Question

find the area of quadrilateral ABCD in which ab is equal to 99 CM BC is equal to 40 cm CD is equal to 28 cm is equal to 15 cm and angle abc is equal to 90 degree​

Answer 1

Correct Question:

Find the area of the quadrilateral ABCD in which AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm and ∆ABC = 90°​.

Solution:

Given:

=> ABCD is a Quadrilateral

=> AB = 9 cm

=> BC = 40 cm

=> CD = 28 cm

=> DA = 15 cm

Construction:

=> Draw a diagonal AC.

To Find:

=> Area of Quadrilateral.

Formula used:

=> Pythagoras theorem

Now, ΔABC is a right angled Δ. So, by using Pythagoras theorem we will find side AC.

=> AC² = AB² + BC²

=> AC² = 9² + 40²

=> AC² = 81 + 1600

=> AC² = 1681

=> AC = 41 cm

Now, we will find area of both ΔADC and ΔABC.

$\sf{So,\;Area\;of\;Triangle\;(ABC) = \dfrac{1}{2}\times Base\times Height}$

$\sf{\implies \dfrac{1}{2}\times 9\times 40}$

$\sf{\implies Area\;of\;Triangle\;(ABC) = 180\;cm^{2}}$

Now, we will find area of ΔADC by using Heron's formula,

So,

$\sf{\implies Heron's\;formula = \sqrt{s(s-a)(s-b)(s-c)}}$

$\sf{\implies So,\;s = \dfrac{a+b+c}{2}}$

Here, a = 15, b = 41 and c = 28

$\sf{\implies So,\;s = \dfrac{15+41+28}{2}}$

$\sf{\implies s = 42\;cm}$

Now, put the values in the formula we get,

$\sf{\implies \sqrt{s(s-a)(s-b)(s-c)}}$

$\sf{\implies \sqrt{42(42-15)(42-41)(42-28)}}$

$\sf{\implies \sqrt{42\times 27\times 1\times 14}}$

$\sf{\implies \sqrt{15876}}$

$\sf{\implies Area\;of\;triangle(ADC) = 126\;cm^{2}}$

Now, Area of quadrilateral (ABCD) = Area of ΔABC + Area of ΔADC

                                                         = 180 cm² + 126 cm²

                                                         = 306 cm²

     

So, area of quadrilateral (ABCD) = 306 cm²

Answer 2

Solution is given in the attachment