Question

☢ QUESTION ☢


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PQR is a triangle, right-angled at P. If PQ = 10 cm
and PR = 24 cm, find QR.

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Answer 1

GIVEN :-

$\small \bold \pink{→~ PQ~ =~ 10 ~CM~ , ~ PR~ = ~24~ CM }$

$\large \underline\bold{LET'S ~QR ~BE~ x ~CM}$

$\small\bold \pink{I ~RIGHT~ ANGLED ~TRAINGLE ~QPR}$

$\bold{( HYPOTENUSE)^2 = ( BASE)^2 + ( PERPENDICULAR)^2 } \: \: \: \: \: \: \: \: \: \bold \pink{[ \: by \: pythagoras \: theorem \:] }$

$\small \bold\red\longmapsto</p><p> \bold \red{( QR)^2 = ( PQ)^2 + ( PR)^2 }$

$\large \red\longmapsto \bold\red{ {x}^{2} = (10 {}^{2}) + (24) {}^{2} }$

$\large \red\longmapsto \bold \red{x {}^{2} = 100 + 576 = 676}$

$\large \orange\longmapsto \bold \orange{ \sqrt{675} = 26 \: cm \: \checkmark}$

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☢ HENCE THE LENGTH OF QR IS 26 CM ☢

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Answer 2

Answer:

Give thanks

Step-by-step explanation:

GIVEN :-

\small \bold \pink{→~ PQ~ =~ 10 ~CM~ , ~ PR~ = ~24~ CM }→ PQ = 10 CM , PR = 24 CM

\large \underline\bold{LET'S ~QR ~BE~ x ~CM}

LET

′

S QR BE x CM

\small\bold \pink{I ~RIGHT~ ANGLED ~TRAINGLE ~QPR}I RIGHT ANGLED TRAINGLE QPR

\bold{( HYPOTENUSE)^2 = ( BASE)^2 + ( PERPENDICULAR)^2 } \: \: \: \: \: \: \: \: \: \bold \pink{[ \: by \: pythagoras \: theorem \:] }(HYPOTENUSE)

2

=(BASE)

2

+(PERPENDICULAR)

2

[bypythagorastheorem]

\small \bold\red\longmapsto < /p > < p > \bold \red{( QR)^2 = ( PQ)^2 + ( PR)^2 }⟼</p><p>(QR)

2

=(PQ)

2

+(PR)

2

\large \red\longmapsto \bold\red{ {x}^{2} = (10 {}^{2}) + (24) {}^{2} }⟼x

2

=(10

2

)+(24)

2

\large \red\longmapsto \bold \red{x {}^{2} = 100 + 576 = 676}⟼x

2

=100+576=676

\large \orange\longmapsto \bold \orange{ \sqrt{675} = 26 \: cm \: \checkmark}⟼

675

=26cm✓

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☢ HENCE THE LENGTH OF QR IS 26 CM ☢

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