Question

Construct a
∆ ABC, in which
ZB = 60°.
ZC = 45°
and AB + BC + CA = 11 cm.
Give
justification
also
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Answer 1

Answer:

Q). Construct a triangle ABC in which ∠B = 60°,∠C = 45° and AB+BC+CA=11cm.

Solution:

Given ∠B=60°,∠C=45° and AB+BC+CA=11cm

Let's construct ∆ABC.

Step to construction:

①Draw a line segment XY equal to AB+AC+BC=11cm.

②Make angle equal to ∠B=60° from point X

Let the angle be ∠LXY.

③Make angle equal to ∠C=45° from point Y

Let the angle be ∠MYX.

④Bisect ∠LXY and ∠MYX.

Let these bisectors intersect at a point A.

⑤Make perpendicular bisector of AX

Let it intersect XY at point B

⑥Make perpendicular bisector of AY

Let it intersect XY point C

⑦Join AB & AC.

∴∆ABC is the required triangle.

Answer 2

Given, in ABC,

$\rm \footnotesize{ \implies \: \angle \: B \: = 60 \degree} \\ \rm \footnotesize{ \implies \: \angle C \: = 45 \degree}$

$\rm \footnotesize{ \implies \: AB + BC + CA = 11 \: cm}$

$\mathbb \purple{ STEPS \: \: OF \: \: CONSTRUCTION}$

$\rm \footnotesize \bold{(step - 1)}\rm \footnotesize{Draw \: a \: line \: segment \: PQ = 11 cm}$

$\rm \footnotesize \bold{(step - 2)}\rm \footnotesize{Draw \: ray \: PL \: making \: an \: \angle \: LPQ \: 60 \degree at \: P} \\ \rm \footnotesize{and \: ray \: QM \: making \: an \: \angle \: MQP = 45 \degree \: at \: Q}$

$\rm \footnotesize \bold{(step - 3)}\rm \footnotesize{Bisect \: the \: \angle \: LPQ \: and \: \angle \: MQP, \: let \: the} \\ \rm \footnotesize{intersection \: point \: of \: these \: Bisectors \: be \: A \: }$

$\rm \footnotesize \bold{(step - 4)}\rm \footnotesize{Draw \: the \: \bot \: of \: ED \: of \: AP \: intersect \: PQ \: at \: \: B}$

$\rm \footnotesize \bold{(step - 5)}\rm \footnotesize{Draw \: another \: \bot \: FG \: of \: AQ \: which \: intersect \: PQ \: at \: C}$

$\rm \footnotesize \bold{(step - 6)}\rm \footnotesize{Join \: AB \: and \: AC, \: Thus \: ABC \: is \: a \: required \: \triangle}$

$\mathbb \purple{JUSTIFICATION}$

Since, B lies on the perpendicular Bisector ED of AP.

$\rm \small{so \: AB = BP \: \: then, \angle \: ABP = \angle \: PAB}$

Similarly, C lies on the perpendicular Bisector FG of AQ.

$\rm \small{ \therefore \: AC = CQ \: then, \: \angle \: AQC = \angle \: QAC}$

$\rm \small{ Now,. BC + CA + AB = BC + CQ + BP = PQ}$

$\rm \small{ \angle \: ABC = \angle \: APB + \angle \: PAB = 2( \angle \: APB)}$

$\rm \small{ \implies \: \angle \: LPQ = 60 \degree}$

$\rm \small{Similarly \: \angle ACB = \angle \: AQC + \angle QAC}$

$\rm \small{ \implies \: 2( \angle \: AQC)}$

$\rm \small{ \implies \: \angle \: MQP = 45 \degree}$

Thus, the construction is justified.

$\tiny \rm{hope \: it \: helps \: uhh}$