Question

In rectangle ABCD, diagonal BD = 26 cm and cotangent of angle ABD = 1.5. Find the area and the perimeter of the rectangle ABCD. 

Answer 1

Consider the diagram: 

cot ∠ABD = `(15)/(10) = (3)/(2)`

i.e.`"base"/"perpendicular" = "AB"/"BD" = (3)/(2)`

Therefore, if length of base = 3x, length of perpendicular = 2x

Since

AB² + AD² = BD²            ...[Using Pythagoras Theorem]

(3x)² + (2x)² = BD²

BD² = 13x²

∴ BD = `sqrt13x` 

Now

BD = 26

`sqrt13x` = 26

x = `(26)/sqrt13`

Therefore

AD = 2x

= 2x `(26)/sqrt13`

= `(52)/sqrt13` cm

AB = 3x

= `3 xx (26)/sqrt13`

=`(78)/sqrt13` cm

Now

Area of rectangle ABCD = AB × AD

= `(78)/sqrt13 x (52)/sqrt13`

= 312 cm² 

Perim of rectangle ABCD = 2 (AB + AD)

= 2 `((78)/sqrt13 + (52)/sqrt13)`

= `(260)/sqrt13`

= 20`sqrt13` cm