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3 years ago

How do you prove this?

Mathematics

1 answer:

frozen [14]3 years ago

3 0

Answer:

Show ΔBCD ≅ ΔGFE, so ∠C ≅ ∠F. Base angle of an isosceles triangle are congruent, so ΔACF is isosceles.

Step-by-step explanation:

Informally, subtract DE from CE and DF. This will show CD ≅ EF.

Then ΔBCD ≅ ΔGFE by the HL theorem for right triangles.

Corresponding parts of congruent triangles are congruent, namely the angles C and F.

Since base angles of ΔACF are congruent, it is isosceles.

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3 years ago

Pentagon TEXAS has vertices at T(1, 8), E(3, 8), X(6, 2), A(3, 2), and S(1, 4). What is the area of Pentagon TEXAS?

Tamiku [17]

The area of the pentagon is 19 square units

The coordinates of the pentagon are given as:

$T = (1, 8)$

$E =(3, 8)$

$X =(6, 2)$

$A=(3, 2)$

$S = (1, 4)$

The area of the pentagon is then calculated using:

$Area = \frac 12|(x_1y_2+x_2y_3+x_3y_4+x_4y_5+x_5y_1) -(y_1x_2+y_2x_3+y_3x_4+y_4x_5+y_5x_1)|$

So, we have:

$Area = \frac 12|(1 \times 8 +3 \times 2+6\times2+3 \times 4 + 1 \times 8) -(8\times3+8\times6+2\times3+2\times1+4\times1)|$

Simplify the expressions in both brackets

$Area = \frac 12|46 -84|$

$Area = \frac 12|-38|$

Remove absolute bracket

$Area = \frac 12 \times 38$

$Area = 19$

Hence, the area of the pentagon is 19 square units