Question

Construct a triangle of sides 4cm,5cm and 6cm and then a triangle similar to it whose sides are2÷3 of the corresponding sides of the first triangle​

Answer 1

Answer:

Let's first construct ∆ABC with sides 4cm,5cm,6cm

Steps to draw ∆ABC

1. Draw base AB of side 4cm.

2. With A as centre , and 5cm as radius, draw an arc.

3. With B as centre , and 6cm as radius ,draw an arc.

4. Let C be the point where two points intersect.

5. Join AC and BC

Thus ∆ABC is the required triangle

Now, lets make the triangle similar to this named as ∆PQR whose sides are $\frac{2}{3}$

of the corresponding sides so the sides will become $\frac{8}{3}$cm, $\frac{10}{3}$cm, 4cm

Steps to draw ∆PQR

1. Draw base PQ of side $\frac{8}{3}$cm.

2. With P as centre , and $\frac{10}{3}$cm

as radius, draw an arc.

3. With Q as centre , and 4cm as radius ,draw an arc.

4. Let R be the point where two points intersect.

5. Join PQ and PR

Thus ∆PQR is the required triangle

Hope it helps you.

Answer 2

Hello(´∩｡• ᵕ •｡∩`)

Answer:~

Let's first construct △ABC with sides 4cm,5cm,6cm

Steps to draw △ABC

1. Draw base AB of side 4cm.

2. With A as centre , and 5cm as radius, draw an arc.

3. With B as centre , and 6cm as radius ,draw an arc.

4. Let C be the point where two points intersect.

5. Join AC and BC

Thus △ABC is the required triangle

Now, lets make the triangle similar to this named as △PQR whose sides are 3/2 of the corresponding sides so the sides will become 8/3 cm, 10/3 cm,4cm

Steps to draw △PQR

1. Draw base PQ of side 8/3 cm.

2. With P as centre , and 10/3 cm. as radius, draw an arc.

3. With Q as centre , and 4cm as radius ,draw an arc.

4. Let R be the point where two points intersect.

5. Join PQ and PR

Thus △PQR is the required triangle

❥Thank You(◍•ᴗ•◍)❤

Itzgirl45⚘