![\bold{\huge{\green{\underline{ Solutions }}}}](https://tex.z-dn.net/?f=%5Cbold%7B%5Chuge%7B%5Cgreen%7B%5Cunderline%7B%20Solutions%20%7D%7D%7D%7D)
<h3>
<u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>
<u>We </u><u>have</u><u>, </u>
![\sf{HM = 5 cm }](https://tex.z-dn.net/?f=%5Csf%7BHM%20%3D%205%20cm%20%7D)
- <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>
![\sf{ = 4 × side }](https://tex.z-dn.net/?f=%5Csf%7B%20%3D%204%20%C3%97%20side%20%7D)
![\sf{ = 4 × 5 }](https://tex.z-dn.net/?f=%5Csf%7B%20%3D%204%20%C3%97%205%20%7D)
![\sf{ = 20 cm }](https://tex.z-dn.net/?f=%5Csf%7B%20%3D%2020%20cm%20%7D)
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>
![\sf{MX = 3.5 cm }](https://tex.z-dn.net/?f=%5Csf%7BMX%20%20%3D%203.5%20cm%20%7D)
- <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>
![\sf{MX = MT/2}](https://tex.z-dn.net/?f=%5Csf%7BMX%20%20%3D%20MT%2F2%7D)
![\sf{MT = 2 * 3.5 }](https://tex.z-dn.net/?f=%5Csf%7BMT%20%3D%202%20%2A%203.5%20%7D)
![\sf{MT = 7 cm}](https://tex.z-dn.net/?f=%5Csf%7BMT%20%3D%207%20cm%7D)
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>
![\sf{\angle{MAT = 90° }}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7BMAT%20%20%3D%2090%C2%B0%20%7D%7D)
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>
![\sf{\angle{MHA = }}{\sf{\angle{ MHT/2}}}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7BMHA%20%20%3D%20%7D%7D%7B%5Csf%7B%5Cangle%7B%20MHT%2F2%7D%7D%7D)
![\sf{\angle{MHA = 90°/2}}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7BMHA%20%3D%2090%C2%B0%2F2%7D%7D)
![\sf{\angle {MHA = 45°}}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%20%7BMHA%20%3D%2045%C2%B0%7D%7D)
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>
![\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}](https://tex.z-dn.net/?f=%5Csf%7B35%C2%B0%20%2B%2090%C2%B0%20%2B%20%7D%7B%5Csf%7B%5Cangle%7B%20OLM%20%3D%20180%C2%B0%7D%7D%7D)
![\sf{\angle{OLM = 180° - 125°}}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7BOLM%20%3D%20180%C2%B0%20-%20125%C2%B0%7D%7D)
![\sf{\angle{ OLM = 55° }}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7B%20OLM%20%3D%2055%C2%B0%20%7D%7D)
<u>Now</u><u>, </u>
![\sf{\angle{OLM = }}{\sf{\angle{OLA}}}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7BOLM%20%3D%20%7D%7D%7B%5Csf%7B%5Cangle%7BOLA%7D%7D%7D)
- <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>
![\sf{\angle{ PLA = 55° }}](https://tex.z-dn.net/?f=%5Csf%7B%5Cangle%7B%20PLA%20%3D%2055%C2%B0%20%7D%7D)
Thus, Angle PLA is 55° .
Hence, Option C is correct