 6.6m

The Girl ➝ EC ➝ 180cm ➝ 1.8m

BC ➝ Distance travelled in 8 seconds from the lamppost.

Shadow ➝ DC = ?

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On comparing both ΔABD and ΔECD we get;

∠ADB = ∠ EDC [Common angle]

∠ABD = ∠ECD [Both the lamppost and the girl stand perpendicular to the ground]

∴ ΔABD ≈ ΔECD using AA similarity. (Angle-Angle similarity)

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We know that the ratio of the corresponding sides of the two triangles are equal to one another, therefore for ΔABD and ΔECD;

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$\sf \dashrightarrow \ \dfrac{AB}{EC} = \dfrac{BD}{CD} = \dfrac{AD}{ED}$

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We're asked to find the length of the shadow (DC), so we'll equate the fraction that holds the value of DC with AB/EC as we know the values of both AB and EC.

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$\sf \dashrightarrow \ \dfrac{AB}{EC} = \dfrac{BD}{CD}$

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$\sf \dashrightarrow \ \dfrac{AB}{EC} = \dfrac{BC + CD}{CD}$

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$\sf \dashrightarrow \ \dfrac{6.6}{1.8} = \dfrac{speed \times time_{(BC)} + CD}{CD}$

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$\sf \dashrightarrow \ \dfrac{6.6}{1.8} = \dfrac{(2 \times 8) + CD}{CD}$

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$\sf \dashrightarrow \ \dfrac{6.6}{1.8} = \dfrac{16 + CD}{CD}$

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$\sf \dashrightarrow \ 6.6 \times CD = (16 + CD) \times 1.8$

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Multiply both sides by 10;

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$\sf \dashrightarrow \ 66 \times CD = (16 + CD) \times 18$

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$\sf \dashrightarrow \ 66CD = (16)18 + (CD)18$

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$\sf \dashrightarrow \ 66CD = 288 + 18CD$

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$\sf \dashrightarrow \ 66CD - 18CD = 288$

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$\sf \dashrightarrow \ 48CD = 288$

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$\sf \dashrightarrow \ CD = \dfrac{288}{48}$

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$\sf \dashrightarrow \ CD = 6m$

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Therefore the length of her shadow after 8 seconds is 6m.

Answer 2

Question:-

• A girl of height 180 cm is walking away from a lamp post of 6.6m at the speed of 2m/s. find the length of her shadow after 8s.

To Find:-

• Find the length of her shadow after 8s.

Solution:-

Here ,

• ∆ABC = ∆ADE

• $\sf \: \dfrac { BC } { DE } =  \dfrac { BA } { DA } = \dfrac { CA } { EA }$

Now ,

$\longrightarrow\sf \: \dfrac { CB } { ED } = \dfrac { BA } { DA }$

$\longrightarrow\sf \: \dfrac { CB } { ED } = \dfrac { BD + DA } { DA }$

$\longrightarrow\sf \: \dfrac { 6.6 } { 1.8 } = \dfrac { 16 + DA } { DA }$

$\longrightarrow\sf \: 6.6DA = 1.8( 16 + DA ) [ Multiply \: both \: sides \: with \: 10 ]$

$\longrightarrow\sf \: 66DA = 228 +18DA$

$\longrightarrow\sf \: 66DA - 18DA = 228$

$\longrightarrow\sf \: 48DA = 228$

$\longrightarrow\sf \: DA = \cancel\dfrac { 228 } { 38 }$

$\longrightarrow\sf \: DA = 6 \: m$

Hence ,

• Length of girl shadow is 6 m