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Georgia [21]
3 years ago
7

Please help and show steps to answer

Advanced Placement (AP)
1 answer:
Rasek [7]3 years ago
6 0
Answer is a

*The correct answer is IJ = JK


•explanation:

Given:
AB is the perpendicular bisector of
IK. ⇒ AB divides the line segment IK in two equal parts i.e. IJ = JK and the angle formed at the point of intersection J is 90° ⇒ ∠AJI = 90°. In ΔAIJ, By angle sum property of a triangle ∠AJI + ∠AIJ + ∠IAJ = 180° ( But ∠AJI = 90° ) ∠AIJ + ∠IAJ = 90° ⇒ ∠IAJ < 90° So, ∠IAJ is not a right angle. Its not given IK is a perpendicular bisector so AJ = BJ need NOT be true. As A does not lie on the line IK so A can not be the mid point of IK.

•Hence, we conclude the correct statement is IJ = JK

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\bold{\huge{\pink{\underline{ Solution }}}}

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

  • <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>

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

  • <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>

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

<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

\bold{=}{\bold{\pink{\dfrac{1}{2}}}}{\bold{\pink{ × Base × Height }}}

<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>

\sf{=}{\sf{\dfrac{1}{2}}}{\sf{ × 7 × 6 }}

\sf{ = 7 × 3 }

\sf{ = 21 cm²}

<h3><u>Therefore</u><u>, </u></h3>

<u>Area </u><u>of </u><u>2</u><u> </u><u>triangle's </u>

\sf{ = 2 × ( Area\:of\:1\:traingle) }

\sf{ = 2 × 21 }

\sf{ = 42 cm² }

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

\bold{\red{ = Length × Breath }}

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

Area of 1 rectangle

\sf{ = 13.1 × 7 }

\sf{ = 91.7 cm² }

<h3><u>Therefore</u><u>,</u></h3>

<u>Area </u><u>of </u><u>3</u><u> </u><u>rectangles </u>

\sf{ = 3 ×( Area\:of\:1\:rectangle)  }

\sf{ = 3 × 91.7 }

\sf{ = 275.1 cm² }

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

\sf{ = 42 + 275.1 }

\bold{ = 317.1 cm²}

Hence, The total surface area of the given figure is 317.1 cm² .

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