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2 years ago

A square is a quadrilateral always sometimes never

Mathematics

2 answers:

lbvjy [14]2 years ago

6 0

A quadrilateral is a shape with 4 sides, regardless of their lengths, angles, or symmetry.

Therefore, a square is always a quadrilateral.

Daniel [21]2 years ago

4 0

Answer with explanation:

→A quadrilateral is a four sided convex polygon . The sum of interior angles of any Quadrilateral is 360 degree.

→As,Square is also a four sided polygon,having measure of each interior angle is 90 degree, so sum of all the interior angles gives 360 degree.

So,→Square satisfies all the properties of Quadrilateral.

So, Yes a square is always a quadrilateral .

Always

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Given: r is inversely proportional to s. If r = 16 when s = 3, then write the formula for the relation between r and s.

riadik2000 [5.3K]

Answer:

Step-by-step explanation:

Given that r is inversely proportional to s then the equation relating them is

r = $\frac{k}{s}$ ← k is the constant of proportion

To find k use the condition r = 16 when s = 3

k = rs = 16 × 3 = 48

r = $\frac{48}{s}$ ← equation of variation → B

3 0

2 years ago

In the diagram below, we have ST parallel to QR. angle P= 40 degrees, and angle Q= 35 degrees. Find the measure of angles STR in

Drupady [299]

Answer:

$m\angle S = 35^\circ, \ m\angle T = 105^\circ,\ m\angle R = 105^\circ$

Step-by-step explanation:

<u>Similar Triangles</u>

Lines ST and QR are parallel. Thus, angles S and Q are congruent, and angles T and R are congruent.

Considering the triangle PQR, the sum of its internal angles must be 180°:

$m\angle P + m\angle Q + m\angle R = 180^\circ$

Substituting the known values:

$40^\circ + 35^\circ + m\angle R = 180^\circ$

Solving for R:

$m\angle R = 180^\circ - 40^\circ - 35^\circ$

$m\angle R = 105^\circ$

Angles S and Q are congruent, thus

$m\angle S = 35^\circ$

Angles T and R are congruent, thus

$m\angle T = 105^\circ$

Summarizing:

$\mathbf{m\angle S = 35^\circ, \ m\angle T = 105^\circ,\ m\angle R = 105^\circ}$

7 0

2 years ago

Read 2 more answers

Umnica [9.8K]

Answer:

320 ....................

4 0

1 year ago

I need help please :,(

Ede4ka [16]

I think it’s the first one

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2 years ago

How do I solve for the missing lengths?

Advocard [28]

Answer:

Part 4) $ER=3\ units$

Part 5) $DF=9\sqrt{10}\ units$

Part 6) $DE=30\ units$

Step-by-step explanation:

Part 4) Find ER

we know that

In the right triangle ERF

Applying the Pythagorean Theorem

$EF^2=ER^2+RF^2$

substitute the given values

$(3\sqrt{10})^2=ER^2+9^2$

solve for ER

$ER^2=(3\sqrt{10})^2-9^2$

$ER^2=90-81\\ER^2=9\\ER=3\ units$

Part 5) Find DF

we know that

In the right triangle DRF

Applying the Pythagorean Theorem

$DF^2=DR^2+RF^2$

substitute the given values

$DF^2=27^2+9^2$

$DF^2=810\\DF=\sqrt{810}\ units$

simplify

$DF=9\sqrt{10}\ units$

Part 6) Find DE

we know that

$DE=DR+RE$ ----> by segment addition postulate

we have

$DR=27\ units\\RE=ER=3\ units$

substitute

$DE=27+3=30\ units$

3 0

2 years ago