(a)  Answer: 1 Mark
(b) Answer: 1 Mark
      Reason: 1 Mark
(a) In a right-angled isosceles triangle, sides other than the hypotenuse are equal in length.
In ABC, AB = BC $\Rightarrow \angle 1=\angle 3$ -------------------I
By angle sum property of triangle,
$\angle 1+\angle 2+\angle 3={180}^{\circ }$
$\angle 1+90+\angle 1={180}^{\circ }$
$2\angle 1={90}^{\circ }$
$\Rightarrow \angle 1={45}^{\circ }=\angle 2$
So, angles are ${45}^{\circ },{45}^{\circ }and{90}^{\circ }$
(b) A triangle ABC with the given side lengths of AB = 12 cm, BC = 8 cm and  AC = 20 cm is not possible because in a triangle the sum of any two sides should be greater than the third side but in this triangle, AB + BC = 20 cm which is equal to the third side AC = 20 cm. AB + BC should have been greater than AC.

Answer: 1 Mark
Justification: 1 Mark
Yes.

The median AD will divide the ∆ABC into 2 triangles ∆ADB and ∆ADC. These two triangles, as shown in the figure will have the same altitude AN as the altitude is the perpendicular distance from the vertex to the base of the triangle.The foot of the perpendicular can be inside as well as outside the triangle as shown in the figure. (Note: The median of the triangle divides it into 2 triangles with equal area.)

(a) Steps: 1 Mark
     Proof: 1 Mark
(b) Steps: 1 Mark
      Result: 1 Mark
(a)

$\mathrm{âˆ}1$, $\mathrm{âˆ}2$ and $\mathrm{âˆ}3$ are angles of triangle ABC and 4 is the exterior angle when BC is extended to D.
$\mathrm{âˆ}1$ + $\mathrm{âˆ}2$ = $\mathrm{âˆ}4$ (Exterior Angle Property)
$\mathrm{âˆ}1$ + $\mathrm{âˆ}2$ + $\mathrm{âˆ}3$ = $\mathrm{âˆ}4$ + $\mathrm{âˆ}3$ (Adding $\mathrm{âˆ}3$ to both sides)
Also, $\mathrm{âˆ}3$ + $\mathrm{âˆ}4$ = 180${}^o$ (Linear Pair of angles)
$∴$ $\mathrm{âˆ}1$ + $\mathrm{âˆ}2$ + $\mathrm{âˆ}3$ = 180${}^o$
(b) 
   
$\mathrm{âˆ}1$ + $\mathrm{âˆ}2$ = $\mathrm{âˆ}4$ (Exterior Angle Property)
$\mathrm{âˆ}1+{60}^{∘}={140}^{∘}$
$\mathrm{âˆ}1={140}^{∘}−{60}^{∘}={80}^{∘}$
Since AE is the angular bisector therefore $\mathrm{âˆ}EAC=\frac{\mathrm{âˆ}1}{2}$
$⇒x=\frac{80}{2}={40}^{∘}$

(a) Steps: 1 Mark
     Correct answer: 1 Mark
(b) Reason: 1 Mark
     Correct answer: 1 Mark
(a)

In the figure, 1, 2 and 3 are the interior angles whereas 4, 5 and 6 are exterior angles. Using linear pair axiom, we can say that:
$\mathrm{âˆ}1+\mathrm{âˆ}4={180}^{∘}$

$\mathrm{âˆ}3+\mathrm{âˆ}6={180}^{∘}$

$\mathrm{âˆ}2+\mathrm{âˆ}3={180}^{∘}$

Adding all of them, we get:

$\mathrm{âˆ}\left(\right(1+4)+(3+6)+(2+5\left)\right)={540}^{∘}$

Rearranging the terms, we get:

$\mathrm{âˆ}\left(\right(1+2+3)+(4+5+6\left)\right)={540}^{∘}$

$\mathrm{âˆ}(1+2+3)={180}^{∘}$ (Angle Sum Property of a triangle)

So, $\mathrm{âˆ}\left(\right(4+5+6\left)\right)={540}^{∘}−{180}^{∘}={360}^{∘}$

So, the sum of the exterior angles of a triangle is ${360}^{∘}$
(b) Suppose such a triangle is possible. Then the sum of the lengths of any two sides would be greater than the length of the third side. Let's check this.
Is 4.5 + 5.8 > 10.2?   Yes
 Is 5.8 + 10.2 > 4.5?   Yes
Is 10.2 + 4.5 > 5.8?   Yes
Therefore, the triangle is possible.

Properties : 1 Mark
Steps : 2 Marks
Result : 1 Mark

In triangle ABC, AC < AB + BC ------------------- 1
In triangle ADE, AD < AE + DE ------------------- 2
Adding 1 and 2,
AC + AD < AB + BC + AE + DE   ----------------3
In triangle ADC, CD < AD + AC ------------------- 4
From 3 and 4, we can write
CD < AB + BC + AE + DE
Adding CD to both sides,
2CD < AB + BC + CD + DE + EA
Or, 2 CD < Perimeter of ABCDE

(a) Steps: 1 Mark
     Correct answer: 1 Mark
(b) Formula: 1 Mark
     Result: 1 Mark
(a)

$âˆOAD=50°(Âgiven)âˆDOCÂ=Â180°−50°=130°(ÂlinearÂpairÂ)âˆDPCÂ=Â160°Â\left(given\right)âˆÂDPOÂ=180°Â−Â160°(ÂLinearÂpair)âˆDPO=Â20°InÂâ–³ÂDOPâˆODPÂ=180°−(130°Â+Â20°)Â(ÂbyÂangleÂsumÂpropertyÂ)âˆODPÂ=30°InÂâ–³ÂDBEâˆDBEÂ=180°−(Â90°Â+Â30°Â)Â(ByÂangleÂsumÂpropertyÂ)âˆDBEÂ=Â60°$
(b)   
     
The wall and the ground are at right angle. Therefore the ladder, ground and the wall make the sides of a right-angled triangle.
$∴a^2+{12}^2={15}^2$
$⇒a^2=225−144=81$
$a=\sqrt{8}1=9\text{Â}m$

(a) Solution: 2 Marks
(b) Proof: 2 Marks
(a)
       

If we draw the path taken by you, it will look something like the figure above. We have to find AC. Applying Pythagoras property,

$A{B}^2$ + $B{C}^2$ = $A{C}^2$

$5^2$ + ${12}^2$ = $A{C}^2$

$A{C}^2$ = 25 + 144

$A{C}^2$ = √169

AC = 13 km
(b) Let ABC be an equilateral triangle.

BC = AC = AB (Length of all sides is same)
 $⇒\mathrm{âˆ}A=\mathrm{âˆ}B=\mathrm{âˆ}C$ ( Angle opposite to equal sides are equal)
Also,
$\mathrm{âˆ}A+\mathrm{âˆ}B+\mathrm{âˆ}C={180}^{∘}$
$⇒3\mathrm{âˆ}A={180}^{∘}$
$⇒\mathrm{âˆ}A={60}^{∘}$
$∴\mathrm{âˆ}A=\mathrm{âˆ}B=\mathrm{âˆ}C={60}^{∘}$
Thus, the angles of an equilateral triangle are ${60}^{∘}$ each.