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3 years ago

9

How do i solve this problem?

1 answer:

Pani-rosa [81]3 years ago

6 0

Area of a triangle = 1/2 x base x height

using the given information: area = 40 and base = 20

you now have 40 = 1/2 x 20 x height

divide both sides by 1/2

80 = 20 x height

divide both sides by 20

height = 80 / 20

height = 4 in.

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A quadrilateral has exactly one right angle and two nonadjacent angles that each measure 105 degrees

LiRa [457]

Answer:

C) Kite

Step-by-step explanation:

The problem states that the polygon is a quadrilateral, meaning it has 4 sides. This rules out D, as pentagons have 5 sides. Our next clue is that this shape has <em>only one </em>right angle, ruling out A and B, because, by definition, rectangles and squares must have four 90 degree angles. We can already see that we are left with only one choice, but to confirm: a kite has 2 opposite angles of 105 degrees, like the problem states.

Therefore, the answer is C) Kite

Hope this helps!

8 0

4 years ago

HELP PLEASE!!!!!!!!!! plz

scoundrel [369]

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22 4 (22,4)
33 6 (33,6)

7 0

3 years ago

Read 2 more answers

Round 48545.5607849 to the nearest ten.

VashaNatasha [74]

The answer is 48.550.0

8 0

3 years ago

25 pints for this I need it it would mean aaaaalooottt ty

pashok25 [27]

Im pretty sure duncans friend is wrong but i may be as well lol sorry hope this helps

8 0

3 years ago

1. Trapezoid KLMN has vertices K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2) .

chubhunter [2.5K]

Answer:

Step-by-step explanation:

(A) The vertices of the trapezoid KLMN are K(1, 3) , L(3, 1) , M(3, 0) , and N(1, −2)

Now, KL= $\sqrt{(3-1)^{2}+(1-3)^{2}}$ =\sqrt{8}[/tex]

LM= $\sqrt{(3-3)^{2}+(0-1)^{2}}=1$

MN= $\sqrt{(1-3)^{2}+(-2-0)^{2}}=\sqrt{8}$

NK= $\sqrt{(1-1)^{2}+(-2-3)^{2}}=5$

Now, as KL and MN are equal, therefore, KLMN is an isosceles trapezoid.

(B) Since m∠XWY=47°, therefore ∠XWZ=47+47=94° and ∠ZYW=18°, therefore ∠XYZ=36°( as diagonals bisect the angles)

In ΔXWO,

∠XWO+∠WXO+∠WOX=180°(Angle sum property)

∠WXO=43°

Also, from ΔXOY,

∠OXY=72° using the angle sum property.

Therefore, ∠WXY=43+72=115°

Now, sum of all the angles of a quadrilateral is equal to 360°, therefore

∠WXY+∠XYZ+∠YZW+∠ZWX=360°

115°+36°+∠YZW+94°=360°

∠YZW=115°

Therefore,∠WZY=115°

(C) Since, AB and CD are the two parallel lines as the slope of both the sides are equal.

LetM be the mid point of AB, therefore M= $(\frac{-2+4}{2}, \frac{4+3}{2})$

= $(1,\frac{7}{2})$

Also, let N be the mid point of DC, therefore,

N= $(\frac{4-2}{2},\frac{-2-5}{2})$

= $(1,\frac{-7}{2})$

Now, length of the mid segment MN= $\sqrt{(1-1)^{2}+(\frac{7}{2}+\frac{7}{2})^{2}}=\sqrt{0+49}=7 cm$

(D)Given: Kite PQRS, TS=6cm and TP=8cm

Then, From triangle TSP, we have

$(SP)^{2}=(TS)^{2}+(TP)^{2}$

$(SP)^{2}=36+64$

$SP=10cm$

8 0

3 years ago