In this section, let us construct a triangle when the following are given :
(i) the base, vertical angle and the median on the base
(ii) the base, vertical angle and the altitude on the base
(iii) the base, vertical angle and the point on the base where the bisector of the vertical angle meets the base.

**Construction of triangle**

We have already learnt in previous class how to construct triangles
when sides and angles are given.

In this section, let us construct a triangle when the following
are given :

(i) the base, vertical angle and the median on the base

(ii) the base, vertical angle and the altitude on the base

(iii) the base, vertical angle and the point on the base where the
bisector of the vertical angle meets the base.

First, we consider the following construction,

**Construction of a segment of a circle on a given line segment
containing an angle** θ

**Construction**

**Step 1: **Draw a line segment

**Step 2: **At *A*, take ∠*BAE* = *θ *Draw *AE*.** **

**Step 3: **Draw, *AF* ┴ *AE* .

**Step 4: **Draw the perpendicular bisector of *AB *meeting* AF *at*
O*.

**Step 5: **With *O* as centre and *OA* as radius draw a circle *ABH.*

**Step 6: **Take any point *C* on the circle, By the alternate segments
theorem, the major arc *ACB *is the required segment of the circle
containing the angle* θ*.

If** ***C*** **_{1}** **,** ***C*_{2},...**
**are points on the circle, then all the triangles** **Δ*BAC*** **_{1}**
**,** **Δ*BAC*_{2},...** **are with** **same base and
the same vertical angle.

**Construction of a triangle when its base, the vertical angle and
the median from the vertex of the base are given.**

Construct a** **Δ*PQR*** **in which** ***PQ*** **=** **8** **cm,** **∠*R*** **=** **60°** **and the median** ***RG *from *R* to *PQ*
is 5.8 cm. Find the length of the altitude from *R* to *PQ*.

**Step 1: **Draw a line segment *PQ* = 8cm.

**Step 2: **At *P*, draw *PE* such that ∠*QPE* = 60°.

**Step3: **At *P*, draw *PF* such that ∠*EPF* = 90°.

**Step 4: **Draw the perpendicular bisector to *PQ*, which intersects *PF*
at *O* and *PQ* at *G.*

**Step 5: **With *O* as centre and *OP* as radius draw a circle.

**Step 6: **From *G* mark arcs of radius 5.8 cm on the circle. Mark them
as *R* and *S* .

**Step 7 : **Join *PR* and *RQ*. Then Δ*PQR *is the required
triangle.

**Step 8 : **From *R* draw a line *RN* perpendicular to *LQ*. *LQ
*meets* RN *at* M *

**Step 9: **The length of the altitude is *RM* = 3.5 cm.

We can get another Δ*PQS* for the given measurements.

**Construct a triangle when its base, the vertical angle and the
altitude from the vertex to the base are given.**

Construct a triangle** **Δ*PQR*** **such that** ***QR*** **=** **5 cm, ∠*P *=* *30°* *and
the altitude from* P *to* QR *is of length 4.2 cm.

**Step 1 : **Draw a line segment** ***QR*** **=** **5 cm.

**Step 2 : **At** ***Q*** **draw** ***QE*** **such that**
**∠*RQE*** **=** **30°.

**Step 3 : **At** ***Q*** **draw** ***QF*** **such that**
**∠*EQF*** **=** **90°.

**Step 4 : **Draw the perpendicular bisector** ***XY*** **to** ***QR***
**which intersects** ***QF*** **at** ***O*** **and** ***QR***
**at** ***G*.

**Step 5 : **With** ***O*** **as centre and** ***OQ*** **as
radius draw a circle.

**Step 6: **From *G* mark an arc in the line *XY* at *M*, such
that *GM* = 4. 2 cm.

**Step 7: **Draw** ***AB*** **through** ***M*** **which
is parallel to** ***QR*.

**Step 8: ***AB*** **meets the circle at** ***P*** **and** ***S*.

**Step 9: **Join** ***QP*** **and** ***RP*. Then** Δ***PQR***
**is the required triangle.

Δ*SQR* is another required triangle for the
given measurements.

**Construct of a triangle when its base, the vertical angle and
the point on the base where the bisector of the vertical angle meets the base**

Draw a triangle** ***ABC*** **of base** ***BC*** **=** **8 cm,** **∠*A*** **=** **60° and the bisector of ∠*A* meets *BC* at *D* such that *BD* = 6 cm.

**Step 1 : **Draw a line segment** ***BC*=** **8 cm.

**Step 2 : **At *B*, draw *BE* such that ∠*CBE* = 60°

**Step 3 : **At *B*, draw BF such that ∠*EBF* = 90° .

**Step 4 : **Draw the perpendicular bisector to *BC*, which intersects *BF*
at *O* and *BC* at *G*.

**Step 5 : **With *O* as centre and *OB* as radius draw a circle.

**Step 6 : **From *B*, mark an arc of 6cm on *BC* at *D*.

**Step 7 : **The perpendicular bisector intersects the circle at I. Joint *ID*.

**Step 8 : ***ID *produced meets the circle at* A*. Now join* AB *and*
AC. *Then Δ*ABC *is the required triangle.

Tags : Geometry | Mathematics , 10th Mathematics : UNIT 4 : Geometry

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10th Mathematics : UNIT 4 : Geometry : Construction of triangle | Geometry | Mathematics

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