Question

(i) Draw a line segment of length 8 cm and divide it internally in the ratio 4 : 5.

(ii) Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8 Measure the two parts.

Answer 1

(i)

1) Draw a line segment AB = 8 cm.

2) Draw a ray AX making an acute angle $\angle$ BAX=60° with AB.

3) Draw a ray BY parallel to AX by making an acute angle $\angle$ ABY=$\angle$ BAX.

4) Mark four points $A_1,A_2,A_3,A_4$ on AX and five points $B_1,B_2,B_3,B_4,B_5$ on BY in such a way that
$A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4.$

5) Join $A_4{B}_5$

6) Let this line intersect AB at a point P

Thus, P is the point dividing the line segment AB internally in the ratio of 4:5.

(ii)

1) Draw a line segment AB = 7.6 cm.

2) Draw a ray AX making an acute angle $\angle$ BAX along with AB.

3) On AX make 5 + 8 i.e. 13 equal parts and mark them as $A_1,A_2,A_3,A_4,...A_{13}$

4) Join B to $A_{13}$. From $A_5$ draw $A_{5}C\parallel A_{13}B$.

C is the required point of division and AC: CB = 5 : 8.

On measuring, we get

AC = 3.1 cm,

CB = 4.5 cm

Justification

$A_{5}C\parallel A_{13}B$

$\frac{A_{5}C}{A_5{A}_{13}}=\frac{AC}{CB}$

[Using basic proportionality theorem]

But $\frac{A_{5}C}{A_5{A}_{13}}=\frac{5}{8}$

This shows that C divides AB in the ratio 5 : 8.