20 cm

10m

26 cm

25 cm

(i) 60°
(ii) 30°
(iii) 140°
(iv) 50°
(v) 35°
(vi) 60°

Let the ladder is represented by AC, wall by BC and AB is the distance of the foot of the ladder from the wall.

\(\because\)\(\angle\)B = 90°
\(\therefore\) \(\triangle\)ABC is a right triangle such that its hypotenuse is AC.
We have AB² + BC² = AC²
[Using Pythagoras theorem]
x² + 8² = 10²\(\Rightarrow\)x² +64= 100
\(\Rightarrow\) x²= 100 - 64 = 36
\(\Rightarrow\) x² ​​​​​​​ = 6² \(\Rightarrow\) X = 6
Thus, the foot of the ladder is 6 m from the wall.

Let us draw a rectangle ABCD, such that BD is a diagonal
\(\therefore\)Each angle of ABCD is a right angle.
\(\therefore\) \(\angle C\)= 90°
\(\Rightarrow\) \(\triangle\)BCD is a right triangle
Since the side opposite to 90° is hypotenuse and (Hypotenuse)² = [Sum of the squares of the legs of the triangle]
\(\therefore\) BD² = BC² + CD²
\(\Rightarrow\) (x)² = (8)² + (6)²
\(\Rightarrow\) x² = 64 + 36
\(\Rightarrow\) x² = 100
x² ​​​​​​​ = 10²\(\Rightarrow\)X = 10
Thus, the required length of the diagonal is 10 cm.

Since\(\angle\) b and \(\angle\) d form a linear pair.
\(\therefore\) \(\angle\) b + \(\angle\) d = 180°
Similarly, \(\angle\)c + \(\angle\)e = 180°
and \(\angle\)a + \(\angle\)f = 180°
Adding the angles on both sides, we get
\(\angle\)b + \(\angle\)c + \(\angle\)a + \(\angle\)d +\(\angle\)e + \(\angle\)f
= 180° + 180° + 180°
\(\Rightarrow\) (\(\angle\)a +\(\angle\)b + \(\angle\)c) + (\(\angle\)d + \(\angle\)e + \(\angle\)f) = 540°
But \(\angle\)d + \(\angle\)e + \(\angle\)f= 180°
[\(\because\) Sum of angles of a triangle is 180°.]
\(\therefore\) (\(\angle\)a + \(\angle\)b + \(\angle\)c) + (180°) = 540°
\(\Rightarrow\) (\(\angle\)a + \(\angle\)b + \(\angle\)c) = 540° - 180°
\(\Rightarrow\)\(\angle\)a + \(\angle\)b +\(\angle\)c = 360°
i.e. Sum of exterior angles of \(\triangle\)ABC is 360°.

(a) PQ + PA > QA
Yes, because sum of two sides of a triangle is always greater than the third side.
(b) PQ + PA > OQ + OA
Yes,because: PQ + PA > QA
PQ + PA >QO + OA [\(\Box \) QA = QO + OA]
(c) PQ + PA +AR > OQ + OA +AR
Yes, because: PQ + PA > OQ + OA
Adding AR in both sides, we get
PQ + PA + AR > OQ + OA + AR
(d) PQ + PR > OQ + OR
Yes, because
PQ + PA > QO + OA                 ...(1)
OA+AR > OR                            ...(2)
Adding (1) and (2), we get
PQ + PA + OA + AR > QO + OA + OR
PQ + PR > OQ + OR
(e) PQ + QR + PR > OP + OQ + OR
Yes, because
PQ + PR > OQ + OR          ...(1)
PQ + QR > OP + OR          ...(2)
PR + PQ > OP + OQ          ...(3)
Adding (1), (2) and (3), wet get
2(PQ + PR + QR) > 2(OP + OQ + OR).