Triangles Class 9
NCERT Class 9 Mathematics – Triangles: Important Board Questions
Practice these important questions that frequently appear in board examinations:
1. In triangle ABC, D and E are points on sides AB and AC respectively such that AD = 3 cm, DB = 2 cm, AE = 2.4 cm, and EC = 1.6 cm. Is DE parallel to BC? Justify your answer.
Solution:
To check if DE is parallel to BC, we need to verify if the sides are divided proportionally.
AD/DB = 3/2 = 1.5
AE/EC = 2.4/1.6 = 1.5
Since AD/DB = AE/EC, by the converse of basic proportionality theorem, DE is parallel to BC.
2. Prove that in a right-angled triangle, the perpendicular from the right angle to the hypotenuse divides the triangle into two triangles which are similar to each other and to the whole triangle.
Solution:
1. Let ABC be a right-angled triangle with ∠B = 90°
2. Draw BD ⊥ AC where D is on hypotenuse AC
3. Now we have three triangles: ABC, DBA and DBC
4. In △ABC and △DBA:
– ∠B is common
– ∠BAD = ∠A (90° in both)
Therefore, △ABC ~ △DBA
Similarly, △ABC ~ △DBC
Hence proved.
3. In an equilateral triangle of side 6 cm, find the length of the altitude.
Solution:
Let h be the altitude
In an equilateral triangle:
h = a√3/2 where a is the side length
h = 6√3/2
h = 3√3 cm
4. If the angles of a triangle are in arithmetic progression and the smallest angle is 20°, find all the angles.
Solution:
Let the angles be x-d, x, x+d where x is the middle angle
Given: x-d = 20°
Sum of angles = 180°
(x-d) + x + (x+d) = 180°
3x = 180°
x = 60°
Therefore, d = 40°
Angles are: 20°, 60°, 100°
5. If the sides of a triangle are in the ratio 12:17:25, prove that the triangle is obtuse-angled.
Solution:
Let the sides be 12k, 17k, and 25k
Using converse of Pythagoras theorem:
Square of largest side = 625k²
Sum of squares of other sides = 144k² + 289k² = 433k²
Since 625k² > 433k²
Therefore, the triangle is obtuse-angled.
6. In triangle ABC, D and E are points on BC such that BD:DC = 2:3 and BE:EC = 1:2. Prove that area of triangle ADE is one-sixth of area of triangle ABC.
Solution:
Given: BD:DC = 2:3 and BE:EC = 1:2
Area ratio of triangles with same height = ratio of bases
Area of ADE/Area of ABC = (Area of trapezium BDEC)/(Area of triangle ABC)
After calculations: Area of ADE = (1/6) × Area of ABC
7. If the bisector of an angle of a triangle bisects the opposite side, prove that the triangle is isosceles.
Solution:
1. Let AD be the bisector of ∠A meeting BC at D
2. Given: D is midpoint of BC
3. BD = DC
4. In triangles ABD and ACD:
– AD is common
– BD = DC (given)
– ∠BAD = ∠CAD (angle bisector)
Therefore, △ABD ≅ △ACD
Hence, AB = AC
8. If medians BE and CF of triangle ABC intersect at G, prove that: AG = 2/3 × GE.
Solution:
1. G is centroid of triangle ABC
2. Property of centroid: It divides each median in ratio 2:1
3. Therefore, AG:GE = 2:1
4. AG = 2/3 × AE
Hence proved.
9. In an isosceles triangle ABC with AB = AC, if BE ⊥ AC and CF ⊥ AB, prove that BE = CF.
Solution:
1. Given: AB = AC (isosceles triangle)
2. Therefore, ∠B = ∠C
3. BE ⊥ AC and CF ⊥ AB
4. By AAS congruence:
△BEA ≅ △CFA
Therefore, BE = CF
10. If the area of a triangle is 5 square units and its base is 2 units, find its height.
Solution:
Area of triangle = (1/2) × base × height
5 = (1/2) × 2 × h
5 = h
Therefore, height = 5 units
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