Points E, F, and D are located on circle C. Circle C is shown. Line segments E C and D F are radii. Lines are drawn from points
E and D to point F to form chords E F and D F. Arc E D is 68 degrees. The measure of arc ED is 68°. What is the measure of angle EFD? 34° 68° 112° 132°
2 answers:
Answer:
Option (1). 34°
Step-by-step explanation:
From the figure attached, CE and CD are the radii of the circle C.
Central angle CED formed by the intercepted arc DE = 68°
Since measure of an arc = central angle formed by the intercepted arc
Therefore, m∠CED = 68°
Since m∠EFD = [Central angle of an intercepted arc measure the double of the inscribed angle by the same arc]
Therefore, m∠EFD =
= 34°
Therefore, Option (1) 34° will be the answer.
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<u>We </u><u>have</u><u>, </u>
- <u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>
<u>The </u><u>perimeter </u><u>of </u><u>square </u>
Thus, The perimeter of square is 20 cm
Hence, Option C is correct .
<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3><u>We </u><u>have</u><u>, </u>
- <u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>
<u>Here</u><u>, </u>
Thus, The MT is 7cm long
Hence, Option C is correct .
<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3><u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>
- <u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>
<u>From </u><u>above </u>
Thus, Angle MAT is 90°
Hence, Option B is correct .
<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3><u>We </u><u>know </u><u>that</u><u>, </u>
- <u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>
<u>Therefore</u><u>, </u>
Thus, Angle MHA is 45°
Hence, Option A is correct
<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>Refer the above attachment for solution
Hence, Option A is correct
<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>Both a and b
- <u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
- <u>The </u><u>diagonals </u><u>are </u><u>congruent </u>
Hence, Option C is correct
<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>In rhombus PALM,
- <u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>
Let O be the midpoint of Rhombus PALM
<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>
<u>Now</u><u>, </u>
- <u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>
<u>Therefore</u><u>, </u>
Thus, Angle PLA is 55° .
Hence, Option C is correct
