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3 years ago

Which statement proves that quadrilateral JKLM is a kite?

Mathematics

2 answers:

horsena [70]3 years ago

8 0

A kite is a flat shape with straight sides. It has two pairs of equal-length adjacent (next to each other) sides.

Main properties of an arbitrary kite:

Two pairs of sides are of equal length.
One pair of diagonally opposite angles is equal.
Only one diagonal is bisected by the other.
The diagonals cross at 90°.

1. Option 1 is false, because sides that are adjacent to the right angle could have different lengths or diagonals can cross not at 90°.

2. Option 2 is correct, because this option is strictly the definition of the kite.

3. Option 3 is false, because MK and LJ can be not perpendicular, and then adjacent sides will not have the same lengths.

4. Option 3 is false, because despite the perpendicularity between MK and LJ, this figure could be rhombus (with all equal sides) or square (with all equal sides and all right angles).

By the way, rhombus and square are partial cases of kite, but in general, an arbitrary kite is not rhombus and is not square.

Answer: correct choice is 2.

Akimi4 [234]3 years ago

4 0

Answer:

second answer choice it rightb

Step-by-step explanation:

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Marty is asked to draw triangles with side lengths of 4 units and 2 units, and a non-included angle of 30°. Select all the trian

777dan777 [17]

Answer:

The drawn in the attached figure

see the explanation

Step-by-step explanation:

<em>First case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\B=30^o$

Applying the law of sines

Find the measure of angle A

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(A)}=\frac{2}{sin(30^o)}$

$sin(A)=1$

$A=90^o$

Find the measure of angle C

In a right triangle

we know that

$B+C=90^o$ ----> by complementary angles

$B=30^o$

therefore

$C=60^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{c}{sin(60^o)}=\frac{2}{sin(30^o)}$

$c=2\sqrt{3}\ units$

therefore

The dimensions of the triangle are

$A=90^o$

$B=30^o$

$C=60^o$

$a=4\ units\\b=2\ units\\c=2\sqrt{3}=3.46\ units$

<em>Second case</em>

In the triangle ABC

Let

$a=4\ units\\b=2/ units\\A=30^o$

Applying the law of sines

Find the measure of angle B

$\frac{a}{sin(A)}=\frac{b}{sin(B)}$

substitute the given values

$\frac{4}{sin(30^o)}=\frac{2}{sin(B)}$

$sin(B)=0.25$

using a calculator

$B=14.48^o$

Find the measure of angle C

we know that

The sum of the interior angles in any triangle must be equal to 180 degrees

$A+B+C=180^o$

$A=30^o\\B=14.48^o$

therefore

$30^o+14.48^o+C=180^o$

$C=135.52^o$

Find the length side c

Applying the law of sines

$\frac{c}{sin(C)}=\frac{a}{sin(A)}$

substitute the given values

$\frac{c}{sin(135.52^o)}=\frac{4}{sin(30^o)}$

$c=5.61\ units$

therefore

The dimensions of the triangle are

$A=30^o$

$B=14.48^o$

$C=135.52^o$

$a=4\ units\\b=2\ units\\c=5.61\ units$

see the attached figure to better understand the problem

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Answer:

Step-by-step explanation:

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