Question

In adjoining figure b, Angle ADB = angle BAC=90°,Ad =3cm, DB=4cm and BC=13cm. calculate the area of shaded region ?​

Answer 1

Please see the attachment

Answer 2

Given:

A figure shows triangle ABC and ADB in which Angle ADB = angle BAC = 90°, AD =3cm, DB=4cm and BC=13cm.

To find:

The area of the shaded region.

Solution:

As we know that Pythagoras theorem states that in a right-angled triangle ABC that has a right angle at B, the hypotenuse AC can be given by:

$AC = \sqrt{{(AB)}^{2} + {(BC)}^{2} }$

So,

in a given right-angled triangle ADB,

${AB} = \sqrt{ {(AD)}^{2} + {(DB)}^{2} }$

$AB = \sqrt{ {3}^{2} + {4}^{2} }$

$AB = \sqrt{9 + 16}$

$AB = \sqrt{25}$

$AB = 5cm \: \: \: (i)$

Now, in right-angled triangle BAC,

${(BC)}^{2} = {(AB)}^{2} + {(AC)}^{2}$

${13}^{2} = {5}^{2} + {(AC)}^{2} \: [from \: (i) \: AB = 5cm]$

$169 - 25 = {(AC)}^{2}$

$AC = \sqrt{144}$

$AC = 12cm$

Now,

area of a triangle is given by:

$= \frac{1}{2} \times base \times height$

Also,

area of the shaded region = area of triangle BAC - the area of triangle ADB

So,

area of the shaded region is

$= (\frac{1}{2} \times 12 \times 5) - ( \frac{1}{2} \times 4 \times 3)$

$= 30 - 6$

$= 24 \: {cm}^{2}$

Hence, the area of the shaded region is 24cm2.