Question

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hey MATES......❤❤❤
Help me​

Answer 1

Answer:

Given:- <B=90°

${cb}^{2} + {ba}^{2} = {ca}^{2}$

${14}^{2} + {ba}^{2} = {50}^{2}$

196+ba^2=2500

ba^2=2500-196

ba^2=2304

ba=√2304

ba =48(by Pythagoras thoream)

We know that bc||Ad

If <b=90°

So, CBAE is a rectangle

CB=AE

AE=14

CE= BA

CE=48 cm

So, In triangle CED

Ce^2+DE^2=x^2

48^2+14^2=x^2

2304+196=x^2

2500=x^2

x=√2500

x=50

Answer 2

Answer:

x = 50

solutions

Given: ∆B = 90°

CB² + BA² = CA²

14² + BA² = 50²

196 + BA² =2500

BA² = 2500 - 196

BA² = 2304

BA = √2304

BA = 48 ( by Pythagoras theorem )

We know that BC||AD

if ∆B = 90°

so, CBAE is a rectangle

CB = AE

AE = 14cm

CE = BA

CE = 48cm

so, In triangle CED

CE² + DE² = x²

48² + 14² = x²

2304 + 196 = x²

2500 = x²

x = √2500

x = 50

silentlover45.❤️