Questions on Geometry: Area and Surface Area answered by real tutors!

In rhombus ADEF, the half of the long diagonal is   =  =  units long.


So, we can find the area of the parallelogram ABCF as  the product of the base FC = 11+5+5 = 21 by the height AG = .


So, the area of the parallelogram ABCF is   = .


To it, we should add the area of triangle FDE   =  = .


Thus, the total combined area is  +  = .


ANSWER.  The total combined area is    square units.

I got  and the AI got . Which of us made a
mistake?  If it was me,  I'll correct it.

Edwin

We calculate GE by using the Pythagorean theorem on right triangle FGE.


AG is also , so diagonal of the rhombus, AE = 
FG = GD = 5 so the other diagonal of the rhombus, FD is 10.

The area of a rhombus is  the product of the diagonals, or,
Area of rhombus ADEF is .


Since ABCF is a parallellogram, 
ABCD is an isosceles trapezoid (or trapezium if you live in the UK).
We draw two green lines at D and C perpendicular to AB and CD. They partition
the trapezoid into two congruent right triangles, and a rectangle in the middle.
Since AH = GD = 5, the area of right triangle AHD is

The area of right triangle BIC is also 
All that's left of isosceles trapezoid ABCD is the area of rectangle HICD which
is 
Area of rectangle = 
Area of right triangle BIC = 
Area of isosceles trapezoid ABCD = 

Area of rhombus ADEF = 
Area of isosceles trapezoid ABCD = 

Area of composite figure = 

That's approximately 422.4 sqruare units.

Edwin

An envelope is pictured below. Solve for the area of the regions
marked XXX.



DE is half of 12 or 6. So we find AD with the Pythagorean theorem.
(Or you may recognize it as a 6-8-10 right triangle, a 3-4-5 right triangle
with all sides doubled):

 

So tha area of right triangle ADE is 
in2.

Triangle BCE is congruent to triangle ADE so it also has area in2. 

Total shaded area (or area marked XXX) = (2)(24) = 48 in2.

Edwin

My drawing is not quite to scale, because GCDE doesn't quite
look like a rhombus.

I labeled two more points than the drawing on the site, H and I. 



Area of rhombus GCDE is half the product of the diagonals.



Area of triangle EFG is half the base times height


 
Now all we need is the area of trapezoid ABCG.

We must find CG. 

The diagonals of a rhombus are perpendicular, and therefore triangle CIG
is a right triangle. GI = DI = 7, and IC = 4, so



The area of a trapezoid is the average base times the height:

The average base of trapezoid ABCG is 

The height of trapezoid is 6, so the area of trapezoid ABCG is



So we add the three areas together:

Area of trapezoid ABCG = 

Area of rhombus GCDE = 56

Area of triangle EFG = 100

Answer = 

Area of rhombus GCDE is half the product of the diagonals.



Area of triangle EFG is half the base times height


 
Now all we need is the area of trapezoid ABCG.

We must find CG. 

The diagonals of a rhombus are perpendicular, and therefore triangle CIG
is a right triangle. GI = DI = 7, and IC = 4, so



The area of a trapezoid is the average base times the height:

The average base of trapezoid ABCG is 

The height of trapezoid is 6, so the area of trapezoid ABCG is



So we add the three areas together:

Area of trapezoid ABCG = 

Area of rhombus GCDE = 56

Area of triangle EFG = 100

Answer =  cm2

That's approximately 222.2 cm2 
 
Edwin

There are three parts to this trapezoid, a rectangle in the middle, and two
right triangles on each end.

On the left, we have a special right triangle, a 6-8-10 right triangle,
whose sides are twice a 3-4-5 right triangle.

To calculate the longer leg 8, if you need to,  

So the area of the triangle on the left is



The area of the rectangle in the middle is



The area of the right triangle on the right:



Adding the 3 parts:



Edwin

We want to know the area of the two segments marked "XXX" below



The triangle is inscribed in a semicircle, i.e., its hypotenuse is a diameter,
so we know it is a right triangle.

Since it has a 30o angle, we know it is a 30-60-90 right triangle.

We know that the longer leg is 

You should learn the following information about the special 30-60-90
special right triangle.  It occurs very often.

The sides of the standard 30-60-90 right triangle:
The shorter leg is 1 unit long. 
The longer leg is  units long.
The hypotenuse is 2 units long.

Let the shorter leg of your 30-60-90 right triangle be x.

You are given the longer leg of the 30-60-90 triangle as 

So make the proportion:





cross-multiply:







So the shorter leg of the 30-60-90 special right triangle is 9 cm.

Therefore the area of the 30-60-90 special right triangle is


 cm2

To get the area of the two segments marked XXX, we must find the radius
of the circle.  Since the hypotenuse is twice the shorter side, the
hypotenuse is (9)(2) = 18 cm. The hypotenuse is a diameter, so the radius is
half that or 9.

The area of the whole circle = 
The area of the semicircle the triangle is inscribed in is half that, or


So we subtract the area of the triangle from the area of the semicircle
and get

 cm2

Factor out ,

 cm2, about 57.1 cm2
 
Edwin

This is a rhombus is made of 4 special right triangles. 
These are 5-12-13 right triangles.

Each of them has shorter leg = 5, hypotenuse = 13, and therefore
longer leg = 

So each 5-12-13 right triangle below has base = 12 and height = 5

So each of the four 5-12-13 right triangles has area

 
 
and since there are 4 of them, the total area =  cm2.

I think you are studying special right triangles, so this is a good
page to look at:

https://flexbooks.ck12.org/cbook/ck-12-precalculus-concepts-2.0/section/4.3/primary/lesson/special-right-triangles-pcalc/

Edwin

This triangle is equilateral, since all three its sides are congruent segments a = 4 cm.


The area of every equilateral triangle with the side length "a" is

    area = =   square units.


In this problem,  the area of the triangle is

    area =  =  = 6.9282 cm^2  (rounded).    ANSWER