practice problems pending

Q1. The two sides of a triangle are 2 cm and 7 cm. Find the number of possible lengths of the third side. (The length of three sides of the triangle is an integer value)

Solution:
a + 2 > 7
a + 7 > 2
7 + 2 > a
=>
a < 9
a > -5
a > 5
=> 5 < a < 9 => a = 6,7,8
Answer: 3

Q2.

Only using triangle inequality
The three sides of a triangle are 9 cm, 7 cm and 12 cm. Which of the following can be a median of the triangle?

12,14,15,16

Solution:
If the median is falling upon side length 12,then the triangles are:
9,6,x(median length)
7,6,x
Only value of x satisfying both triangle inequalities here is: 12(from the given options).

Similarly,
If the median is falling upon side length 9,then the triangles are:
12,4.5,y(median length)
7,4.5,y
From the given options, none satisfies both.

Similarly,
If the median is falling upon side length 7,then the triangles are:
12,3.5,z(median length)
9,3.5,z
From the given options, 12 satisfies both.
So answer is 12.

Though it's not correct in realty.
Using the Apollonius's Theorem:
median length is 1/2.sqrt(2.a^2 + 2.b^2 - c^2)

which will not come out as 12 for any case.

Q3. In triangle XYZ the median XM is longer than half of YZ. Angle YXZ will be-

Solution:

XM falls on YZ.
Let angle YXM = a and angle MXZ = b
=> angle YXZ = a + b
In triangle MXZ XM > MZ (as given) => angle MZX > angle MXZ = b
In triangle MXY XM > MY (as given) => angle MYX > angle MXY = a
=> angle MZX(=YZX) + angle MYX(=ZYX) > a + b = YXZ
=> In triangle XYZ
angle XZY + angle ZYX > angle YXZ
let's write that as:
z + y > x
But
x + y + z = 180
=> 180 - x > x => 180 > 2x => 90 > x => x is acute = Answer

Q4. In convex quadrilateral MNOP, you are given that MN + NP <  MO + OP
then which of the following is true:
(a) MN > MO
(b) MN < MO
(c) MN = MO
(d) none

Solution:
One way to solve it is to think of an isosceles trapezium MNOP where MN || OP and MN < OP.
Also MP = NO.
In this construction, it's clear that the given condition is satisfied which is MN + NP <  MO + OP.
And visually we can see that the smaller side (MN) is less than a diagonal which is MO.
So most likely the answer is MN < MO.

We can prove it for a general quadrilateral as well.
Let the diagonals intersect at X.
Using triangular inequality:
MX + NX > MN
OX + PX > OP
Add:
MN + OP < MX + NX + OX + PX = (MX + OX) + (NX + PX) = MO + NP
MN + OP < MO + NP
But given that:
MN + NP < MO + OP
Add these 2 to get:
2MN + OP + NP < 2MO + OP + NP
=> 
2MN < 2MO
=> MN < MO = Answer