Question

$\huge\fcolorbox{blue}{pink}{QUESTION}$


Tʜᴇ sɪᴅᴇs ᴏғ ᴀ ʀɪɢʜᴛ ᴀɴɢʟᴇᴅ ᴛʀɪᴀɴɢʟᴇ ᴀʀᴇ (8x-1) ᴄᴍ (3x+2) ᴄᴍ ᴀɴᴅ (4x+9) ᴄᴍ ɪғ ᴛʜᴇ ᴘᴇʀɪᴍᴇᴛᴇʀ ᴏғ ᴛʜᴇ ᴛʀɪᴀɴɢʟᴇ ɪs 40ᴄᴍ. Fɪɴᴅ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ʜʏᴘᴏᴛᴇɴᴜsᴇ.


⚔ Sᴘᴀᴍᴍɪɴɢ = I'ᴅ REPORT ​

Answer 1

Answer:

$\huge\fcolorbox{blue}{pink}{☃αղsաҽɾ☃}$

$\bold{⇒ \: perimerter \: = \: sum \: of \: all \: sides \: } \\ \\ ⇒ \: 40 = (8x - 1) +( 3x + 2) + (4x + 9) \\ \\ ⇒ \: 40 = 8x - 1 + 3x + 2 + 4x + 9 \\ \\ ⇒ \: 40 = 15x + 10 \\ \\ ⇒ \: 40 - 10 = 15x \\ \\ ⇒ \: 30 = 15x \\ \\ ⇒ \: x = \frac{30}{15} \\ \\ ⇒ \: x = 2$

$\bold \pink{sides \: of \: the \: triangle \: are \: \ \: : - }$

$⇒ \: (8x - 1) = 8(2) - 1 = 15 \\ \\ ⇒ \: (3x + 2) = 3(2) + 2 = 8 \\ \\ ⇒ \: (4x + 9) =4 (2) + 9 = 17$

$\bold {As \: the \: hypotenues \: is \: the \: longest \: side \: of \: any \: triangle }$

$\bold{so \: the \: hypotenues \: is \: 17 \: cm}$

Answer 2

Answer:

The sides of a right angled triangle are (8x - 1) cm (3x + 2) cm and (4x + 9) cm if the perimeter of the triangle is 40 cm.

To Find :-

What is the length of the hypotenuse.

Solution :-

Let,

\mapsto \rm{\bold{First\: side\: =\: (8x - 1)\: cm}}↦Firstside=(8x−1)cm

\mapsto \rm{\bold{Second\: side =\: (3x + 2)\: cm}}↦Secondside=(3x+2)cm

\mapsto \rm{\bold{Third\: side =\: (4x + 9)\: cm}}↦Thirdside=(4x+9)cm

As we know that :

\begin{gathered}\bigstar\: \: \sf\boxed{\bold{\pink{Perimeter\: of\: triangle =\: Sum\: of\: all\: sides}}}\\\end{gathered}

★

Perimeteroftriangle=Sumofallsides

Given :

Perimeter = 40 cm

According to the question by using the formula we get,

\implies \sf (8x - 1) + (3x + 2) + (4x + 9) =\: 40⟹(8x−1)+(3x+2)+(4x+9)=40

\implies \sf 8x - 1 + 3x + 2 + 4x + 9 =\: 40⟹8x−1+3x+2+4x+9=40

\implies \sf 8x + 3x + 4x - 1 + 2 + 9 =\: 40⟹8x+3x+4x−1+2+9=40

\implies \sf 15x + 1 + 9 =\: 40⟹15x+1+9=40

\implies \sf 15x + 10 =\: 40⟹15x+10=40

\implies \sf 15x =\: 40 - 10⟹15x=40−10

\implies \sf 15x =\: 30⟹15x=30

\implies \sf x =\: \dfrac{\cancel{30}}{\cancel{15}}⟹x=

15

30

\implies \sf \bold{\purple{x =\: 2}}⟹x=2

Hence, the required sides are :

\leadsto⇝ First side :

\longrightarrow \sf (8x - 1)\: cm⟶(8x−1)cm

\longrightarrow \sf \{8(2) - 1\}\: cm⟶{8(2)−1}cm

\longrightarrow \sf (16 - 1)\: cm⟶(16−1)cm

\longrightarrow \sf\bold{\red{15\: cm}}⟶15cm

\leadsto⇝ Second side :

\longrightarrow \sf (3x + 2)\: cm⟶(3x+2)cm

\longrightarrow \sf \{3(2) + 2\}\: cm⟶{3(2)+2}cm

\longrightarrow \sf (6 + 2)\: cm⟶(6+2)cm

\longrightarrow \sf\bold{\red{8\: cm}}⟶8cm

\leadsto⇝ Third side :

\longrightarrow \sf (4x + 9)\: cm⟶(4x+9)cm

\longrightarrow \sf \{4(2) + 9\}\: cm⟶{4(2)+9}cm

\longrightarrow \sf (8 + 9)\: cm⟶(8+9)cm

\longrightarrow \sf\bold{\red{17\: cm}}⟶17cm

As we know that :

\leadsto⇝ Length of Hypotenuse :

Hypotenuse is the longest side of a right-angled triangle.

\therefore∴ The length of hypotenuse of a right-angled triangle is 17 cm .

\begin{gathered}\\\end{gathered}

VERIFICATION :-

\implies \tt{(8x - 1) + (3x + 2) + (4x + 9) =\: 40}⟹(8x−1)+(3x+2)+(4x+9)=40

By putting x = 2 we get,

\implies \tt{\{8(2) - 1\} + \{3(2) + 2\} + \{4(2) + 9\} =\: 40}⟹{8(2)−1}+{3(2)+2}+{4(2)+9}=40

\implies \tt{ (16 - 1) + (6 + 2) + (8 + 9) =\: 40}⟹(16−1)+(6+2)+(8+9)=40

\implies \tt{15 + 8 + 17 =\: 40}⟹15+8+17=40

\implies \tt{\bold{\pink{40 =\: 40}}}⟹40=40