Question

14. Find the area of each of the following figures.(Answer. 76 m²)
Please answer it correctly, I'll mark you brainliest.
Thank you.(in advance) ​

Answer 1

Answer:

76 sq. m

Step-by-step explanation:

Draw diagonal EC. AECB is a square and EDC is an isosceles triangle.

Area of the figure = Area of AECB +  Area of EDC

Area of AECB =  8 * 8 = 64 sq. m

Area of EDC Using Heron's Formula

$s=\frac{a+b+c}{2}$

here, a=8, b=5, c=5

$s= \frac{8+5+5}{2}$

s=9m

Heron's Formula

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{9(9-8)(9-5)(9-5)} \\=\sqrt{9*1*4*4} \\=3*4$

= 12 sq. m

Area of the figure = 64 + 12 = 76 sq. m

Answer 2

Sure, Here Is The Solution.

We See That The Question Compromises 2 Shapes, Square And A Triangle.

Now If We Split The Triangle From The Square, It Will Give Us 2 Shapes, Right? Yes.

Now, We Apply The Formula (Area Of The Square) Which Is a^2, Because Equal Length Of Size!

A^2 = 8^2 = 64 m^2.

Since, The Length Of The Square Is Same From All Sides, We Can Assume That The Separation Of Triangle Will Give A Base Of 8 m.. EC = AB, Meaning The Length Of AB = 8 m, So EC = 8 m.

Now That Is The Base, Of The Triangle.

1/2 * B * H Is The Formula, But We Know The Base Is 8 m.. But THE HEIGHT IS NOT 5 m... The Height Is ALWAYS PERPENDICULAR.

So, Applying Pythagoras Theorem..

The 8m Will Be Halved, Which Is 8/2 = 4..

Square Root ( 5^2 - 4^2 ) = 3m

Now Apply The Formula..

1/2 * 8 * 3 + 8^2 = 76 m^2