Question

ABCD is a quadrilateral whose diagonal BD is perpendicular to the side AB.If AB = 4.5cm, AD = 7.5CM AND the distance of C from BD IS 1.5cm, then the area of quadrilateral is

Answer 1

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Answer 2

Hence the area of quadrilateral is 18 $cm^2$.

Step-by-step explanation:

Given:

Consider the quadrilateral ABCD, such that diagonal BD is perpendicular to side AB and CM is the perpendicular to BD from C.

Also, we are given that AB = 4.5 cm,  AD=7.5 cm and CM=1.5 cm.

Now, the triangle ABD ia a right angled triangle, by Pythagorean theorem:

$AB^2+BD^2=AD^2$

⇒ $BD^2=AD^2-AB^2$ = $(7.5)^2-(4.5)^2$ = 56.25−20.25 = 36 ⇒ $BD =\sqrt{36}$

⇒ BD = 6 cm.

Area of right angled triangle ABD

$A_1=\frac{1}{2} (BD)(AB)$                           [$Area= \frac{1}{2} |times base \times altitude$]

⇒ $A_1=\frac{1}{2} (6)(4.5)$

⇒ $A_1= 13.5 cm^2.$

Area of triangle BCD

$A_2=\frac{1}{2} (BD)(CM)$                            [$Area= \frac{1}{2} |times base \times altitude$]

⇒$A_2=\frac{1}{2} (6)(1.5)$

⇒$A_2=4.5 cm^2$.

Now, the area of the quadrilateral ABCD

= Area of triangle ABD+ Area of the triangle BCD

= $A_1+A_2$

= 13.5+4.5

=$18 cm^2$

Hence the area of quadrilateral is $18 cm^2$.