Which is not a pair of congruent angles in the diagram below?

Mathematics

2 answers:

alexgriva [62]3 years ago

7 0

Answer:

B. Angles ABD and BAD

Step-by-step explanation:

Markings are different

kvv77 [185]3 years ago

3 0

Answer:

B. Angles ABD and BAD

Step-by-step explanation:

The way that you tell which angles are congruent, are based on the angle marks. Angle A and C are congruent because they both contain 3 of the congruency marks, meaning they would equal the same degree. The answer is B because starting with angle BDA, B is your starting point, whereas D is the angle and A is the point you end on. With angle BAD, you start at point B and end at D, making angle A the one you're looking at. Because angle A contains 3 congruency marks, while angle B only contains one, this makes the two angles NOT congruent. Hope this helps. :)

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The triangle T has vertices at (-2, 1), (2, 1) and (0,-1). (It might be an idea to

Firdavs [7]

Rewrite the boundary lines y = -1 - x and y = x - 1 as functions of y :

y = -1 - x ==> x = -1 - y

y = x - 1 ==> x = 1 + y

So if we let x range between these two lines, we need to let y vary between the point where these lines intersect, and the line y = 1.

This means the area is given by the integral,

$\displaystyle\iint_T\mathrm dA=\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy$

The integral with respect to x is trivial:

$\displaystyle\int_{-1}^1\int_{-1-y}^{1+y}\mathrm dx\,\mathrm dy=\int_{-1}^1x\bigg|_{-1-y}^{1+y}\,\mathrm dy=\int_{-1}^1(1+y)-(-1-y)\,\mathrm dy=2\int_{-1}^1(1+y)\,\mathrm dy$

For the remaining integral, integrate term-by-term to get

$\displaystyle2\int_{-1}^1(1+y)\,\mathrm dy=2\left(y+\frac{y^2}2\right)\bigg|_{-1}^1=2\left(1+\frac12\right)-2\left(-1+\frac12\right)=\boxed{4}$

Alternatively, the triangle can be said to have a base of length 4 (the distance from (-2, 1) to (2, 1)) and a height of length 2 (the distance from the line y = 1 and (0, -1)), so its area is 1/2*4*2 = 4.