Question

Triangle ABC is right angled at B, D is the midpoint of BC. Prove that, AC² = 4AD²– 3AB²
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

[FIGURE IS IN THE ATTACHMENT]

GIVEN:

ABC is a right angled at B and D is the midpoint of BC.

In ∆ABD

AD²= AB²+BD²…………(1)
[By Pythagoras theorem]

In ∆ ABC
AC²= AB²+BC²…………(2)

From equation (1)

AD²= AB²+ (BC/2)²
[D is the midpoint of BC]

AD²= AB²+ BC²/4
AD²= AB²/1+ BC²/4
4AD²= 4AB²+ BC²
BC²= 4AD²-4AB²…………….(3)
Using the value of BC² in eq. 2

AC²=AB²+4AD²-4AB²

AC²=AB²-4AB+4AD²
AC²= 4AD²-3AB²

HENCE PROVED….

HOPE THIS WILL HELP YOU….

Answer 2

heya folk!!
.it's too easy ..

given --A ∆ABC in which <B=90° and D is midpoint of BC

to proove- AC^2=(4AD^2-3AB^2)

proof -In ∆ABC,<B=90°

so, AC^2=AB^2+BC^2 ( by Pythagoras theorm)

=)AB^2+(2BD)^2 [BC=2BD】

=)AB^2+4BD^2

=)AB^2+4(AD^2-AB^2). 【AB^2+BD^2=AD^2】

=)(4AD^2-3AB^2)

Hence ,AC^2=4AD^2-3AB^2

hope it. help you.☺

@rajukumar☺