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zzz [600]
3 years ago
9

Quadrilateral ABCD has vertices A(-3, 4), B(1, 3), C(3, 6), and D(1, 6). Match each set of vertices of quadrilateral EFGH with t

he transformation that shows it is congruent to ABCD.

Mathematics
1 answer:
Sati [7]3 years ago
8 0
Each transformation will give these following coordinates

Translation 7 units right gives E(4, 4), F(8, 3), G(10, 6), and H(8, 6) as shown in brown in the diagram below

Reflection on the y-axis gives E(3, 4), F(-1, 3), G(-3, 6) and H(-1, 6) as shown in green in the diagram below

Reflection on the x-axis gives E(-3, -4), F(1, -3), G(3, -6) and H(1, -6) as shown in red in the diagram below

Translation 5 units down gives E(-3, -1), F(1, -2, G(3, 1) and H(1, 1) as shown in purple in the diagram below 

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{\bold{\red{\huge{\mathbb{QUESTION}}}}}

The semicircle shown at left has center X and diameter W Z. The radius XY of the semicircle has length 2. The chord Y Z has length 2. What is the area of the shaded sector formed by obtuse angle WXY?

\bold{ \red{\star{\blue{GIVEN }}}}

RADIUS = 2

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RADIUS --> XY , XZ , WX

( BEZ THEY TOUCH CIRCUMFERENCE OF THE CIRCLES AFTER STARTING FROM CENTRE OF THE CIRCLE)

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THE AREA OF THE SHADED SECTOR FORMED BY OBTUSE ANGLE WXY.

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AREA COVERED BY THE ANGLE IN A SEMI SPHERE

AREA = ANGLE   \: \: IN  \: \:  RADIAN  \times RADIUS

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Total Area Of The Semi Sphere:-

AREA =   \pi \times radius  \\  \\ AREA = \pi \times 2 = 2\pi

Area Under Unshaded Part .

Given a triangle with each side 2 units.

This proves that it's is a equilateral triangle which means it's all angles r of 60° or π/3 Radian

So AREA :-

AREA =  \frac{\pi}{3}  \times radius \\  \\ AREA =  \frac{\pi}{3}  \times 2 \\  \\ AREA =  \frac{2\pi}{3}

\green{Now:- } \\  \green{ \: Area  \: Under \:  Unshaded \:  Part }

Total Area - Area Under Unshaded Part

Area= 2\pi -  \frac{2\pi}{3}  \\ Area =  \frac{6\pi - 2\pi}{3}   \\ Area =  \frac{4\pi}{3}  \:  \: ans

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It would help the chain run smoother and work better and faster.
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