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oksian1 [2.3K]
3 years ago
5

Raju walks from home to school at a distance of 3km 650m everyday what is the distance he walked in ten days ​

Mathematics
1 answer:
rjkz [21]3 years ago
5 0

Step-by-step explanation:

3km 650m = 3650m

3650 × 10 = 36500m = 36km 500m

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The expression 27.50+ 0.11x models the monthly cost in dollars of a phone plan. In the
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4 years ago
Read 2 more answers
Find the area of quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21)
sasho [114]

Answer:

The area of quadrilateral ABCD is 139 unit^2.

Step-by-step explanation:

Given:

Quadrilateral ABCD whose vertices are A(1,1) B(7,-3) C(12,2) D(7,21).

Now, to find the area.

The coordinates of the quadrilateral are A(1,1), B(7,-3), C(12,2), D(7,21).

So, to find the area we need to bisect the quadrilateral ABCD and get the triangles ABC and ADC and then calculate their areas:

In Δ ABC:

A(x_1,y_1)=(1,1)\:,\:B(x_2,y_2)=(7,-3)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ABC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-3-2)+(7)(2-1)+12(1--3)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(-5)+(7)(1)+12(4)\right|

On solving we get:

Area\,of\,triangle\,=25.

In Δ ADC:

A(x_1,y_1)=(1,1)\:,\:D(x_2,y_2)=(7,21)\:and\:C(x_3,y_3)=(12,2)

Now, to get the area of triangle ADC:

Area\,of\,triangle\,=\,\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

Area\,of\,triangle\,=\,\frac{1}{2}\left|1(21-2)+(7)(2-1)+12(1-21)\right|

On solving it by same process as above we get:

Area\,of\,triangle\,=114

Now, to get the area of the quadrilateral we add the areas of the triangles ABC and ADC:

25+114\\=25+114\\=139\ unit^2

Therefore, area of quadrilateral ABCD is 139 unit^2.

4 0
3 years ago
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