Question

PLEASE SOLVE THESE 3 Que's .....

Answer 1

5)
$3x + 7x + 20 = 180 \\ 10x = 180 - 20 \\ x = \frac{160}{10} \\ x = 16 \\ putting \: the \: value \: of \: x \\ 3x = 3 \times 16 = 48 \\ 7x + 20 = 7 \times 16 + 20 = 112 + 20 = 132$

Answer 2

4.
In this given question.
QP=PS

In triangle APS
There's an exterior angle i.e angle ASP
so we can say that,
∠ASP= ∠APS + ∠SAP
And ∠APS =90°
So,
∠ASP = 90° + ∠SAP

As it's forming an obtuse angle in triangle ASR, and there could only be only one obtuse angle in a triangle which forms the largest angle.
And as we know that side opp. To largest angle is also the longest in the triangle.
Hence,
AR > SR
And
AR > AS -------(i)

In triangle AQP amd APS
∠ APQ=∠APS (both 90°)
AP= AP(common)
QP=PS ( given)

Hence these both triangles are congruent, bu SAS congurence criteria.
Therefore,
AQ=AS

Substitute for AS in eq. (i) and TADA!!
you'll get
AR > AQ.



5.

Use linear pair axiom
Therefore,
3x + 7x+20 = 180°
10x = 160°
x = 16


6.
Let the no. Be x
So,

{x + (4 no. ) }/ 5 =30
x + 4no. = 150
x = 150 - 4no. ------(i)

Now,
Given that
4no./4 = 28
4no = 112
Substitute for those 4no. In (i) and you'llget

150-112
= 38.
Hence, the excluded no. Is 38