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Construct an isosceles triangle whose base is $8\; cm$ and altitude $4\; cm$ and then another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the isosceles triangle

Answers (1)

First we need to construct an isosceles $\Delta ABC$ where ba is $8\: cm$ and altitude $=4\: cm$

Then we have to attend the $\Delta ABC$ to make a triangle which is $\frac{3}{2}$ of $\Delta ABC$

Steps for construction =

(i) Draw a line of $8\: cm$ using scale name it $BC$

(ii) Draw a $\perp$ direction using compars.

(iii) now take $O$ as centre, make $4\; cm$ on the direction like.

(iv) join the point wih $B$ and $C$ .

(V) we will get isosceles $\Delta ABC$

Scale factor $=\frac{3}{2}$

(Vi) Now to make a $\Delta$ which is $\frac{3}{2}\Delta ABC,$ draw a ray $BX$ which is making an acute angle with $BC$

(Vii) Make 3 points on $BX$ $(B_{1},B_{2},B_{3})$ such that $BB_{1}=B_{1}B_{2}=B_{2}B_{3}$

(viii) Join $B_{2}C$ and draw a line $B_{3}$ parallel to $B_{2}C$ to intersect $BC$ extended at $C'$ .

(ix) Draw a line through $C'$ parallel to the line $AC$ to intersect $AB$ extended at $A'$ .

(x) We will get $\Delta AB'C'$ which is $\frac{3}{2}\; \Delta ABC$

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Safeer PP

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