Q. 4 D5.0( 1 Vote )

# Show that the fol

Formula used: (10, 0), (18, 0) and (10, 15)

Let the points be A (10, 15), B (10, 0) and C (18, 0)

Distance of AB

AB = √ ((10 – 10))2 + (0 – 15)2)

AB = √ ((0)2 + (–15)2)

AB = √ (0 + 225)

AB = √225

Distance of BC

B C= √ ((18 – 10)2 + (0 – 0)2)

BC = √ ((8)2 + (0)2)

BC = √ (64 + 0)

BC = √ 64

Distance of AC

AC = √ ((18 – 10)2 + (0 – 15))2)

AC = √ ((8)2 + (–15)2)

AC = √ (64 + 225)

AC = √289

i.e. AB2 + BC2

= (√225)2 + (√64)2

= 225 + 64

= 289 = (AC)2

Hence, ABC is a right–angled triangle. Since the square of one side is equal to sum of the squares of the other two sides.

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