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3 years ago

in quadrilateral ABCD, AE=x+10, and BE=3x-18. for what value of x is ABCD a rectangle? a.24 b.14 c.18 d.16

Mathematics

1 answer:

Evgen [1.6K]3 years ago

4 0

Answer:

for x = 14, the given ABCD is a RECTANGLE.

Step-by-step explanation:

Given : ABCD is a quadrilateral.

E is the intersection points of the diagonal AC and BD.

AE = x+10, and BE= 3x -18

Now, as we know:

IN A RECTANGLES , DIAGONALS BISECT EACH OTHER.

⇒ E is the mid point of AC and BD respectively.

⇒ AE = BE

⇒ x+10 = 3 x - 18

or, 10 + 18 = 3 x - x

or, 2 x = 28

or, x = 28/2 = 14 ⇒ x = 14

Hence, for x = 14, the given ABCD is a RECTANGLE.

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2 years ago

Read 2 more answers

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Minchanka [31]

Step-by-step explanation:

<h3>Given,</h3>

GRAM is a parallelogram
m<GMA = 66°
m<GRP = 32°
m<PAM = 41°

<h3>Solutions:</h3>

i) Since,

Angles on the same side of a transversal in a parallelogram, there sum is 180° so,

<RGM + <GMA = 180°

=> <RGM = 180° - <GMA

=> <RGM = 180° - 66°

=> <RGM = 114° (Ans)

ii) Since,

Opposite angles of a diagonal as a transversal (here RM) are equal so,

<GRP = <PMA = 32°

As we have <GMA = 66° so,

=> <GMP = <GMA - <PMA

= 66° - 32°

=> <GMP = 34° (Ans)

iii) Since,

Opposite angles of a parallelogram are equal so,

<RGM = <RAM = 114°

As we have <PAM = 41° so,

=> <GAR = <RAM - <PAM

= 114° - 41°

=> <GAR = 73° (Ans)

iv) Since,

We got <PMA = 32° and <PAM = 41°, so by angle sum property of a triangle PMA,

=> <MPA = 180° - 32° - 41°

=> <MPA = 107° (Ans)

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