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Rus_ich [418]
3 years ago
12

Given: PRST is a square

Mathematics
1 answer:
xxTIMURxx [149]3 years ago
5 0

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

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