Question

i stood 420m away from the tallest building in Singapore.I held a piece of wood 40 cm long at arms length ,60 cm away from my eye.The piece of wood,held vertically,just blocked the building from my view .Use similar triangles to calculate the height ,h meters of the building​

Answer 1

Answer:

Height of the building = 280 m

Step-by-step explanation:

Before we begin with tackling the question directly or calculation work, Let's draw its diagram to understand deeply what the question's statement speaks out. 

Following are the steps which represent the graphical analysis of the question. (It should be done roughly to get it done fast) 

1. Draw a line segment of length 420 m and mark it as AC.

2. Draw a vertical line segment named 'H' from point C to a suitable size and name it as BC. 

3. Join the point B with A in such a way that it makes a slope line. 

4. Take a line measuring 0.6 m away from the point A. 

5. Now, Draw a vertical line measuring 0.4 m from the point 'Y' to the point 'X'. (It's a perpendicular line)

Once we are done with drawing its diagram, we see a triangle coming out named ∆ABC which consists of an another triangle named ∆AXY. 

Now, Let's head to the question.

Since it's given in the question that similar triangles properties would work to solve the question, we would first prove the triangles to be similar. 

[Now, Refer to the attachment to have better understanding.]

Given: ∆ABC in which angle C represents 90° & XY ⊥ BC.

To Prove: ∆ABC ~ ∆AXY 

Proof:

From ∆ABC & ∆AXY, we have the following points to be recognized. 

∠ACB = ∠AYX   (90° & Perpendicular segment)

∠BAC= ∠XAY    (Common)  

∴ ∆ABC ~ ∆AXY (By AA similarity condition) 

Therefore, from the above similarity criterion, we can have the following resultant. 

∆ABC ~ ∆AXY

Note: Corresponding sides of two similar triangles are always equal in proportion.

 ⟹ BC/XY = AC/AY .... (i)

Now, we would put the given values in the equation (i) 

BC/XY = AC/AY

⟹ H/0.4 = 420/0.6

⟹ H/0.4 = 420/0.6

⟹ 0.6H = 420 × 0.4

⟹ H = 420 ×0.4/0.6

⟹ H = 168/0.6

⟹ H = 280 m 

Hence, The height of the building is 280 metre.

Answer 2

❍ Let's Consider BC be the Height ( h ) of the Building in Singapore , AC be the Distance from the tallest Building in Singapore , XY be the piece of wood of & AY be the distance between the eye and price of wood .

⠀⠀⠀⠀⠀▪︎⠀⠀Now Draw a reference image using Assumed ( or Considered ) things .

⠀⠀⠀⠀⠀⠀⠀⠀⠀[ For This see attachment ( image ) . ]

Given : In $\triangle$ ABC right angled at B & In $\triangle$ AXY : XY $\perp$ AC .

Exigency To Find : Height of Building using similar triangles [ Prove : $\triangle\: ABC \: \cong \:\triangle AXY$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Proving : $\sf \triangle\: ABC \:$ ~ $\sf \:\triangle AXY$

⠀⠀⠀⠀⠀▪︎⠀⠀In $\bf \triangle\: ABC \: \& \:\triangle AXY$ :

• $\angle$ BAC = $\angle$ XAY ⠀⠀⠀⠀⠀[ Common Angle ]
• $\angle$ ACB = $\angle$ AYX ⠀⠀⠀⠀⠀ [ Both angle are 90⁰ , Perpendicular]

⠀⠀⠀⠀⠀▪︎⠀⠀Hence , By ANGLE ANGLE [ AA ] similarity Rule ( or Condition) is proved that $\bf \triangle\: ABC \:$ ~ $\bf \:\triangle AXY$ .

[ ANGLE ANGLE : If two angles of triangle are congruent to two angles of another angle then , the two triangles are similar. ]

⠀⠀⠀⠀⠀Now ,

⠀⠀⠀We know that :

$\qquad \bf \triangle\: ABC \:$ ~ $\bf \:\triangle AXY$ .

⠀⠀⠀⠀⠀Here ,

$\qquad \dag\:\:\bigg\lgroup \sf{ \dfrac{BC}{XY}\:\:=\:\: \:\dfrac{AC}{AY}\:\: }\bigg\rgroup$

$\qquad:\implies \sf \dfrac{BC}{XY}\:\:=\:\: \:\dfrac{AC}{AY}$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values\:\:( Using\: Image\:) \::}}$

$\qquad:\implies \sf \dfrac{BC}{XY}\:\:=\:\: \:\dfrac{AC}{AY}$

$\qquad:\implies \sf \dfrac{h}{\dfrac{40}{100}}\:\:=\:\: \: \dfrac{420}{\dfrac{60}{100}}$

$\qquad:\implies \sf \dfrac{h}{\cancel {\dfrac{40}{100}}}\:\:=\:\: \:\dfrac{420}{\cancel {\dfrac{60}{100}}} \qquad \bigg\lgroup \sf{ 1\:metres\:=\:100\:cm\:\: }\bigg\rgroup$

$\qquad:\implies \sf \dfrac{h}{0.4}\:\:=\:\: \:\dfrac{420}{0.6}$

$\qquad:\implies \sf 0.6 \times h \:\:=\:\: \:420 \times 0.4$

$\qquad:\implies \sf 0.6 \times h \:\:=\:\: \:168$

$\qquad:\implies \sf 0.6 h \:\:=\:\: \:168$

$\qquad:\implies \sf h \:\:=\:\: \dfrac{\:168}{0.6}$

$\qquad:\implies \sf h \:\:=\:\: \cancel {\dfrac{\:168}{0.6}}$

$\qquad:\implies \bf h \:\:=\:\: 280$

$\qquad :\implies \pmb{\underline{\purple{\: h \:\:=\:\: 280\:m\: }} }\:\:\bigstar$

⠀⠀⠀⠀⠀▪︎⠀⠀Here , h denotes Height of Building which is 280 m

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Hence ,\:Height \:of\:Building \:is\:\bf{280\:m}}}}$