All categories

3 years ago

To prove part of the triangle midsegment theorem using

Mathematics

2 answers:

timama [110]3 years ago

4 0

Answer:

Step-by-step explanation:

bonufazy [111]3 years ago

3 0

Answer:the length of GH is half the length of KL

Step-by-step explanation: Just took it on edg and was correct.

You might be interested in

What is the value for x? Enter your answer in the box. x = ____

vlada-n [284]

180 - 73-73 =34

6x+4 = 34

6x = 30

x = 30/6

x = 5

7 0

3 years ago

Look at question attatched

Ratling [72]

Answer:

If the discs were measured to the nearest milimetre, it might be possible, that they are actually 1.64cm wide instead of only 1.64cm as shown.

Therefore the box would fill up much quicker and there will not be enough space.

Step-by-step explanation:

1.6cm * 16 discs = 25.6cm (would fit)

1.64cm * 16 discs = 26.24cm (would not fit)

5 0

3 years ago

Read 2 more answers

HELP asap plz Solve 2y^2 – 3y = 2

sergejj [24]

Are you solving for x or y?

4 0

4 years ago

Find the domain and range of the graph

olga_2 [115]

Domain: [1 ; 6]
Range: [1 ; 7]

6 0

3 years ago

1. Quadrilateral ABCD has vertices A(-1, 1), B(2, 3), C(6, 0) and D(3, -2). Determine using coordinate geometry whether or not t

deff fn [24]

Answer:

The Conclusion is

Diagonals AC and BD,

a. Bisect each other

b. Not Congruent

c. Not Perpendicular

Step-by-step explanation:

Given:

[]ABCD is Quadrilateral having Vertices as

A(-1, 1),

B(2, 3),

C(6, 0) and

D(3, -2).

So the Diagonal are AC and BD

To Check

The diagonals AC and BD

a. Bisect each other. B. Are congruent. C. Are perpendicular.

Solution:

For a. Bisect each other

We will use Mid Point Formula,

If The mid point of diagonals AC and BD are Same Then

Diagonal, Bisect each other,

For mid point of AC

$Mid\ point(AC)=(\dfrac{x_{1}+x_{2} }{2},\dfrac{y_{1}+y_{2} }{2})$

Substituting the coordinates of A and C we get

$Mid\ point(AC)=(\dfrac{-1+6}{2},\dfrac{1+0}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Similarly, For mid point of BD

Substituting the coordinates of B and D we get

$Mid\ point(BD)=(\dfrac{2+3}{2},\dfrac{3-2}{2})=(\dfrac{5}{2},\dfrac{1}{2})$

Therefore The Mid point of diagonals AC and BD are Same

Hence Diagonals,

a. Bisect each other

B. Are congruent

For Diagonals to be Congruent We use Distance Formula

For Diagonal AC

$l(AC) = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} )}$

Substituting A and C we get

$l(AC) = \sqrt{((6-(-1))^{2}+(0-1)^{2} )}=\sqrt{(49+1)}=\sqrt{50}$

Similarly ,For Diagonal BD

Substituting Band D we get

$l(BD) = \sqrt{((3-2))^{2}+(-2-3)^{2} )}=\sqrt{(1+25)}=\sqrt{26}$

Therefore Diagonals Not Congruent

For C. Are perpendicular.

For Diagonals to be perpendicular we need to have the Product of slopes must be - 1

For Slope we have

$Slope(AC)=\dfrac{y_{2}-y_{1} }{x_{2}-x_{1} }$

Substituting A and C we get

$Slope(AC)=\dfrac{0-1}{6--1}\\\\Slope(AC)=\dfrac{-1}{7}$

Similarly, for BD we have

$Slope(BD)=\dfrac{-2-3}{3-2}\\\\Slope(BD)=\dfrac{-5}{1}$

The Product of slope is not -1

Hence Diagonals are Not Perpendicular.

6 0

4 years ago