Question

Can any Genius solve this.

It's a brilliant question from IMO (INTERNATIONAL MATHS OLYMPIAD)

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Answer 1

$\huge\mathcal{\boxed{\fcolorbox{red}{aqua}{SOLUTION;-}}}$

✨ First of all we find the area of the rectangle .

☃️ Area of rectangle = (24 × 15) cm²

→ Area of rectangle = 360 cm² .

✴️ It is clearly that,

✔️ Area of the shaded region = Area of the rectangle - [Area of (∆ADF + ∆ABE + ∆CEF)]

(i) Area of ∆ADF ;-

• ∆ADF is a right angled triangle .

☃️ Area of ∆ADF = 1/2 × AD × DF

→ Area of ∆ADF = 1/2 × 15 × 12

→ Area of ∆ADF = 15 × 6

→ Area of ∆ADF = 90 cm²

(ii) Area of ∆ABE ;-

• ∆ABE is a right angled triangle .

• BE = BC - CE = 15 - 8 = 7 cm

☃️ Area of ∆ABE = 1/2 × AB × BE

→ Area of ∆ABE = 1/2 × 24 × 7

→ Area of ∆ABE = 12 × 7

→ Area of ∆ABE = 72 cm²

(iii) Area of ∆CEF ;-

• ∆CEF is a right angled triangle .

• CF = DC - DF = 24 - 12 = 12 cm²

☃️ Area of ∆CEF = 1/2 × CE × CF

→ Area of ∆CEF = 1/2 × 8 × 12

→ Area of ∆CEF = 4 × 12

→ Area of ∆CEF = 48 cm²

⚡ Therefore,

Area of the shaded region = 360 - (90 + 72 + 48)

→ Area of the shaded region = 360 - 210

→ Area of the shaded region = 150 cm²

∴ The area of the shaded region is '150cm²' .

Answer 2

♣ Qᴜᴇꜱᴛɪᴏɴ :

• Find the area of shaded region, if ABCD is a rectangle ?

(Figure in attachment)

★═════════════════★

♣ ᴀɴꜱᴡᴇʀ :

Area of Shaded Region = 138.516 cm²

★═════════════════★

♣ ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴꜱ :

Area of Shaded region is a Area of triangle , we need to find area of triangle AFE

Area of Triangle = 1/2 × Base × Height

_______________________________

First we need to find Area of Rectangle :

$\setlength{\unitlength}{0.25cm}\begin{picture}(5,5)\thicklines\multiput(0,0)(24,0){2}{\line(0,1){15}}\multiput(0,0)(0,15){2}{\line(1,0){24}}\put(10,-2){\sf{24\ cm}}\put(25,7){\sf{15\ cm}}\end{picture}$

Area of Rectangle = Length × Breadth

Area of Rectangle = 24 cm × 15 cm

Area of Rectangle = 360 cm²

_______________________________

Now let's solve for Base of Triangle :

The base of triangle AFE is FE (or) The base of triangle AFE is the hypotenuse of triangle ECF

$\setlength{\unitlength}{.4in} \begin{picture}(7,5)(0,0) \linethickness{1pt} \put(0,0){\line(1,0){4}} \put(4,0){\line(0,1){3}} \put(0,0){\line(4,3){4}} \put(2,-.25){\makebox(0,0){12\:\:cm}} \put(4.25,1.5){\makebox(1,0){8\:\:cm}} \put(2,2){\makebox(0,0){Base}} \end{picture}$

Using Pythagoras Theorem :

Hypotenuse² = Side² + Side²

In this case Base is Hypotenuse, then :

Base² = 8² + 12²

Base² = 64 + 144

Base² = 208

√Base² = √208

Base = 14.422 cm

_______________________________

Now let's solve for Height of Triangle :

The height of triangle AFE is AF (or) The height of triangle AFE is the hypotenuse of triangle ADF

Using Pythagoras Theorem :

Hypotenuse² = Side² + Side²

In this case Height is Hypotenuse, then :

Height² = 15² + 12²

Height² = 225 + 144

Height² = 369

√Height² = √208

Height = 19.209 cm

_______________________________

Now Let's find Area of Shaded region or Area of Triangle AFE :

Area of Triangle = 1/2 × Base × Height

Area of Triangle = 1/2 × 14.422 cm × 19.209 cm

Area of Triangle = 14.422 cm/2 × 19.209 cm

Area of Triangle = 7.211 cm × 19.209 cm

Area of Triangle = 138.516 cm²

Area of Shaded Region = 138.516 cm²