Each lunar cycle has one full moon, in which the relative positions of Earth, the sun, and the moon form a straight line. Which
1 answer:
You might be interested in
- <u>We </u><u>have </u><u>given </u><u>one </u><u>irregular </u><u>figure </u><u>which </u><u>is </u><u>composed </u><u>2</u><u> </u><u>triangle's </u><u>and </u><u>3</u><u> </u><u>rectangles </u>
- <u>The </u><u>height </u><u>and </u><u>base </u><u>of </u><u>the </u><u>triangles</u><u> </u><u>is </u><u>6cm </u><u>and </u><u>7</u><u> </u><u>cm</u><u> </u><u>each </u>
- <u>The </u><u>length </u><u>and </u><u>breath </u><u>of </u><u>the </u><u>rectangles</u><u> </u><u>is </u><u>1</u><u>3</u><u>.</u><u>1</u><u> </u><u>cm </u><u>and </u><u>7cm </u><u>each </u>
- <u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>total </u><u>surface </u><u>area </u><u>of </u><u>the </u><u>given </u><u>figure</u><u>? </u>
<u>We </u><u>have</u><u>, </u>
- 2 triangles of height 6 cm and base 7cm each
- 3 rectangles of length 13.1 cm and 7cm each
<u>Therefore</u><u> </u><u>,</u>
<u>We </u><u>know </u><u>that</u><u>, </u>
Area of triangle
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
<u>Area </u><u>of </u><u>1</u><u> </u><u>triangle </u>
<u>Area </u><u>of </u><u>2</u><u> </u><u>triangle's </u>
Thus, The area of 2 triangles is 42 cm²
<h3><u>Now</u><u>, </u></h3><u>We </u><u>know </u><u>that</u><u>, </u>
Area of rectangle
<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>
Area of 1 rectangle
<u>Area </u><u>of </u><u>3</u><u> </u><u>rectangles </u>
Thus, The area of 3 rectangle is 275.1 cm²
<h3><u>So</u><u>, </u></h3>Total area of the given figure
- = Area of 2 triangles + Area of 3 rectangles
Hence, The total surface area of the given figure is 317.1 cm² .
