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Mathematics

2 answers:

jok3333 [9.3K]3 years ago

7 0

I’m pretty sure parallelograms are squares, rectangles and rhombuses. So you could take up the needed amount of space with a rectangles and rhombus. Sorry if this isn’t that much help.

Margarita [4]3 years ago

4 0

Answer: I’m pretty sure parallelograms are squares, rectangles and rhombuses. So you could take up the needed amount of space with a rectangles and rhombus. Just make one of the parallelograms listed and you should be fine!

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The lengths of the

Dmitrij [34]

Answer:

$S_{\Delta ABC}$ = 4· $S_{\Delta EDC}$

Step-by-step explanation:

From the midpoint theorem, which states that the line that a line drawn such that it joins the midpoints of two sides of a triangle, is parallel to the third side of the triangle and is equal to half the length of the third side

Therefore, the lengths of the sides of ΔDEF, drawn by joining the midpoints of ΔABC is equal to half the length of and parallel to the corresponding side of ΔABC

We therefore, have that the corresponding sides of ΔABC and ΔDEF have a common ratio and a pair of sides in each triangle form same angles, therefore;

ΔDEF is similar to ΔABC by Side, Side, Side SSS similarity.

The length of the perimeter of ΔABC, $S_{\Delta ABC}$ = 2 × The length of the perimeter of triangle ΔEDC, $S_{\Delta EDC}$

$S_{\Delta ABC}$ = 2 × $S_{\Delta EDC}$

∴ $S_{\Delta ABC}$ ≠ 4 × $S_{\Delta EDC}$

The statement which is incorrect is therefore;

$S_{\Delta ABC}$ = 4 × $S_{\Delta EDC}$ .