Can You Find the Area of This Figure?

Use a right triangle and the Pythagorean Theorem to find the area of the entire figure.

[Grizzly]

Without some additional information on the diagram, it is an assumption that the top 2 squares are centred on the bottom 3 squares.
A set of 'equal length' hashes on the left and right 'shelves' would suffice.

[rodrigomarinho]

I know BY EYE the vertical line divides the square in half, but nothing is said or proved. So such information was necessary in order to solve it properly, otherwise it's just an assumption.

[werhsdnas]

This problem is unsolvable unless you know that the vertices of the top squares intersect with the midpoints of the bottom square. Hate it when videos leave out critical information to make the problem seem harder than it is

[santiagoarosam]

If we call “L” next to the component squares, the area of the initial figure is the same as that of the rectangle with dimensions 2.5Lx2L ⇒ (BC)²=41=2.5²L²+2²L² ⇒ 10.25L²=41 ⇒ L²=4 ⇒ L=2 ⇒ Area of the initial figure = 2.5Lx2L=2.5x2x2x2=20

[اممدنحمظ]

تمرين جميل . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين

[honestadministrator]

Side of square tiles be 'a', then
(2 a)^2 + ( 5 a /2) ^2 = 41 sqr unit
41 a ^2 /4 = 41 sqr unit
Desired area = 5 a^2 = 20 sqr unit

[treforjones]

41 = 25 +16 so 41 = 4^2 +5^2, thus the triangle is vert height 4, base is 5, hyp is sqrt41. If height is 4, then each squart is side 2. 5 squares of 2*2 = 20 sq units for the figure

[鈞齊]

(2x)²+(5x/2)²= 41, find 5x²
=> 4x²+25x²/4= 41
=> 5x²= 41×(4×5)/41= 20

[denisrenaldo]

Don’t understand why you need to find the side of each square. As you are looking for an area which is the area of 5 identical squares, you only need to find the square of this side and multiply it by 5 to solve the problem. If you call it a**2, you have BC**2 = a**2 (2**2 + 2.5**2)=41, so 5 a**2 = 5*41/10.25 = 20. That’s it !

[kibeterick]

Congratulations my sister ! Your using creativity of the highest order

[Nikioko]

(2, 5x)² + (2x)² = √41²
10, 25x² = 41
x² = 41/10, 25
5x² = 205/10, 25 = 20 SU

[opytmx]

4x^2 + 6.25x^2 = 41 or 10.25x^2 = 41 give x^2 = 4. Thus x = 2 as is has to be positiv. So each square has 2 * 2 = 4 square units, and 5 * 4 = 20 - nice riddle!

[himo]

side of each square = x
(2x)²+(5x/2)²=(√41)² 4x²+25x²/4=41 41x²/4=41 x²=4
x²*5=4*5=20
area of this figure : 20

[bandarusatyanandachary]

Units of sides were not specified. How can we assume all base sides and vertical sides are equal?

[devondevon]

20
Turn the figure into a rectangle by removing half of the square at the bottom left
and place it on the top right to form a rectangle
Let the length of each square = 2N
Hence 2N + 2N + N = 5N the length of the rectangle And
2N + 2N = 4N width of the rectangle. Hence
(5N)^2 + (4N)^2 = (sqrt 41)^2 (Pythag)
25N^2 + 16N^2 = 41
41N^2 =41
N^2 = 1
N =1
Hence the dimension of the rectangle = 4 (1) by 5(1) or 4 by 5
Hence area = 20

[RövşənHəbibov-bg]

Soldakı yarım kvadratı kəsib sağayerləşdirsək tərələri 4 və 5 olan duzbucaqlı alınar.

[rcmcjeb]

Tout à fait d'accord avec le commentaire précédent au sujet l'information préalable manquante. En considérant cette information acquise de manière empirique, j'ai pu voir que 41 était la somme de deux carrés parfaits (25 + 16)... Après c'est easy 😊!

[benjaminchang]

One who knows math shall get answer within 1 min.

[alamshaikhahmad]

2×2=4+2.5×2.5=10.25sc root3.2015621187