Question

please answer with solution guys ​

Answer 1

Step-by-step explanation:

in ∆ABC right angled at B and also AB = BC = 4cm

using Pythagoras theorem

(AC)^2 = (AB)^2 + (BC)^2

(AC)^2 =( 4)^2 + (4)^2

(AC)^2 = 16 + 16

AC = √32 = 4√2 cm

now AB is perpendicular to BC

so area of ∆ABC = 1/2 (base ×height)

= 1/2 (4×4 ). = 8 (cm)^2

now draw BD perpendicular AC

now area of ∆ABC = 1/2 ( AC×BD)

1/2 (4√2 ×BD ) = 8 (cm)^2

2√2 ×BD = 8

BD = 8/2√2

BD = 4/√2

on rationalising

BD = 4×√2 / √2×√2

BD = 4√2/ 2

BD = 2√2 cm

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