Question

Construct a triangle with sides $5cm,6cm$ and $7cm$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle.

Answer 1

Constructing the first triangle:

Step 1: Draw base $AB$ of length $7cm$

Step 2: Place a compass on $A$ and draw an arc of radius $5cm$

Step 3: Place the compass on $B$ and draw an arc of radius $6cm$ such that it intersects the arc drawn in step $2$.

Step 4: Mark the intersection of the two arcs as $C$ and join $A-C$ and $B-C$

$△ABC$ is the given triangle.

Let $△PAQ$ be the required triangle.

Step 1: Draw a ray $AX$ from point $A$ on the opposite side of $C$, making an acute angle with $AB$.

Step 2: Mark $7$ equidistant points $A_1,A_2,A_3,...,A_7$ on ray $AX$.

Step 3: Join $A_5$ to $B$.

Step 4: Draw $A_{7}Q$ parallel to $A_{5}B$ and mark the point on the extension of line $AB$ as $Q$.

Step 5: Draw $PQ$ parallel to $BC$ and mark the intersection point on the extension of $AC$ as $P$.

$△PAQ$ is the required triangle.