Question

Draw a triangle ABC with AB=4. 5 cm, AC=7. 5 cm and angle B =90. Construct a similar triangle whose sides are 4/5 cm the sides of the triangle ABC.

Answer 1

Answer:

Step-by-step explanation:

Draw a base 4.5 cm

Mark angle 90 at angle b

Using compass draw 7.5 cm AC

Extend side A less than 90

Cut the 5arcs on extended side by using compass taking the radius of your own

Join a5 and B

Place the compass at A5 and cut the arc and without changing the radius place it on A4 and cut the arc

By measuring the distance between E andF ( shown in fig)in compass draw the co incide of arc by placing where it was started

Join this line to the base of triangle mark ot as B dash

Take the radius of your cjoice place it on b and cut arc

With same measurement place it on bdash and cut the arc

By measuring the distance of pand q as shown in fig draw the arc from starting point of arc join it to AC

Answer 2

$\boxed{GIVEN}$

✔️ AB = 4.5cm
✔️ AC = 7.5 cm
✔️Angle B = 90°

We need to construct a similar triangle whose sides are 4/5 cm to the sides of the ΔABC.

$\boxed{STEPS\: OF\: CONSTRUCTION}$

⏩ First draw a line of measurement 4.5cm and name the line as AB.

⏩ As B as center, draw 90°.

⏩ Take 7.5cm on compass. Draw an arc on 90° as B as center. And name the point as C.

⏩ Draw a ray opposite to the vertex B. Name the ray as BX.

⏩ Locate 5 points on BX, (because 5 > 4 in 4/5) Name the point as B1,B2,...B5. So that B1=B2.........B4=B5.

⏩ Join B5 (since the denominator of 4/5 is 5) to the vertex A.

⏩ Draw a line parallel to B5 through B4(since the numerator of 4/5 is 4) intersecting in line the line AB at A'.

⏩ Draw a line parallel to AC through A' which intersect BC at C'.

Hence ΔA'BC' is the required triangle.

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