Question

triangle PQR is constructed similar to triangle ABC with scale factor 2÷3 Find triangle PQR

Answer 1

ΔPQR is constructed = 2/3 of ΔABC

Step-by-step explanation:

Let say Δ ABC is already constructed

Now to Draw Δ PQR  with Scale factor of 2 /3

Mark A as P

Step 1  : Draw an acute angle at point A on AB  opposite to vertex C side

Step 2: Take 3 Equal segment Consecutively one by one at angle Drawn in step 1

Step 3: join 3rd point with B

Step 4: Draw a line parallel to line drawn in step 3 such that it Passes through 2nd point on line drawn in step 2 & intersect AB / PB  at Q

Step 5 : Now draw a line parallel to BC passing through Q such that intersect AC/PC at R

ΔPQR is constructed = 2/3 of ΔABC

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Answer 2

Given: $\triangle PQR$ is constructed similar to $\triangle ABC$ with scale factor $\frac{2}{3}$.

To Find: Find $\triangle PQR$.

Step-by-step explanation:

Following Steps:

• Draw any $\triangle ABC$.
• Draw a line $BX$, as any acute $\angle XBC$. Now draw three equal radius arc on line $BX$ that intersect at $B_{1} ,B_{2}\ and\ B_{3}$.

        As: $BB_{1}=B_{1}B_{2}=B_{2}B_{3}$

• Join $B_{3}$ to $C$ and than draw a line from $B_{2}$ that is parallel to $B_{3}C$ that parallel line intersect our line $BC$ at $Q$.

        So, $B_{2}Q\parallel B_{3}C$

• Draw line From $Q$ as parallel to $AC$ that intersect line $AB$ at $R$.  

        Here vertex $A$ is same as vertex $P$.

Now, We can see that $\triangle PQR$ is smaller than $\triangle ABC$.