Question

answer in copy please ​

Answer 1

Answer:

$\huge\bold\pink\star\purple{ANSWER :-}$

Step by step explanation:

$\huge\bold\blue\star\green{1st \: step}$

Split the figure into two ∆ i.e, ∆BCD and ∆ADB.

Find the area of the ∆BCD :-

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

Here, s = semi perimeter i.e, 18cm.so,:-

$\sqrt{18(18 - 15)(18 - 12)(18 - 9)}$

$\sqrt{18(3)(6)(9)} = \sqrt{2916} = 54 {cm}^{2}$

The area of the ∆ BCD is 54cm^2.

$\huge\bold\green\star\purple{2nd \:\: step}$

Find the Area of ∆ABD :-

here AB is can be find out with the help of Pythagoras theorem.

So,

${17}^{2} = {(15)}^{2} + {ab}^{2}$

$ab = 8cm$

Now,

The area of the∆ABD using HERON'S formula:-

$\sqrt{20(20 - 17)(20 - 15)(20 - 8)}$

= 20cm.

So,

Total area of the figure is 54 + 20= 74 cm^2.

$\bold\pink\star\purple{HOPE\: YOU \: LIKE \: THE \: ANSWER }$