Question

please answer this fast I will mark
him/her as brainlist​

Answer 1

Answer:

$\huge\purple\star\underline\mathfrak\pink{Solution :-}\purple\star$

Step-by-step explanation:

✧Split Figure :-

Split the terms into two parts as shown in the figure :

Now, we got a ∆ and a trapezium.

Let's find the area of ∆ first :-

Here, we are unknown about the third

side of the ∆

Let's find out by Pythagorean theorem:-

${h}^{2} = {60}^{2} + {11}^{2}$

${h}^{2} = 3600 + 121$

$h = \sqrt{3721} = 61$

Area of the ∆ :-

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

Here semi - perimeter will be 66 i.e, s

$\green { \underline \bold{area \: of \: the \: triangle:- }} \\ \tt \ \sqrt{66(66 - 61)(66 - 60)(66 - 11)} \\ \tt \\= \sqrt{66 \times 5 \times 6 \times 11} \\ = \sqrt{108900} = 330 {cm}^{2}$

Area of the trapezium :-

$\frac{a + b}{2} \times h = \frac{26 + 61}{2} \times 9 = 783$

$\purple\star\underline\mathfrak\pink{Therefore\: total \: area \: of\: the \: figure \: is \: = 1113cm^2 :-}\purple\star$