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1 year ago

7

A vertical plane intersects the three-dimensional object in the image and divides the object into equal halves. What is the shap

e of the cross section?
Thank you :)

2 answers:

Crank1 year ago

7 0

If you were to take a knife and cut this shape down the middle, starting from the top, what will the cross section look like?
It will look a lot like how the object looks from the side.
The best answer is D, the last image.

dlinn [17]1 year ago

6 0

Answer:

Option 4 is correct .

Step-by-step explanation:

Given : A three dimensional figure

To Find : What is the shape of the cross section?

Solution :

If we intersects the three-dimensional object in the image and divides the object into equal halves by vertical plane then we can see the attached figure what the output will be .

We can see that if we see the cross section from the side then it looks like the fourth one

So, Option 4 is correct .

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Answer:

Here is the question attached with.

$m\angle CEA =90 \ (deg)$

$m\angle BEF=135\ (deg)$

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$\angle AEF$ is a right angled triangle.

Options $1,4,5,6$ are correct answers.

Step-by-step explanation:

⇒As $\ ray\ AE$ is ⊥ $FEC$ so it will forms right angled triangle then $m\angle CEA =90\ (deg)$ .

⇒Measure of $\angle BEF =135\ (deg)$ as $\angle BEF =\angle AEB +\angle AEF = (45+90)=135\ (deg)$ as $\angle AEB$ is the bisector of $\angle AEC$ ,meaning that $\angle AEB$ is half of $\angle AEC$ so $\angle AEB = 45\ (deg)$ .

⇒ $\angle CEF$ is a straight line as the angles measure over it is $180\ (deg)$ .

⇒Measure of $\angle AEF = 90\ (deg)$ from linear pair concept.

As $\angle CEA + \angle AEF = 180\ (deg)$ ,plugging the values of $m\angle CEA =90\ (deg)$ we have $\angle AEF = 90\ (deg)$ .

The other two options are false as:

$m\angle CEF=m\angle CEA + m\angle BEF = (90+135)=225$

it is exceeding $180\ (deg)$ whereas $\angle CEF$ is a

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And $m\angle CEB=2(m\angle CEA)$ is not true.

As $\angle CEA = 90\ (deg)$ and $\angle CEB=45\ (deg)$

So we have total $4$ answers.

The correct options are $1,4,5,6$ .

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