Question

Identify the type of triangle if AB = 3(x-1),BC = 2(x+1) ,AC = x+7 where x=5cm.give answer of this​

Answer 1

Answer:

$\huge\underline{\bf\red{ANSWER}}$

We are given that in △ABC,∠=90°

type sides of △ABC is given by

AB=x cm

BC=(4x+4)cm

AC=(4x+5) cm

From the pythagoras theorem,

+BC2 So,(4x+5)2−(x)

2+(4x+4) 2 16x 2 +40x+25=x 2 +16x

2 +22z−16

∴ 16x

2 +40x+25=17x

2+32x+16

+32x+16∴ x

2 −8x−9=0∴ (x−9)(x+1)=0∴ x=9 or x=−1

but distance cannot be negative

So, x=9

Now, sides are:

AB=x=9 cm

BC=4(9)+4=40 cm

Now, sides are

AB=x=9 cm

BC=4(9)+4=40 cm

AC=4(9)+5=41 cm

Answer 2

Answer:

We are given that in △ABC,∠=90°

type sides of △ABC is given by

AB=x cm

BC=(4x+4)cm

AC=(4x+5) cm

From the pythagoras theorem,

+BC2 So,(4x+5)2−(x)

2+(4x+4) 2 16x 2 +40x+25=x 2 +16x

2 +22z−16

∴ 16x

2 +40x+25=17x

2+32x+16

+32x+16∴ x

2 −8x−9=0∴ (x−9)(x+1)=0∴ x=9 or x=−1

but distance cannot be negative

So, x=9

Now, sides are:

AB=x=9 cm

BC=4(9)+4=40 cm

Now, sides are

AB=x=9 cm

BC=4(9)+4=40 cm

AC=4(9)+5=41 cm

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Step-by-step explanation:

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