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3 years ago

Find the area of the figure. Round to the nearest tenth if necessary.

Mathematics

1 answer:

rosijanka [135]3 years ago

5 0

Answer:

63 sq ft

Step-by-step explanation:

The formula for the area of a trapezoid is

A = h(a + b)/2 where a and b are the lengths of the parallel sides and h is the distance between them

A = 7(11 + 7)/2

= 7(18)/2

= 63 sq ft

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Answer:

B, you are correct

Area of the bigger trapezoid = (6 + 2) x 1/2 x 6 = 8 x 1/2 x 6 = 4 x 6 = 24

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3 years ago

The two-dimensional net of a rectangular prism is

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3 years ago

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According to the distributive property, 6 x (5 + 3) =

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3 years ago

Which best describes the triangle or the triangles, if any, that can be formed with two sides that measure 5 inches and an angle

Arte-miy333 [17]

One isosceles triangle

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3 years ago

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Given: KLMN is a trapezoid, m∠N=m∠KML, FD=8, LM KN = 3 5 F∈ KL , D∈ MN , ME ⊥ KN KF=FL, MD=DN, ME=3 5 Find: KM

denis23 [38]

Answer:

The length of side KM is $\sqrt{109}$ units.

Step-by-step explanation:

Given information: KLMN is a trapezoid, ∠N= ∠KML, FD=8, LM:KN=3:5, F∈ KL, D∈ MN , ME ⊥ KN KF=FL, MD=DN, $ME=3\sqrt{5}$ .

From the given information it is noticed that the point F and D are midpoints of KL and MN respectively. The height of the trapezoid is $3\sqrt{5}$ .

Midsegment is a line segment which connects the midpoints of not parallel sides. The length of midsegment of average of parallel lines.

Since LM:KN=3:5, therefore LM is 3x and KN is 5x.

$\frac{3x+5x}{2}=8$

$\frac{8x}{2}=8$

$x=2$

Therefore the length of LM is 6 and length of KN is 10.

Draw perpendicular on KN form L and M.

$KN=KA+AE+EN$

$10=6+2(EN)$ (KA=EN, isosceles trapezoid)

$EN=2$

$KE=KN-EN=10-2=8$

Therefore the length of KE is 8.

Use pythagoras theorem is triangle EKM.

$Hypotenuse^2=base^2+perpendicular^2$

$(KM)^2=(KE)^2+(ME)^2$

$(KM)^2=(8)^2+(3\sqrt{5})^2$

$KM^2=64+9(5)$

$KM=\sqrt{109}$

Therefore the length of side KM is $\sqrt{109}$ units.

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3 years ago

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