Question

In ΔABC, ∠B=90°, AB=12, BC=16. Let D,E be points on AB and BC respectively so that DE=10. Then AE2 + CD2
equals.

Answer 1

Answer:

ΔABE is a right triangle, right angled at B

AB²+BE² = AE²……………..(1)

(by the Pythagoras theorem)

ΔDBC is a right triangle, right angled at B

DB²+BC² = CD²…………….(2)

(by the Pythagoras theorem

Adding eq 1 & 2

AE²+CD²= (AB²+BE²)+(BD²+BC²)

AE²+CD²= (AB²+BC²)+(BE²+BD²)......(3)

[Rearranging the terms]

ΔABC is a right triangle,

AB²+BC² = AC²…………….(4)

(by the Pythagoras theorem)

Δ DBE is a right triangle

DB²+BE² = DE²………………(5)

(by the Pythagoras theorem)

AE²+CD²= (AB²+BC²)+(BE²+BD²)

AE²+CD²= AC²+DE²

[ From equation 4 and 5]

[FIGURE IS THE ATTACHMENT FIGURE IS THE ATTACHMENT]

Step-by-step explanation:

This Answer is Right Answer  

Please Make Me Brainlist  

Have a Nice Day  

Keep Distance in infected Person

Thank You…..

Answer 2

Answer:

AC^2 + DE^2 = 500

Step-by-step explanation:

Given: AB=12, BC=16, DE=10

Therefore , DE^2 = 100.... (equation 1)

Now, ∆ABC is a right angle traingle ,

∵ AB^2 + BC^2 = AC^2...(By Pythagoras theorem)

∵AC^2 = AB^2 + BC^2

= (12)^2 + (16)^2

= 144 + 256

= 400

∵ AC^2 = 400....(equation 2 )

Now,

From equation (1) and equation (2), we get,

AC^2 + DE^2 = 400 + 100

= 500