Question

Draw a triangle ABC with side base BC=8cm and altitude 4cm and then construct another triangle whose sides are 5 by 3 times the
corresponding sides of the isosceles triangle ABC
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Answer 1

Step-by-step explanation:

arcs do with compass ok.

Answer 2

$\triangle A^{'}BC^{'}$ is the triangle whose sides are $\frac{5}{3}$ to the $\triangle ABC$

Step-by-step explanation:

Given,

$\triangle ABC$ where base $BC=8\ cm$ and altitude $4\cm$ and construct another triangle whose sides are $\frac{5}{3}$ time of the sides of $\triangle ABC$

Following steps:

• Draw a base line $BC=8\ cm$ using scale.
• Draw a perpendicular line from mid point of the line $BC$ like $XD\perp BC$
• $D$ as center and $4\ cm$ as radius draw an arc on line $DX$ which is known as $A$
• Join $AB$ and $AC$
• Below the line $BC$ draw any angle like $\angle CBY=\theta$
• Divide the line $BY$ into $8$ equal parts like $B_{1},B_{2},......,B_{8}$

• Join $B_{3}C$

• Draw a point $B_{8}$ draw a parallel line with $B_{3}C$ to intersect $BC$ extended at $C^{'}$

• Draw a line through $C^{'}$ parallel to the line $AC$ to intersect $AB$ extended at $A^{'}$

Now,  $\triangle A^{'}BC^{'}$ is the triangle whose sides are $\frac{5}{3}$ to the $\triangle ABC$