Question

Draw a right angled triangle in which the sides are of length 3cm and 4 cm.
Then construct another triangle whose sides are $ \dfrac{5}{3} $ times the corresponding sides of the given triangle.

Answer 1

Hint: In the above question we have to construct a right triangle.
So, here we will leave how to make a right angle, so that we can construct a triangle from it.
1. Draw a line sequent of given length PQ
2. Taking P as a centre and some radius draw an arc which cuts PQ at B.
3. Place point of compass at B cut an arc of radius PB that cuts the arc drawn in step 2, at $ {O_1} $
4. With the point of compass on $ {O_1} $ draw an arc of radius PB such that it cuts the arc drawn in step 2, at $ {O_2} $
5. Now from point $ {O_1} $ and $ {O_2} $ respectively and radius PQ cut arcs and name the intersection points as O.
6. Join O and P.
7. $ \angle OPQ = 90^\circ $

Complete step-by-step answer:
First we draw a rough sketch

Now, by using the following steps we will construct the above triangle.
1. Draw a line segment (Base) BC of 3 cm.
2. Draw an angle of $ 90^\circ $ at point B, $ \angle B = 90^\circ $
3. Taking B as a centre and 4 cm as radius drawn an arc, let the point where arc intersects the ray be point A
4. Join AC
5. $ \Delta ABC $ , where $ AB = 4cm,BC = 3cm $ and $ \angle ABC = 90^\circ $ is the required triangle.

Now we need to make a triangle which is $ \dfrac{5}{3} $ times of its size.
Steps –
1. Draw a ray BX making an acute angle with BC on the side opposite to vertex A.

2. Mark 5 points (greater of $ \dfrac{5}{3} $ is 5) i.e., $ {B_1},{B_2},{B_3},{B_4} $ and $ {B_5} $ on BC. So that $ B{B_1} = B{B_2} = B{B_3} = B{B_4} = B{B_5} $
3. Join $ {B_3} $ and C (smaller of $ \dfrac{5}{3} $ is 3) and draw a line through $ {B_5} $ parallel to $ {B_3}C $ to intersect BC extended at C’
4. Draw a line through C’ parallel to AC to intersect AB extended at A’.
5. Thus A’BC’ is the required triangle.

Note: 1. In every question related to construction we should make a rough figure. It will make our construction easier.
2. To check whether the constructed triangle is right or not we can use Pythagoras theorem. First measure the side by scale and secondly we will find the unknown side by hypotenuse.
3. To check both the triangles are in the ratio of $ \dfrac{5}{3} $ . We can use the similarity of triangles: the ratio of sides of both the triangles should be in the ratio of $ \dfrac{5}{3} $ .