Question 1: Identify the triangle with an altitude outside its boundaries.

Problem 1: Which of the following figures will have its altitude outside the triangle?

- Solution:
- The altitude of a triangle is a perpendicular line segment from a vertex to the opposite side (or its extension). For an acute triangle, all altitudes lie inside the triangle. For a right triangle, one altitude lies on the triangle itself (the leg perpendicular to the hypotenuse), and the other two are inside the triangle. For an obtuse triangle, at least one altitude lies outside the triangle.
- In the given options:
- (a) Acute triangle: All altitudes are inside.
- (b) Right triangle: One altitude is on the triangle, and the others are inside.
- (c) Obtuse triangle: At least one altitude is outside.
- (d) Acute triangle: All altitudes are inside.
- Therefore, the correct answer is (c).

Problem 2: Fill up the blanks:

(i) Every triangle has at least ___________ acute angles.
- Solution:
- A triangle can have either three acute angles (acute triangle), one right angle and two acute angles (right triangle), or one obtuse angle and two acute angles (obtuse triangle). In all cases, a triangle has at least two acute angles.
- Answer: two.

(ii) The longest side of a right-angled triangle is called its ___________.
- Solution:
- In a right-angled triangle, the side opposite the right angle is the longest side and is called the hypotenuse.
- Answer: hypotenuse.

(iii) Median is also called ___________ in an equilateral triangle.
- Solution:
- In an equilateral triangle, each median is also an altitude, angle bisector, and perpendicular bisector. Therefore, the median is also called an altitude.
- Answer: altitude.

(iv) The line segment joining a vertex of a triangle to the mid-point of its opposite side is called its ___________.
- Solution:
- The line segment joining a vertex of a triangle to the midpoint of the opposite side is called a median.
- Answer: median.

Problem 3: If one angle of a triangle is 60° and the other two angles are in the ratio 1:2, then find the angles.

- Solution:
- Let the other two angles be \( x \) and \( 2x \). Since the sum of the angles in a triangle is 180°, we have:
\[
60^\circ + x + 2x = 180^\circ
\]
\[
60^\circ + 3x = 180^\circ
\]
\[
3x = 120^\circ
\]
\[
x = 40^\circ
\]
Therefore, the angles are \( 60^\circ \), \( 40^\circ \), and \( 80^\circ \).
- Answer: 60°, 40°, 80°.

Problem 4: In the figure, find the value of \( \angle A + \angle B + \angle C + \angle D + \angle E + \angle F \).

- Solution:
- The given figure is a hexagon with a star-like pattern. The sum of the interior angles of a hexagon is:
\[
(n-2) \times 180^\circ = (6-2) \times 180^\circ = 4 \times 180^\circ = 720^\circ
\]
However, the angles \( \angle A, \angle B, \angle C, \angle D, \angle E, \angle F \) are the exterior angles of the smaller triangles formed by the star. Each of these angles is supplementary to the interior angle of the hexagon at that vertex. Since the sum of the exterior angles of any polygon is always 360°, we have:
\[
\angle A + \angle B + \angle C + \angle D + \angle E + \angle F = 360^\circ
\]
- Answer: 360°.

Problem 5: Two poles of 8m and 14m stand upright on a plane ground. If the distance between the tops is 10m, find the distance between their feet.

- Solution:
- Let the distance between the feet of the poles be \( d \). The situation forms a right triangle where the difference in height of the poles is one leg, the distance between the feet is the other leg, and the distance between the tops is the hypotenuse.
- The difference in height is \( 14 - 8 = 6 \) meters. Using the Pythagorean theorem:
\[
10^2 = d^2 + 6^2
\]
\[
100 = d^2 + 36
\]
\[
d^2 = 64
\]
\[
d = 8
\]
- Answer: 8m.

Problem 6: Mohini walks 1200m due East and then 500m due North. How far is she from her starting point?

- Solution:
- This forms a right triangle where the legs are 1200m and 500m, and the hypotenuse is the distance from the starting point. Using the Pythagorean theorem:
\[
\text{Distance} = \sqrt{1200^2 + 500^2}
\]
\[
= \sqrt{1440000 + 250000}
\]
\[
= \sqrt{1690000}
\]
\[
= 1300
\]
- Answer: 1300m.

Problem 7: Find the value of \( x \) and \( y \).

(i) Here \( CD \parallel AB \)
- Solution:
- Since \( CD \parallel AB \), the alternate interior angles are equal. Therefore, \( \angle BAC = \angle ACD = 40^\circ \) and \( \angle ACB = 70^\circ \). The angle \( y \) is the exterior angle for the triangle \( \triangle ABC \) at vertex \( C \), so:
\[
y = \angle BAC + \angle ABC = 40^\circ + 70^\circ = 110^\circ
\]
The angle \( x \) is the same as \( \angle BAC \) because \( CD \parallel AB \):
\[
x = 40^\circ
\]
- Answer: \( x = 40^\circ \), \( y = 110^\circ \).

(ii)
- Solution:
- In the given figure, \( \angle QPS = 60^\circ \) and \( \angle PSR = 50^\circ \). The angle \( x \) is the exterior angle for the triangle \( \triangle PQS \) at vertex \( Q \), so:
\[
x = \angle QPS + \angle PSR = 60^\circ + 50^\circ = 110^\circ
\]
The angle \( y \) is the same as \( \angle QPS \) because \( QR \parallel PS \):
\[
y = 60^\circ
\]
- Answer: \( x = 110^\circ \), \( y = 60^\circ \).

Final Answers:

1. (c)
2. (i) two, (ii) hypotenuse, (iii) altitude, (iv) median
3. 60°, 40°, 80°
4. 360°
5. 8m
6. 1300m
7. (i) \( x = 40^\circ \), \( y = 110^\circ \)
(ii) \( x = 110^\circ \), \( y = 60^\circ \)

\[
\boxed{(c), \text{two, hypotenuse, altitude, median}, 60^\circ, 40^\circ, 80^\circ, 360^\circ, 8m, 1300m, x = 40^\circ, y = 110^\circ, x = 110^\circ, y = 60^\circ}
\]