Question

Draw an isosceles triangle ABC in which AB=AC=6 cm and BC=5 cm. Construct a triangle PQR similar to AABC in which PQ=8 cm. Also justify the construction.​

Answer 1

The New Triangle is Similar but $\frac {4}{3}$ of ABC

Step-by-step explanation:

we assume that,

1.) With one base, any two similar triangle (1 Vertex) are to be made.

In ∆ABC & ∆PQR: vertex B = Q, Hence there is requirement of Scale factor.

2.) ∆ABC (Sides $\frac {PQ}{AB}$) of respective ∆ABC are similar, let's construct them now.

$\frac {PQ}{AB} = \frac {8}{6} = \frac {4}{3}$ is the scale factor between the respective sides of similar triangles, ∆PQR and ∆ABC

• Constructional Steps are-

1. BC=5 cm is the line segment being drawn initially.

2. OQ is the perpendicular and bisector also of BC at P'

3. After bisection of BC with B & C (Centers) draw two arcs of (6 cm each) intersecting at A.

4.  Joining BA and CA, results in ∆ABC (isosceles)

5. ∠CBX is the acute angle being drawn after drawing BX ray from B.

6. Locate four points ($B_1,B_2,B_3\ and\ B_4$ on BX such that $BB_1=B_1B_2=B_2B_3=B_3B_4$)

7. Draw line $B_4R$ ||$B_3C$ from $B_4$ after joining $B_3C$, which intersects BC being extended at R.

8. Draw RP||CA (from R) is drawn which meets BA being extended at P.  

∆PBR triangles.

Justification:

Let $BB_1 = B_1B_2 = B_2B_3 = B_3B_4 = x$

As per the construction $B_4R$ ||$B_3C$

$\frac {BC}{CR} = \frac {BB_3}{B_3B_4} = \frac {3x}{x} = \frac {3}{1}$

$\frac {BC}{CR} = \frac {3}{1}$

Now,

$\frac {BR}{BC} = \frac {BC + CR}{BC} = \frac {BC}{BC} + \frac {CR}{BC} = 1 + \frac {1}{3} = \frac {4}{3}$

Also, from the construction RP||CA

Therefore rABC is congruent to rPBR

(by the AA criteria, as ∠ PBR = ∠ ABC, also as RP||CA, ∠ ACB is corresponding to ∠ PRB so ∠ ACB = ∠ PRB)

and

$\frac {PB}{AB} = \frac {RP}{CA} = \frac {4}{3}$

Hence, the new triangle rPBR is similar to the given isosceles triangle rABC and its sides are $\frac {4}{3}$ times of the corresponding sides of rABC.

Answer 2

Given : an isosceles ΔABC in which AB=AC=6cm and BC=5cm

To Find : Construct Triangle

and similar triangle PQR with PQ = 8 cm

Solution:

Step 1 : Draw  a line segment BC  = 5 cm

Step 2 : Using compass width = 6 cm and taking  B as center draw an arc

Step 3 : Keeping compass width same taking C as center draw an arc intersecting arc drawn in previous step at A

Step 4 : Join AB  & AC

Draw  a line segment PQ  = 8 cm

Using suitable compass width Taking A as center draw an arc intersecting

AC at X & AB at Y

Keeping compass width same taking P as center Draw an arc intersecting PQ at M

open compass = XY and taking M as center intersect arc drawb in previous step at N

draw a ray PZ passing passing thorugh N

Take a length PR = 8 cm

Join QR

ΔPQR ≈ ΔABC

SAS criteria of similarity Used

Learn More:

construct a right angle triangle in which base if two times of the ...

https://brainly.in/question/14146138

construct a triangle ABC with base BC=4.2 cm, A=45° and altitude ...

https://brainly.in/question/13345207