In a Trapezium ABCD, if AB || CD, then (A C^2+B D^2)=?

Download App

MCQ

In a Trapezium $ABCD,$ if $\text{AB || CD},$ then $(A C^2+B D^2)=?$

A
$B C^2+A D^2+2 B C \times A D$
B
$\mathrm{AB}^2+C \mathrm{D}^2+2 \mathrm{AB} \times \mathrm{CD}$
C
$A B^2+C D^2+2 A D \times B C$
✓
$\mathrm{BC}^2+\mathrm{AD}^2+2 \mathrm{AB} \times \mathrm{CD}$

✓

Answer

Correct option: D.

$\mathrm{BC}^2+\mathrm{AD}^2+2 \mathrm{AB} \times \mathrm{CD}$

Given: $ABCD$ is a trapezium with $\text{AB || CD}$
Construction: Draw $DE$ and $CF$ $\bot$ to $AB.$
Then in $\triangle\text{ABC}$
$\angle\text{BAC}$ is acute
$\therefore B C^2=A C^2+A B^2-2 A F: A B \ldots(1)$
and in $\triangle \mathrm{BDA}$
$\angle \mathrm{DBA}$ is acute
$\therefore A D^2=B D^2+A B^2-2 B E: A B \ldots(2)$
Adding $(1)$ and $(2)$ we get
$B C^2+A D^2=A C^2+B D^2+2 A B^2-2 A F \cdot A B-2 B E \cdot A B$
$\Rightarrow A C^2+B D^2=B C^2+A D^2-2 A B[A B-A F-B E)$
$=B C^2+A D^2-2 A B[A B-(A E+E F)-(B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A E+E F+B F+E F)]$
$=B C^2+A D^2-2 A B[A B-(A B+C D)](\therefore E F=D C)$
$=B C^2+A D^2-2 A B[-(C D)]$
$=A D^2+B C^2+2 A B \times C D$