Please answer soon!

Mathematics

2 answers:

MA_775_DIABLO [31]3 years ago

7 0

Answer:

Answer is No.

Step-by-step explanation:

To construct a quadrilateral uniquely, five measurements are required. A quadrilateral can be constructed uniquely if the lengths of its four sides and a diagonal are given or if the lengths of its three sides and two diagonals are given.

Just given two angles we cannot construct a unique quadrilateral. There may be an infinite number of quadrilaterals having atleast two right angles

Examples:

All squares with varying sides

All trapezoids with two right angles

All rectangles with different dimensions

and so on.

Answer is

No.

pav-90 [236]3 years ago

3 0

The figure is NOT unique.
Imagine the following quadrilaterals:
Rectangle
Square
We know that:
Both quadrilaterals have at least two right angles.
However, they are not unique because they depend on the lengths of their sides.
Answer:
The figure described is not unique.

You might be interested in

What is the following product? root(3, 4) * sqrt(3)

gladu [14]

Answer:

It’s option C

Step-by-step explanation: just took the quiz on edge

3 0

3 years ago

indicate whether the measures 8, 14, and 16 can be the side lengths of a triangle. if they can, classify the triangle.

Aleksandr-060686 [28]

Answers:

A triangle is possible.

The triangle is acute and scalene

=============================================================

Explanation:

We use the triangle inequality theorem to see if a triangle is possible with these side lengths.

Consider the side lengths a, b, and c. A triangle is possible if and only if the following 3 conditions hold true

a+b > c
a+c > b
b+c > a

Basically pick any two sides. If the sum is larger than the third side, then a triangle is possible.

-----------------

In this case we have: a = 8, b = 14, c = 16. The order of the a,b,c values doesn't really matter.

Then:

a+b = 8+14 = 22 which is larger than c = 16. So a+b > c is true.
a+c = 8+16 = 24 is larger than b = 14. So a+c > b is true.
b+c = 14+16 = 30 is larger than a = 8. So b+c > a is true.

All three inequalities mentioned are true; therefore, a triangle is possible with these side lengths.

-----------------

We'll use those a,b,c values in the converse of the pythagorean theorem to determine what kind of triangle this is. So far we know it's scalene because all three side lengths are different. It would be isosceles if it had exactly two equal sides, and equilateral if all 3 sides were the same length.

$a^2+b^2 = c^2\\\\8^2+14^2 = 16^2\\\\260 = 256\\\\$