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e-lub [12.9K]
3 years ago
15

What is the area of this irregular figure?

Mathematics
1 answer:
Ahat [919]3 years ago
4 0
The answer is 180mm^2. (B)

First, we can decompose the shape into a rectangle and a parallelogram. Let’s start with the rectangle.

Formula for a rectangle: bh

The base is 6 while the height is 12. You multiply like so. 6*12=72.

The area of the rectangle is 72.

The formula of a parallelogram is the same as the rectangle. (Bh)

To identify the base in this parallelogram, you would subtract the whole base length, 15 by half of the base length, 6.

15 - 6 = 9.

So the other half base length is 9.

You can simply transform a parallelogram into a rectangle by decomposing the parallelogram into two right triangles and a rectangle. You would take the right triangle at the bottom of the figure and move it to the top of the parallelogram in the photo. Thus, becoming a rectangle.

Now, we can identify the base and the height. The base of the parallelogram is 9 while the height is 12.

You would then multiply 9 by 12 like so to get 108.

After finding the area of the rectangle and parallelogram, you would add them up.

72 + 108 = 180mm^2

Therefore, making the answer choice B.
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Calculate the length of sides triangle pqr and determine weather or not triangle is a right angled. P(-4,6) q(6,1) r(2,9)
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\bold{\huge{\underline{ Solution }}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

  • We have given the coordinates of the triangle PQR that is P(-4,6) , Q(6,1) and R(2,9)

<h3><u>To</u><u> </u><u>Find </u><u>:</u><u>-</u></h3>

  • <u>We </u><u>have </u><u>to </u><u>calculate </u><u>the </u><u>length </u><u>of </u><u>the </u><u>sides </u><u>of </u><u>given </u><u>triangle </u><u>and </u><u>also </u><u>we </u><u>have </u><u>to </u><u>determine </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not </u>

<h3><u>Let's </u><u>Begin </u><u>:</u><u>-</u></h3>

<u>Here</u><u>, </u><u> </u><u>we </u><u>have </u>

  • Coordinates of P =( x1 = -4 , y1 = 6)
  • Coordinates of Q = ( x2 = 6 , y2 = 1 )
  • Coordinates of R = ( x3 = 2 , y3 = 9 )

<u>By </u><u>using </u><u>distance </u><u>formula </u>

\pink{\bigstar}\boxed{\sf{Distance=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2\;}}}

<u>Subsitute </u><u>the </u><u>required </u><u>values </u><u>in </u><u>the </u><u>above </u><u>formula </u><u>:</u><u>-</u>

Length of side PQ

\sf{ = }{\sf\sqrt{ (6 - (-4))^{2} + (1 - 6)^{2}}}

\sf{ = }{\sf\sqrt{ (6 + 4 )^{2} + (- 5)^{2}}}

\sf{ = }{\sf\sqrt{ (10)^{2} + (- 5)^{2}}}

\sf{ = }{\sf\sqrt{ 100 + 25 }}

\sf{ = }{\sf\sqrt{ 125 }}

\sf{ = 5 }{\sf\sqrt{ 5 }}

Length of QR

\sf{ = }{\sf\sqrt{(2 - 6)^{2} + (9 - 1)^{2}}}

\sf{ = }{\sf\sqrt{(- 4 )^{2} + (8)^{2}}}

\sf{ = }{\sf\sqrt{16 + 64 }}

\sf{ = }{\sf\sqrt{80 }}

\sf{ = 4 }{\sf\sqrt{5 }}

Length of RP

\sf{ = }{\sf\sqrt{ (-4 - 2 )^{2} + (6 - 9)^{2}}}

\sf{ = }{\sf\sqrt{ (-6 )^{2} + (-3)^{2}}}

\sf{ = }{\sf\sqrt{ 36 + 9 }}

\sf{ = }{\sf\sqrt{ 45 }}

\sf{ = 3}{\sf\sqrt{ 5 }}

<h3><u>Now</u><u>, </u></h3>

We have to determine whether the triangle PQR is right angled triangle

<h3>Therefore, </h3>

<u>By </u><u>using </u><u>Pythagoras </u><u>theorem </u><u>:</u><u>-</u>

  • Pythagoras theorem states that the sum of squares of two sides that is sum of squares of 2 smaller sides of triangle is equal to the square of hypotenuse that is square of longest side of triangle

<u>That </u><u>is</u><u>, </u>

\bold{ PQ^{2} + QR^{2} = PR^{2}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>,</u>

\bold{  125 + 80 = 45 }

\bold{  205  = 45 }

<u>From </u><u>above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

  • The triangle PQR is not a right angled triangle because 205 ≠ 45 .
6 0
3 years ago
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