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2 years ago

Letters in the word mathematics

Mathematics

1 answer:

liubo4ka [24]2 years ago

3 0

Answer:

Answer is given below .

Step-by-step explanation:

Letters in the word "MATHEMATICS "

M,A,T,H,E,I,C,S.

 

THANKS FOR GIVING ME THE OPPORTUNITY TO ANSWER YOUR QUESTION. 

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Which transformation will map an isosceles trapezoid onto itself? rotation by 180° about its center, reflection across a diagona

alukav5142 [94]

Answer:

The correct option is 4.

Step-by-step explanation:

The non parallel sides of an isosceles trapezoid are congruent.

The image of an isosceles trapezoid is same as the preimage of isosceles trapezoid if

1. Reflection across a line joining the midpoints of parallel sides.

2. Rotation by 360° about its center.

3. Rotation by 360° about origin.

If we rotate the trapezoid by 180° about its center, then the parallel sides will interchanged.

If we reflect the trapezoid across a diagonal, then the resultant figure will be a parallelogram.

If we reflect across a line joining the midpoints of the nonparallel sides, then the parallel sides will interchanged.

After rotation by 360° about the center, we always get an onto figure.

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Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are given the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

Finding  $\angle DFE$ :

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$

Finding  $\angle E$ :

Property of external angle: External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external $\angle DFG$ = $\angle E + \angle EDF$

$\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2} -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)$

Comparing equations (1) and (2):

It can be clearly seen that:

$\angle DFE = \angle E =90-\dfrac{y}{2}$

The two angles of $\triangle EDF$ are equal hence $\triangle EDF$ is isosceles.

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