Answer: Option B. -> False
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B
A median connects a vertex of a triangle to the mid point of the opposite side.
An altitude is a perpendicular on one of the sides from the opposite vertex.
As OB is the altitude, B may or may not be the midpoint of CE.

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Explanation: 2 Marks
Three types of triangles categorized by sides are:
(i) If all the three sides of the triangle have unequal length, the triangle is called a Scalene triangle.
(ii) If two out of three sides are equal in length, it's called an Isosceles triangle.
(iii) If all three sides of a triangle are equal in length, it's called an Equilateral triangle.

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(a) Each sub-part: 1 Mark
(b) Each angle: 1 Mark
(a)
(i)(ii)
(b)

$\angle BAC={92}^{\circ }$
$\angle ACB=x^{\circ }$ (Equal sides have equal angles opposite to them.)
$\angle ACB+\angle ABC+{92}^{\circ }={180}^{\circ }$
$2x+{92}^{\circ }={180}^{\circ }$
$x+{46}^{\circ }={90}^{\circ }$
$x={90}^{\circ }-{46}^{\circ }$
$x={44}^{\circ }$
$\angle ACB+y={180}^{\circ }\Rightarrow x+y={180}^{\circ }$
$\Rightarrow {44}^{\circ }+y={180}^{\circ }$
$\Rightarrow y={180}^{\circ }-{44}^{\circ }$
$\Rightarrow y={136}^{\circ }$

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Answer withReason:1 Mark each
(i) Triangle ABC is not possible. This is because, for any triangle, the sum of two sides should be greater than the third but AB + BC < AC.
(ii) Triangle ABC is not possible. Sum of all angles in a triangle should be 180${}^o$ but here,
$\angle A$$\left({60}^o\right)$ + $\angle B$$\left({120}^o\right)$ + $\angle C$$\left({30}^o\right)$ = 210${}^o$
(iii)Triangle ABC is possible because it satisfies the condition for the formation of a triangle i.e. the sum of two sides should be greater than the third.
(iv)Triangle ABC is possible. Sum of all angles in a triangle should be 180${}^o$ and here,
$\angle A$$\left({60}^o\right)$ + $\angle B$$\left({90}^o\right)$ + $\angle C$$\left({30}^o\right)$ = 180${}^o$

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Properties: 2 Marks
Steps: 1 Marks
Result: 1 Mark
In the given triangle,
Two sides are equal and the angle made by joining them is 90${}^o$.
Let the other angles be y ($\because$ equal sides have equal angles opposite to them)
For the given triangle,
Sum of all angles = 180${}^o$
Or, y + y + 90${}^o$ = 180${}^o$
Or, 2y = 180${}^o$ - 90${}^o$
Or, y = $\frac{{90}^o}{2}$ = 45${}^o$
$\because$ The two horizontal lines are parallel to each other.
$\therefore$ X = 45${}^o$ + y ($\because$ interior alternate angles are equal)
Or, X = 45${}^o$ + 45${}^o$ = 90${}^o$

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(a)Steps: 1 Mark
Result: 1 Mark
(b) Answer with reason: 1 Mark
(a) According to the question, the three sides of the triangle are equal.
Hence, all angles will also be equal (say x).
Sum of angles of a triangle = 180${}^o$
Or, x + x + x = 180${}^o$
Or, 3x = 180${}^o$
Or x = $\frac{{180}^o}{3}$ = 60${}^o$
(b) In a triangle, the side opposite to the angle with the largest value is the longest.
Here the biggest angle is the right angle, hence QR is the longest side in $△$QPR, right angled at P.

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Correct answer: 1 Mark
Reason: 1 Mark
Steps: 2 Marks
From the figure, we can see that
BO is perpendicular to AC and AO = OC
$\because$ BO is cutting the base into half and is also perpendicular to it.
$\therefore$ BO is acting both as an altitude and median.
An altitude acts as a median only in an Equilateral or Isosceles triangle with AB and BC being equal in length.
For the triangle ABC,
Sum of all angles = 180${}^o$
Or, $\angle A$ + $\angle B$ + $\angle C$ = 180${}^o$
Or, $\angle A$ + $\angle C$ + 60${}^o$ = 180
Or, 2$\angle A$ = 180${}^o$ - 60${}^o$ ($\because$ angles opposite to equal sides are equal)
Or, $\angle A$ = $\frac{{120}^o}{2}$ = 60${}^o$
$\because$ $\angle A$ = $\angle B$ = $\angle C$ = 60${}^o$
$\therefore$ The above triangle is an Equilateral triangle.

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Types of triangle: 2 Marks
The three types of triangles based on angles are:
(i) A triangle is called Obtuse angled when one of the angles has an angle greater than 90${}^o$.
(ii) If all angles of the triangle are less than 90${}^o$, it's called an Acute angled triangle.
(iii) If one of the angles is equal to 90${}^o$, it's called a Right angled triangle.

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$\angle A$ will be formed by the sides BA and AC. Hence, the side opposite to $\angle A$ will be BC.

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Each difference: 1 Mark
1.Median is a line segment formed between a vertex and the mid-point of the opposite side.
Altitude is the perpendicular distance of a vertex from its opposite side.
2.Median always bisects the opposite side of the vertex irrespective of the type of triangle.
Altitude may or may not bisect the opposite side of the vertex depending on the type of triangle.
In the figure below, CH is the altitude from vertex C on side AB and CM is median from C on side AB.