Question

Draw triangle abc right angled at b such that ab is equal to 3 cm bc is equal to 4 cm now construct a triangle pbq similar to triangle abc each of whose side is 7 upon 4 times the corresponding sides of the triangle abc

Answer 1

$\triangle abc$ is the required triangle and $\triangle pbq$ is the another triangle whose each sides $\frac{7}{4}$ times of sides of the $\triangle abc$.

Step-by-step explanation:

Draw $\triangle abc$ where $b$ is right angle,$ab=3\ cm$ and $bc=4\ cm$

Draw another $\triangle pbq$ whose each sides $\frac{7}{4}$ times of sides of the $\triangle abc$

Following steps:

• Draw line segments $bc=4\ cm$ using scale.
• At $b$, draw an angle $90\ degree$ using protector like $\angle cbx=90\ degree$
• $b$ as center and $3\ cm$ as radius draw an arc on line $bx$ name as $a$
• Join $ac$

Hence $\triangle abc$ is the required triangle.

• Divide side $bc$ into $7$ different parts name as $b_{1},b_{2},.....,b_{6},b_{7}$ as shown in the figure.

• Join $cb_{7}$ and draw parallel line with $cb_{7}$ through point $b_{4}$ which intersect the extended line $bc$ at $q$
• Draw another parallel line with $ac$ through point $q$ which intersect the extended line $bx$ at $p$

Hence, $\triangle pbq$ is the another triangle whose each sides $\frac{7}{4}$ times of sides of the $\triangle abc$.