JEE (Based on NTA Guidelines-IIT Engg.) Mains Chemistry Coaching Programs
πΉ Video Course 2024 (0 Lectures [0 Mins]): Offline Support
Click Here to View & Get Complete Material
Rs. 100.00
1 Month Validity (Multiple Devices)
β³ π― Online Tests (1 Tests [30 Questions Each]): NTA Pattern, Analytics & Explanations
Click Here to View & Get Complete Material
Rs. 100.00
3 Year Validity (Multiple Devices)
π Study Material (159 Notes): 2024-2025 Syllabus
Click Here to View & Get Complete Material
Rs. 350.00
3 Year Validity (Multiple Devices)
π― 144 Numeric, 2994 MCQs (& PYQs) with Full Explanations (2024-2025 Exam)
Click Here to View & Get Complete Material
Rs. 650.00
3 Year Validity (Multiple Devices)
NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.3 β Part 3
Altitude of triangle
In geometry, an altitude of a triangle is a line segment through a vertex and perpendicular to (i.e.. forming a right angle with) a line containing the base (the opposite side of the triangle) . This line containing the opposite side is called the extended base of the altitude.
Q-4 BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles
Solution:
Given, BE and CF are two equal altitudes.
- In and ,
(Altitudes)
(Common)
(Common)
- Therefore, by RHS congruence condition.
- Now,
(by Corresponding Parts of Congruent Triangles)
Thus, as sides opposite to the equal angles are equal.
Q-5 ABC is an isosceles triangle with . Draw to show that
Solution:
- Given,
- AB = AC
- In
(AP is altitude)
(Given)
(Common line)
- Therefore, by Right Angle-Hypostenuse-Side congruence condition.
- Thus, (by Corresponding Parts of Congruent Triangles)