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Name: 1 Mark
Theorem: 1 Mark
Answer: 1 Mark
Since $\mathrm{âˆ}C$ is ${90}^{∘}$, the following is a right-angled triangle.
The theorem used is known as Pythagoras Theorem.
In a right angled triangle,
Base${}^2$ + perpendicular${}^2$ = hypotenuse${}^2$ (Pythagaros's Theorem)
Or, AB${}^2$ = BC${}^2$ + AC${}^2$
Or, AB${}^2$ = 16 + 9 = 25
Or, AB = $\sqrt{25}$ = 5

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(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.

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(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.

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(a) Solution: 1 Mark
Type of triangle: 1 Mark
(b) Solution: 1 Mark
Correct answer: 1 Mark
(a) For a triangle, sum of all angles = 180${}^o$
Or, (2x - 15${}^o$) + (x + 20${}^o$) + (x + 15${}^o$) = 180${}^o$
Or, 2x + x + x - 15${}^o$ + 20${}^o$ + 15${}^o$ = 180${}^o$
Or, 4x + 20${}^o$ = 180${}^o$
Or, 4x = 160${}^o$
Or, x = $\frac{160}{4}$ = 40${}^o$
The angles of the triangle will be, (2x - 15${}^o$ = 65${}^o$), (x + 20${}^o$ = 60${}^o$) and (x + 15${}^o$ = 55${}^o$)
Since all angles are less than 90${}^o$, it's an 'acute angled triangle' and also all the angles are of different values therefore all the sides will be of different length, hence it is a 'scalene triangle'.
(b

In triangle ABD
$⇒2x+4+90+30=180$
$⇒2x=180−124$
$⇒x=\frac{56}{2}$
$⇒x=28$
In triangle ADC
$⇒4x−k+90+10=180$
$⇒4x−k=180−100$
$⇒4\left(28\right)−k=80$ (Substituting the value of $x=28$)
$⇒k=4\left(28\right)−80$
$⇒k=112−80$
$⇒k=32$
$∴$ The angles are ${40}^{∘},{60}^{∘}and{80}^{∘}$ for the given triangle.

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Calculation of angle: 1 Mark
Type of triangle: 1 Mark
In scalene triangles, all the angles are different. For $\angle ABC$ to be maximum, the other two angles must be minimum, but different (since scalene triangle).
The minimum value of the other angles will, therefore, be $1^{\circ }$ and $2^{\circ }$ as the angles have an integral value. Hence, for a scalene triangle the maximum integral value of$\angle ABC={177}^{\circ }$ [Angle sum property of a triangle].
The following triangle will be an Obtuse angled triangle , because one of the angle is greater than$90°$

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Equilateral Triangle; Only in an equilateral triangle, all the altitudes and median are same lines. So, point A will be same as point N.

Answer: Option A. -> Fig 1 and Fig 3 only
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A
An obtuse angled triangle is the triangle in which one of the angles is greater than ${90}^{∘}.$
An isosceles triangle is the triangle in which two sides are equal.
1. Fig1:
$\mathrm{Δ}PQR$ is isosceles $[âˆ\mu PQ=PR]$
$⇒\mathrm{âˆ}Q=\mathrm{âˆ}R$
[$âˆ\mu$ angles opposite to equal sides of a triangle are equal]
$\mathrm{âˆ}P+\mathrm{âˆ}Q+\mathrm{âˆ}R={180}^{∘}$
[angle sum property of a triangle]
$\mathrm{âˆ}P+{25}^{∘}+{25}^{∘}={180}^{∘}$
$⇒\mathrm{âˆ}P={180}^{∘}−{50}^{∘}={130}^{∘}$
$\mathrm{Δ}PQR$is an obtuse angled triangle as one of the angles measures 130°.
2. Fig 2:
$\mathrm{Δ}ABC$is isosceles$[âˆ\mu AB=AC]$
Similarly as above,we can find the angles of this triangle.
$\mathrm{âˆ}A={35}^{∘},\mathrm{âˆ}B=\mathrm{âˆ}C={72.5}^{∘}$
Since all angles are less than ${90}^{∘}$, $\mathrm{Δ}ABC$ is an acute angledtriangle.
3. Fig3:
$\mathrm{Δ}XYZ$is an isosceles as well as an obtuse angled triangle as angleY measures 110°.
4. Fig4:
$\mathrm{Δ}MNO$is an isosceles as well as a right angled triangle.
Hence, only Fig1and Fig3are isosceles as well as obtuse angledtriangles.

Answer: Option A. -> True
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A
The third side of a triangle must be greater than the difference between the other two sides.

Answer: Option B. -> False
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B
A triangle with all the three angles less than ${60}^{\circ }$ is not possible, as thesum of all the angles of a triangle is ${180}^{\circ }$​.

Answer: Option B. -> False
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B
Sum of lengths of any two sides of a triangle should always be greater than the third side.
Here, 2 + 3 = 5
Therefore, this triangle is not possible.