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1 year ago

Given: PRST square

Mathematics

2 answers:

Zigmanuir [339]1 year ago

4 0

Answer:

7a²/16

Step-by-step explanation:

Area of the triangle PTS

½ × a × a

a²/2

Length of PS:

sqrt(a² + a²)

asqrt(2)

Length of MS:

¼asqrt(2)

Triangles MCS and TPS are similar

With sides in the ratio:

¼asqrt(2) : a

sqrt(2)/4 : 1

Area of triangle SMC:

A/(a²/2) = [(sqrt(2)/4)]²

2A/a² = 1/8

A = a²/16

Area of PTMC

= a²/2 - a²/16

= 7a²/16

Step-by-step explanation:

Ann [662]1 year ago

3 0

This is a much more interesting problem than what we usually see here.

Let's just start by assuming a=1 so we have two unit squares.

PMC and PTC are two congruent triangles, each right triangles with one leg of length 1 and a shared hypotenuse PC.

We'll put it on the grid at P(0,0), T(1,0).

Angle TPM is 45 degrees and PM=1 so

M(cos 45, sin 45) = M(1/√2, 1/√2)

MK has slope -1, thru M, so is

y = -x + √2

C's on that line at x=1, so C's x coordinate is 1, so its y coordinate is √2 - 1

C(1, √2 - 1)

OK, that's enough to compute the area.

P(0,0), T(1,0), C(1, √2 - 1), M(1/√2, 1/√2)

area(PMCT) = area(PTC) + area(PCM)= 2 area(PTC)

The area of a right triangle is half the product of its legs.

area(PMCT) = 2 (1/2) (1) ( √2 - 1)

area(PMCT) = √2 - 1

When we have sides a, the area scales by a²,

Answer: a²(√2 - 1)

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Attached file

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