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1.

Find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm.

Answer» We can create a rectangle ABCD with the given details.
One side of rectangle `AB = 40` cm
Diagonal of rectangle `AC = 41` cm
So, other side of rectangle, `BC = sqrt(41^2-40^2)cm`
`BC = sqrt(1681-1600) = sqrt81 = 9cm`
So, Perimeter of rectangle ABCD, `P = 2**(AB+BC)`
`P = 2**(40+9) = 2**49 = 98cm`
Discussion
2.

`ABCD` is quadrilateral. Is `AB+BC+CD+DA < 2(AC+BD) ?`

Answer» `In/_AOB`
OA+OB>AB-(1)
`In/_AOD`
OA+OD>AD-(2)
`In/_BOC`
OB+OC>BC-(3)
`In/_COD`
OD+OC>CD-(4)
From Equation 1,2,3and 4
2(AC)+2(BD)>AB+BC+CD+DA
2(AC+BD)>AB+BC+CD+DA.
Discussion
3.

`ABCD` is a quadrilateral. Is `AB+BC+CD+DA > AC+BD ?`

Answer» `In/_ABC And /_ACD`
AB+BC>AC
CD+DA>AC
From these equations
AB+BC+CD+DA>2AC-(1)
Similarly,
AB+BC+CD+DA>2BD-(2)
From equation 1 and 2
2(AB+BC+CD+DA)>2(AC+BD)
AB+BC+CD+DA>AC+BD.
Discussion
4.

The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Answer» We know, sum of 2 sides of the triangles is always greater than 3rd side.
third side<12+15
third side<27cm
12+third side>15
third side>3
3cm
Discussion
5.

The lengths of two sides of a triangle are 6 cm and 8 cm. Between whichtwo numbers can length of the third side fall?

Answer» In a triangle, length of the third side should be more than the difference of two sides
and should be less than the sum of the other two sides.
In our case, two given sides are `6cm` and `8cm`.Then,
The third side should be more than `2` cm and less than `14` cm.
Discussion
6.

Is there a triangle whose sides have lengths 10.2 cm, 5.8 cm and 4.5 cm?

Answer» We know that the sum of any 2 sides of triangle is always greater than the 3rd side.
Let a=10.2,b=5.8,c=4.5
a+b=10.2+5.8=16>4.5=c
b+c=5.8+4.5=10.3>10.2=a
c+a=4.5+10.2=14.7>5.8=b.
Discussion
7.

Determine whether the triangle whose lengths of sides are 3 cm, 4 cm,5 cm is a right-angled triangle.

Answer» We know, every right angle follows Pythagoras theoram
lengt`h^2`+breat`h^2`= hypotenus`e^2`
LHS=`3^2+4^2`=9+16=25
RHS=`5^2`=25
LHS=RHS
So, this is a right angled triangle.
Discussion
8.

`Delta ABC` is right-angled at `C.` If `AC = 5 cm and BC = 12 cm` find the length of `AB.`

Answer» According to Pythagoras theorem
`AB^2=BC^2+CA^2`
`AB^2=5^2+12^2`
`AB^2=25+144`
`AB^2=169`
`AB=13`.
Discussion