Question

construct a triangle ABC in which BC = 7cm and <B = 75⁰ and AB + AC = 13 cm.​​

Answer 1

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R.E.F. Image.

Given,

to construct a ΔABC where

BC=7cm,∠B=75

∘

& AB+AC=13cm.

construction steps :-

(i) Draw a line segment BC = 7 cm as a base.

(ii) Now mark ∠B=75

∘

and extend it till x.

(iii) Now, by taking radius 13 cm draw an arc from 'B' on the line of angle just projected.

(iv) Now mark it as 'D', & join C to D.

(v) Draw ⊥

lar

bisectors for DC and mark them as P & Q.

(vi) Now extend a line from 'B' to the ⊥

lar

bisector that cuts on C & D line.

(vii) Now mark that point as 'A'.

(viii) Thus, required ΔABC has been formed.

(ix) Now, try taking exactly 1/2 of the angle and mark an arc on B.

Conclusion :-

∴ The required ΔABC will BC = 7 cm,

∠B=45

∘

& AB + AC = 13 cm is constructed.

Answer 2

Answer:

Given : Base BC = 7cm, angle B = 75° and sum of two sides AB + AC = 12 cm.

Required : To Construct ∆ABC    

STEPS OF CONSTRUCTION :    

1. Draw a ray BX and cut off a line segment BC = 7cm; from it.    

2. At B; construct angle YBX = 75°    

3. With B as centre and radius = 12 cm (because AB + AC = 12cm) draw an arc to meet BY at D.    

4. Join CD    

5. Draw Perpendicular bisector PQ of CD intersecting BD at A.    

6. Join AC  

Then ABC is the required triangle.  

A lies on perpendicular bisector of CD.  

Therefore, AC = AD    

=> AB = BD - AD    

=> AB = BD - AC

=> AB + AC = BD = 12cm,    

Which is true as given.

Step-by-step explanation:

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