Question

Draw a triangle with sides 5 cm, 6 cm and 6 cm. Draw a square having the same area as the triangle.

Answer 1

(A)
Steps of Construction for triangle:
1. Draw a line segment BC of length 5cm.
2. Measure distance of 6 cm in compass. Taking B as centre draw an arc.
3. Taking C as centre with same distance and draw another arc.
4. Both the arcs intersect each other; name that point of intersection as A.
5. Required triangle ABC is drawn.
Area of $\Delta$ABC can be calculated using Heron's formula.

$s=\frac{5+6+6}{2}$

$=\frac{17}{2}cm$

Therefore, Area $△ABC=\sqrt{\frac{17}{2}(\frac{17}{2}-5)(\frac{17}{2}-6)(\frac{17}{2}-6)}$

$=\sqrt{\frac{17}{2}\times \frac{7}{2}\times \frac{5}{2}\times \frac{5}{2}}$

$=13.6c{m}^2$

Therefore, Area of Square $=(side{)}^2$

$\Rightarrow 13.63=(side{)}^2$

$\therefore side=3.7cm$

(B)
Now, Steps of Construction for square:

1.Draw segment BC of length 3.7cm.
2.Draw a line perpendicular to BC through B.
Measure a distance of 3.7 cm in compass and cut this line at point A.
3.Draw a line perpendicular to BC through C.
Measure a distance of 3.7 cm in compass and cut this line at point D.
4.Join AB, CD, and AD.
5.Required square ABCD is drawn.