Question

027. 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.

(A) 356 (B) 500

(C) 544 (D) 578​

Answer 1

option number is

(A) 356

hope it helps you

Answer 2

Answer:

hey here is your solution

pls mark it as brainliest

refer the reference diagram uploaded above

Step-by-step explanation:

$to \: find = \\ AE {}^{2} + CD {}^{2} \\ \\ so \: in \: \: △ABC \\ ∠ABC = 90 \\ by \: \: pythagoras \: theorem \\ \\ AB {}^{2} + BC {}^{2} = AC {}^{2} \\ \\ = (12) {}^{2} + (16) {}^{2} \\ \\ = 256 + 144 \\ \\ = \sqrt{400} \\ \\ AC = 20$

$moreover \: here \\ DE = 10 \\ \\ thus \: now \: \: dividing \: DE \: and \: AC \\ \\ \frac{DE}{ AC } = \frac{10}{20} \\ \\ \frac{DE}{AC} = \frac{1}{2} \\ \\ DE = \frac{1}{2} \: AC \\ \\ so \: by \\ \: converse \: of \: midpoint \: theorem \\ we \: can \: conclude \: that \\ \\ D \: and \: E \: are \: midpoints \: of \: AB \: and \: BC \\ respectively \\ and \: \: DE \: || \: AC \\ \\ so \: hence \: then \\ \\ AD=DB = \frac{1}{2} \: AB \\ \\ = \frac{1}{2} \times 12 \\ \\ AD=DB = 6.$

$similarly \: then \\ \\ BE=EC = \frac{1}{2} BC \\ \\ = \frac{1}{2} \times 16 \\ \\ BE=EC = 8.$

$now \: \: thus \: then \\ considering \: △ DBC \\ as \: ∠DBC = 90 \\ by \: pythagoras \: theorem \\ \\ DB {}^{2} +BC {}^{2} = DC {}^{2} \\ \\ = (16) {}^{2} + (6) {}^{2} \\ \\ = 256 + 36 \\ \\ DC {}^{2} = 292 \: \: \: \: \: \: \: \: (1)$

$similarly \: applying \: \\ pythagoras \: theorem \: on \: △ABE \\ \\ AB {}^{2} + BE {}^{2} = AE {}^{2} \\ \\ = (12) {}^{2} + (8) {}^{2} \\ \\ = 144 + 64 \\ \\ AE {}^{2} = 208 \: \: \: \: \: \: (2)$

$applying \: now \\ (1) + (2) \\ \\ AE {}^{2} +CD {}^{2} = 292 + 208 \\ \\ AE {}^{2} +CD {}^{2} = 500$