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USPshnik [31]
2 years ago
3

the length of two parallel sides of a trapezium are 91 cm and 51 cm and the length of two other side 37 CM and 13 cm respectivel

y determine the area of trapezium​

Mathematics
1 answer:
vlabodo [156]2 years ago
7 0

Answer:

903 sq. cm.

Step-by-step explanation:

It is given that the length of two parallel sides of a trapezium are 91 cm and 51 cm and the length of two other side 37 CM and 13 cm respectively.

Now draw a dotted line as shown in the below figure. So, we have a triangle with sides 13 cm, 37 cm, 40cm and a parallelogram with sides 51 cm, 13 cm.

Area of parallelogram is

A_1=base\times height

A_1=51\times 13

A_1=663

Using heron's formula, area of triangle is

A=\sqrt{s(s-a)(s-b)(s-c)}

where, s=\dfrac{a+b+c}{2}.

The sides of triangle are 13cm, 37 cm, and 40 cm.

s=\dfrac{13+37+40}{2}=45

Area of triangle is

A=\sqrt{45(45-13)(45-37)(45-40)}=240

The area of trapezium is

A=A_1+A_2=663+240=903

Therefore, the area of trapezium is 903 sq. cm.

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To solve this problem, you simply have to do base 1 * height 1 = base 2 * height 2. Plugging our numbers in gives us 24 * 4 = 5 * height 2. Solving it out, we get 19.2 as our missing height. 
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A rectangular football field is 64 meters wide and 100 meters long. A player runs from one corner of the field in a diagonal lin
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5 0
2 years ago
Martin's car had 86,456 miles on it. Of that distance, Martin's wife drove 24,901 miles,
navik [9.2K]

Answer:

Martin drove 53,558 miles.

Step-by-step explanation:

If you add up the amount of miles his son and wife drove, you get 32,898. To get the answer, you would subtract that from the total number of miles to get 53,558.

3 0
2 years ago
Find the distance from (4, −7, 6) to each of the following.
LenKa [72]

Answer:

(a) 6 units

(b) 4 units

(c) 7 units

(d) 9.22 units

(e) 7.21 units

(f) 8.06 units

Step-by-step explanation:

The distance d from one point (x₁, y₁, z₁) to another point (x₂, y₂, z₂) is given by;

d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

Now from the question;

<em>(a) The distance from (4, -7, 6) to the xy-plane</em>

The xy-plane is the point where z is 0. i.e

xy-plane = (4, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, -7, 0)</em>

d = √[(4 - 4)² + (-7 - (-7))² + (0 - 6)²]

d = √[(0)² + (0)² + (-6)²]

d = √(-6)²

d = √36

d = 6

Hence, the distance to the xy plane is 6 units

<em>(b) The distance from (4, -7, 6) to the yz-plane</em>

The yz-plane is the point where x is 0. i.e

yz-plane = (0, -7, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 6)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 6)²]

d = √[(4)² + (0)² + (0)²]

d = √(4)²

d = √16

d = 4

Hence, the distance to the yz plane is 4 units

<em>(c) The distance from (4, -7, 6) to the xz-plane</em>

The xz-plane is the point where y is 0. i.e

xz-plane = (4, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 6)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 6)²]

d = √[(0)² + (-7)² + (0)²]

d = √[(-7)²]

d = √49

d = 7

Hence, the distance to the xz plane is 7 units

<em>(d) The distance from (4, -7, 6) to the x axis</em>

The x axis is the point where y and z are 0. i.e

x-axis = (4, 0, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(4, 0, 0)</em>

d = √[(4 - 4)² + (-7 - 0)² + (6 - 0)²]

d = √[(0)² + (-7)² + (6)²]

d = √[(-7)² + (6)²]

d = √[(49 + 36)]

d = √(85)

d = 9.22

Hence, the distance to the x axis is 9.22 units

<em>(e) The distance from (4, -7, 6) to the y axis</em>

The x axis is the point where x and z are 0. i.e

y-axis = (0, -7, 0).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, -7, 0)</em>

d = √[(4 - 0)² + (-7 - (-7))² + (6 - 0)²]

d = √[(4)² + (0)² + (6)²]

d = √[(4)² + (6)²]

d = √[(16 + 36)]

d = √(52)

d = 7.22

Hence, the distance to the y axis is 7.21 units

<em>(f) The distance from (4, -7, 6) to the z axis</em>

The z axis is the point where x and y are 0. i.e

z-axis = (0, 0, 6).

Therefore, the distance d is from <em>(4, -7, 6) </em> to <em>(0, 0 6)</em>

d = √[(4 - 0)² + (-7 - (0))² + (6 - 6)²]

d = √[(4)² + (-7)² + (0)²]

d = √[(4)² + (-7)²]

d = √[(16 + 49)]

d = √(65)

d = 8.06

Hence, the distance to the z axis is 8.06 units

5 0
2 years ago
Victor is enlarging a poster for a school baseball match. The graph below shows the size y of the poster after x enlargements: g
Norma-Jean [14]
For this case we have the following function:
 y = 1.8 ^ x
 The intersection with the y axis occurs when x = 0.
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 In this case, the intersection with the y-axis represents the original size of the poster.
 Answer:
 
the y-intercept of the graph represents:
 
the original size of the poster.
3 0
2 years ago
Read 2 more answers
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