Embibe Experts Solutions for Chapter: Constructions, Exercise 1: Exercise

To construct a triangle similar to a given $△\mathrm{ABC}$ with its sides $\frac{3}{7}$ of the corresponding sides of $△\mathrm{ABC}$, first draw a ray $\mathrm{BX}$ such that $\angle \mathrm{CBX}$ is an acute angle, and $\mathrm{X}$ lies on the opposite side of $\mathrm{A}$ with respect to $\mathrm{BC}$. Then locate points ${\mathrm{B}}_1, {\mathrm{B}}_2, {\mathrm{B}}_3, ...$ on $\mathrm{BX}$ at equal distances and next step is to join

To construct a triangle similar to a given $△\mathrm{ABC}$ with its side $\frac{8}{5}$ of the corresponding sides of $△\mathrm{ABC}$, we draw a ray $\mathrm{BX}$ such that $\angle \mathrm{CBX}$ is an acute angle, and $\mathrm{X}$ is on the opposite side of $\mathrm{A}$ with respect to $\mathrm{BC}$. The minimum number of points to be located at equal distances on ray $\mathrm{BX}$  is

To divide a line segment $\mathrm{AB}$ in the ratio $5:6$, draw a ray $\mathrm{AX}$ such that $\angle \mathrm{BAX}$ is an acute angle, then a ray $\mathrm{BY}$ parallel to $\mathrm{AX}$ and the points ${\mathrm{A}}_1, {\mathrm{A}}_2, {\mathrm{A}}_3, ...$ and ${\mathrm{B}}_1, {\mathrm{B}}_2, {\mathrm{B}}_3, ...$ are located at equal distances on ray $\mathrm{AX}$ and $\mathrm{BY}$ respectively. Then the point joined are:

In the figure given below, $\mathrm{AM}: \mathrm{MC}= 3 : 4$,  $\mathrm{BP}: \mathrm{PM}= 3 : 2$ and $\mathrm{BN} = 12 \mathrm{cm}$. Then, $\mathrm{AN}$ equals to:

Which of the following steps of construction is INCORRECT while dividing a line segment $\mathrm{A}\mathrm{B}$ of length $7.6\mathrm{ }\mathrm{c}\mathrm{m}$ in the ratio of $5:8$ internally.

Steps of Construction:

Step I: Draw line segment $\mathrm{A}\mathrm{B}=7.6\mathrm{ }\mathrm{c}\mathrm{m}$

Step II: Construct an acute $\mathrm{\angle }\mathrm{B}\mathrm{A}\mathrm{X}$ .

Step III: Locate $13\left(=5+8\right)$ points ${\mathrm{A}}_1,{\mathrm{A}}_2,\mathrm{ }{\mathrm{A}}_3,{\mathrm{A}}_4.\mathrm{ }.\mathrm{ }.\mathrm{ }.\mathrm{ }{\mathrm{A}}_{13}$ on $\mathrm{AX}$ so that

${\mathrm{AA}}_1={\mathrm{A}}_1{\mathrm{A}}_2={\mathrm{A}}_2{\mathrm{A}}_3.\mathrm{ }.\mathrm{ }.\mathrm{ }={\mathrm{A}}_{12}{\mathrm{A}}_{13}$

Step IV: Join $\mathrm{B}$ to ${\mathrm{A}}_8.$ From ${\mathrm{A}}_5$ draw ${\mathrm{A}}_5\mathrm{C}$ parallel to ${\mathrm{A}}_8\mathrm{B}$ . Point $\mathrm{C}$ divides $\mathrm{A}\mathrm{B}$ internally in the ratio $5:8$ .

Thus, $\mathrm{A}\mathrm{C}:\mathrm{C}\mathrm{B}=5:8.$ Now $\mathrm{C}$ is the point dividing line segment $\mathrm{A}\mathrm{B}$ of $7.6\mathrm{ }\mathrm{c}\mathrm{m}$ in the required ratio of $5:8$ .

Arrange the following steps of construction while dividing a line segment length $12 \mathrm{cm}$ internally in the ratio $2:5$.

Steps of construction:

Step $1:$ Locate $7\left(=m+n\right)$ points ${\mathrm{A}}_1, {\mathrm{A}}_2,......., {\mathrm{A}}_7$ on $\mathrm{AX}$ such that ${\mathrm{AA}}_1={\mathrm{A}}_1{\mathrm{A}}_2=...........={\mathrm{A}}_6{\mathrm{A}}_7.$

Step $2:$ Draw a line segment $\mathrm{AB}$ of length $12 \mathrm{cm}$.

Step $3:$ Through ${\mathrm{A}}_2\left(m=2\right)$, draw a line parallel to ${\mathrm{A}}_7\mathrm{B}$ (by making an angle equal to ${\mathrm{AA}}_7\mathrm{B}$) at ${\mathrm{A}}_2$ so as to intersect $\mathrm{AB}$ at the point $\mathrm{C}$.

Step $4:$ Join ${\mathrm{BA}}_7$.

Step $5:$ Draw any ray $\mathrm{AX}$, making an acute angle with $\mathrm{AB}$.

Divide the line segment $\mathrm{AB}=10 \mathrm{cm}$ in the ratio $2:3$ geometrically.

Construct a triangle of sides $6 \mathrm{cm}, 4 \mathrm{cm} \text{and} 7 \mathrm{cm}$ then construct another triangle whose corresponding sides are $\frac{3}{4}$ of the sides of the first triangle.