area of a shaded part of the rectangle

[sufinaz]

one more?

 

[firemath]

Please follow the rules of posting on this forum as enunciated by "Read before posting."
Remember that the area of a triangle is A=bh(1/2).

 

[sufinaz]

the complete base is 16m, what's it the base of the triangle? If I knew that I could find the area?

 

[firemath]

Can I see the entire problem?

 

[sufinaz]

I did attach the pic you can see from there, can't draw here

 

[pka]

I did attach the pic you can see from there, can't draw here

But the image in small and sideways making it unreadable.

 

[firemath]

If I'm not mistaken, this is the way to solve it.

Since the triangles all have a base that comprises the 16m base, you can think of all the triangles as one large triangle as far as the equation goes.
16x6=96
96/2=?
Am I right, pka?

 

[sufinaz]

can you actually do that? they are two separate triangles in a rectangle

 

[Dr.Peterson]

You have to be a little creative here. They seem to give you too little information, but what they leave out doesn't affect the answer.

One approach to take is to see what fraction of the rectangle is shaded.

Another is to define the base of each triangle by a variable, and determine the shaded area in terms of those variables, hoping that they will cancel out (as they will).

 

[firemath]

can you actually do that? they are two separate triangles in a rectangle

Think about it. If you took half of the large white triangle and one of the large grey triangles, their area would be the same.

 

[sufinaz]

Triangle 1

eyed length= 13
13x6=78
78/2=39

triangle 2

eyed length=3
3x6=18
18/2=9

39+9=48
rectangle area=96
96-48=48

Is that right?

 

[Dr.Peterson]

You can't really assume a particular length; but this can count as experimentation to see whether the answer depends on the individual lengths.

Try making the lengths x and y, and see what you get.

Then try drawing in vertical lines through the apex of each white triangle, and observe that you have divided the region into four rectangles, each of which is half white and half black.

 

[sufinaz]

ok, when you draw two vertical lines from the triangles apex, you'll have two equal square, their area is 36 each.
the remaining base is 4m ,
area of remaining rectanglesarea is 4x6=24
all have half shaded that means 36+36+24=96/2=48 still 48 m square is that right?

 

[Dr.Peterson]

48 is the correct answer; I've been trying to help you see alternative ways to see the answer, not to correct the answer itself.

Here is the picture I described:



There are four rectangles; each is divided exactly in half by a diagonal. So the shaded triangles are exactly half of the entire figure; their area is therefore exactly half of 6*16, and 6*16/2 = 48m². We don't need to know how wide each individual rectangle is. No guessing is needed.

 

[Deleted member 4993]

ok, when you draw two vertical lines from the triangles apex, you'll have two equal square, their area is 36 each.
the remaining base is 4m ,
area of remaining rectanglesarea is 4x6=24
all have half shaded that means 36+36+24=96/2=48 still 48 m square is that right?

Your answer is correct but your logic escapes me!

Post #12 gave the hint for correct solution. If you draw vertical lines from the apexes of every "inside" triangles (three of those - two white and one black), you'll divide the rectangle into 8 right-angled triangles. If observe carefully, you will be able to see congruent white and black triangles.

Thus the sum of the areas of the black triangles is equal to the sum of the areas of the black triangles (shaded area).

Hence shaded area = 96/2 = 48 m²

 

[Steven G]

Suppose the base of the left triangle is $\displaystyle b_1$ Then the area of the left triangle is $\displaystyle \dfrac{1}{2}b_1\cdot6$

Suppose the base of the right triangle is $\displaystyle b_2$ Then the area of the right triangle is $\displaystyle \dfrac{1}{2}b_2\cdot6$

Note that $\displaystyle b_1 + b_2 = 16$

Now the total area of the triangles is $\displaystyle \dfrac{1}{2}b_1\cdot6$ + $\displaystyle \dfrac{1}{2}b_2\cdot6$ = $\displaystyle \dfrac{1}{2}(b_1 + b_2)\cdot6$ =$\displaystyle \dfrac{1}{2}\cdot 16\cdot6 = 48$

 

[Steven G]

Suppose the base of the left triangle is $\displaystyle b_1$ Then the area of the left triangle is $\displaystyle \dfrac{1}{2}b_1\cdot6$

Suppose the base of the right triangle is $\displaystyle b_2$ Then the area of the right triangle is $\displaystyle \dfrac{1}{2}b_2\cdot6$

Note that $\displaystyle b_1 + b_2 = 16$

Now the total area of the triangles is $\displaystyle \dfrac{1}{2}b_1\cdot6$ + $\displaystyle \dfrac{1}{2}b_2\cdot6$ $\displaystyle \dfrac{1}{2}(b_1 + b_2)\cdot6$ =$\displaystyle \dfrac{1}{2}\cdot 16\cdot6 = 48$

Meant to write $\displaystyle \dfrac{1}{2}b_1\cdot6$ + $\displaystyle \dfrac{1}{2}b_2\cdot6$ = $\displaystyle \dfrac{1}{2}(b_1 + b_2)\cdot6$ =$\displaystyle \dfrac{1}{2}\cdot 16\cdot6 = 48$