Question

Construct a triangle with 3 cm base and sum of other two sides is 8 cm and one base angle is ${60}^o$.

Answer 1

Given the base BC of the triangle ABC be 3 cm, one base angle $\angle B={60}^o$ and the sum of the other two sides be 8 cm i.e, AB + AC = 8 cm.
STEPS :
(i) Draw BC = 3 cm
(ii) At point B, draw PB so that $\angle PBC={60}^o$
(iii) From BP, cut BC = 8 cm.
(iv) Join D and C.
(v) Draw perpendicular bisector of CD, which meets BD at point A.
(vi) Join A and C.

Thus, ABC is the required triangle.
Proof : Since, OA is perpendicular bisector of CD
$\Rightarrow$OC = OD
$\Rightarrow \angle$AOC = $\Rightarrow \angle$AOD = 90°
Also, OA = OA [Common]
$\therefore △$AOC $\cong △$ AOD [By SAS]
$\Rightarrow$ AC = AD
$\therefore$ BD = BA + AD
= BA + AC
= Given sum of the other two sides
Thus, base BC and $\angle$B are draw as given and BD = AC.
Hence Proved