[tex]\bold{\huge{\green{\underline{ Solutions }}}}[/tex]
Answer 11 :-
We have,
[tex]\sf{HM = 5 cm }[/tex]
- In square all sides of squares are equal
The perimeter of square
[tex]\sf{ = 4 × side }[/tex]
[tex]\sf{ = 4 × 5 }[/tex]
[tex]\sf{ = 20 cm }[/tex]
Thus, The perimeter of square is 20 cm
Hence, Option C is correct .
Answer 12 :-
We have,
[tex]\sf{MX = 3.5 cm }[/tex]
- In square, diagonals are equal and bisect each other at 90°
Here,
[tex]\sf{MX = MT/2}[/tex]
[tex]\sf{MT = 2 * 3.5 }[/tex]
[tex]\sf{MT = 7 cm}[/tex]
Thus, The MT is 7cm long
Hence, Option C is correct .
Answer 13 :-
We have to find the measure of Angle MAT
- All angles of square are 90° each
From above
[tex]\sf{\angle{MAT = 90° }}[/tex]
Thus, Angle MAT is 90°
Hence, Option B is correct .
Answer 14 :-
We know that,
- All the angles of square are equal and 90° each
Therefore,
[tex]\sf{\angle{MHA = }}{\sf{\angle{ MHT/2}}}[/tex]
[tex]\sf{\angle{MHA = 90°/2}}[/tex]
[tex]\sf{\angle {MHA = 45°}}[/tex]
Thus, Angle MHA is 45°
Hence, Option A is correct
Answer 15 :-
Refer the above attachment for solution
Hence, Option A is correct
Answer 16 :-
Both a and b
- The median of isosceles trapezoid is parallel to the base
- The diagonals are congruent
Hence, Option C is correct
Answer 17 :-
In rhombus PALM,
- All sides and opposite angles are equal
Let O be the midpoint of Rhombus PALM
In ΔOLM, By using Angle sum property :-
[tex]\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}[/tex]
[tex]\sf{\angle{OLM = 180° - 125°}}[/tex]
[tex]\sf{\angle{ OLM = 55° }}[/tex]
Now,
[tex]\sf{\angle{OLM = }}{\sf{\angle{OLA}}}[/tex]
- OL is the bisector of diagonal AM
Therefore,
[tex]\sf{\angle{ PLA = 55° }}[/tex]
Thus, Angle PLA is 55° .
Hence, Option C is correct
