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4 years ago

Brad cuts a triangular prism with a plane parallel to the base of the prism what will the shape of the cross section be?

Mathematics

2 answers:

devlian [24]4 years ago

4 0

the answer will be triangular/

Hope I helped. Mark as brainliest

juin [17]4 years ago

4 0

Type of prism is decided by the name of base. Like it can be triangular, square, pentagon etc.

Given that prism is triangular so that means it's base is a triangle.

Given that Brad cuts the given triangular prism with a plane parallel to the base of the prism.

Now we have to find the shape of the cross section of that plane.

It is obvious that the plane obtaind after cutting the ptrism parallel to the base will be of same shape as base.

Hence required plane will be a triangle.

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Evaluate the following expression 5(1 3)-3(2 -1)

BARSIC [14]

Answer:

Step-by-step explanation:

This problem tests your understanding of order of operations rules.

Here you must do any work inside parentheses first, followed by multiplication, followed by subtraction:

5(13) - 3(2 -1) would be 65 -3(1), or 62.

If you actually meant 5(1 - 3) - 3(2 - 1), then you'd have

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Please compare the problem you've shared with the original problem, to ensure that these match.

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3 years ago

Find the sum of -10x^2-x+6−10x 2 −x+6 and 10x^2-510x 2 −5.

irina [24]

Answer:

1. -10x^2-x+6-10x 2-x+6 = -10x^2 - 22x + 12

2. 10x^2-510x 2-5 = 10x^2 - 1020x - 5

Step-by-step explanation:

hope this helps

if i did something wrong let me know so i can fix my answer

#Virgo2007

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2 years ago

Read 2 more answers

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Len [333]

X would equal 81 because the square root of 81 is 9.

9 x 9 = 81

7 0

3 years ago

How many ways can first, second, and third place be awarded in a race of 20 people?

Volgvan

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3 0

3 years ago

Pls answer the question attached

barxatty [35]

$\huge\mathfrak{\underline{answer:}}$

$\large\bf{\angle ACD = 105°}$

__________________________________________

$\large\bf{\underline{Here:}}$

BCD is an isosceles right triangle , right angled at D
ABC is an equilateral triangle

$\large\bf{\underline{To\:find:}}$

∠ ACD

$\large\bf{In\: triangle\:ABC:}$

❒ Sum of all angles is 180° , since it is an equilateral triangle all the three angles would be same

$\large\bf{\underline{So}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = \frac{180}{3}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle ACB = 60°}$

$\large\bf{In\: triangle\:BDC}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle D = 90°}$

$\bf{⟼\angle DBC =\angle DCB }$ (Isosceles triangle)

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DBC + \angle BDC + \angle DCB = 180° }$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB + 90° = 180°}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼2\angle DCB = 180 -90}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = \frac{90}{2}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟼\angle DCB = 45°}$

$\large\bf{\underline{Therefore,}}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = \angle ACB + \angle DCB}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\bf{⟹\angle ACD = 60+45}$

‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎‎‎ ‎ ‎ $\large\bf{⟹\angle ACD = 105°}$

6 0

3 years ago