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emmasim [6.3K]
3 years ago
13

I need help on this :,)​

Mathematics
1 answer:
masha68 [24]3 years ago
3 0

Answer:

a. Figure B and Figure C

b. Figure C

c. Figure B and Figure C

Step-by-step explanation:

Parallelograms need at least one pair of sides to be parallel.

Squares have to have four sides of equal length

Rectangles need to have all interior angles equal to 90 degrees.

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Given: PRST is a square
xxTIMURxx [149]

Answer:

(1-\sqrt{2})a^2

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a

Thus,

MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a

Consider isosceles triangle MSC. In this triangle

MS=MC=(\sqrt{2}-1)a.

The area of this triangle is

A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}

Consider right triangle PTS. The area of this triangle is

A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2

5 0
3 years ago
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