Board | Jammu & Kashmir Board of School Education [JKBOSE] |
Class | 10th/X |
Subject | Mathematics/Maths |
Download | Question Bank |
Document Type | |
Official Website | https://jkbose.nic.in/index.html |
JKBOSE Class 10th Maths Question Bank
Download Jammu & Kashmir Board of School Education [JKBOSE] Class 10th Maths Question Bank Online
Download JKBOSE Class 10th Maths Question Bank
JKBOSE Class 10th Maths Question Bank Chapter 1 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 2 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 3 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 4 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 5 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 6 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 7 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 8 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 9 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 10 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 11 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 12 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 13 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 14 | Download Here |
JKBOSE Class 10th Maths Question Bank Chapter 15 | Download Here |
Important Constructions Of A Triangle In Different Cases
Before constructing a triangle similar to a given triangle say ∆ ABC, we have to construct the ∆ ABC. For this, let us review some constructions in different cases which are given below:
Case – 1 : When three sides are given.
We use the following steps of construction:
Step – 1 : First draw base BC.
Step – 2 : By taking B and C as centers, draw two arcs of radius CA and BA respectively which intersects each other at A.
Step – 3 : Join AB and AC.
Then ABC is a required triangle.
Case – 2 : When base and alternate of a triangle are given.
Let the base of ∆ ABC be BC = m cm and Alternate is n cm. Then, we use the following steps.
Step – 1 : Draw the base BC = m cm and draw its perpendicular bisector OQ (say) which intersects BC at P (say).
Step – 2 : By taking P as center, draw an arc of radius n cm which intersects the line segment PO or extended line segment of PO at A.
Step – 3 : Join AB and AC.
Then ABC is a required triangle.
Case – 3 : When two sides and angle between them are given. Let the two sides of ∆ ABC be AB = a cm, BC = b cm and angle between them i.e. <ABC = x0. Then, we use the following steps.
Step – 1 : Draw base “AB” = “a” cm
Step – 1 : Draw a ray “BX” making an angle x0 at B and cut off BC = b cm from BX.
Step – 3 : Join “AC”. Thus ∆ ABC is required triangle
CASE – 3 : When one side and two angles are given.
Let one side of triangle ABC be AB = a cm and < A = x 0 and < B = y0
Step – 1 : Draw base AB = a cm
Step – 2 : Draw two rays AX and BY making an angle x0 at “A” and y0 at B which intersects each other at C.
Step – 3 : Join AC and CB.
Then ABC is a required triangle.
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