Construction of triangles - class-VIII
Description: Construction of triangles | |
Number of Questions: 16 | |
Created by: Girish Goud | |
Tags: geometry constructions construction of triangles maths geometrical constructions constructions of triangles |
Construct a triangle $PQR$, whose perimeter is $22 cm$ and whose sides are in the ratio $2 : 4 : 5$. Measure the sides of the triangle.
Construct a $\triangle ABC$ in which $AB= 5.4\ cm, \angle CAB= 45^{\circ}$ and $AC + BC= 9\ cm.$Then, $m\angle ACB$ is:
For constructing a triangle whose perimeter and both base angles are given, the base length is equal to:
Choose the correct statement:
For constructing a triangle when the base, one base angle and the difference between lengths of other two sides are given, the base length is equals to:
The construction of a $\Delta ABC$ in which $BC=6$ $cm$ and $\angle B=50^\circ$, is not possible when $(AB-AC)$ is equal to:
The construction of $\Delta EFG$ when $FG=3$ $cm$ and m$\angle G=60^\circ$ is possible when difference of $EF$ and $EG$ is equal to:
For constructing a triangle whose perimeter and both base angles are given, the first step is to:
The construction of $\triangle ABC$ in which $AB = 6\ cm, \angle A = 30^\circ$, is not possible when $AC+BC = $
The construction of $\triangle ABC$ in which $AB = 5\ cm, \angle A = 45^\circ$, is possible when $AC+BC = $
Construct a $\triangle ABC$ in which:
$AB= 5.4\ cm$, $\angle CAB= 45^{0}$ and $AC\, +\, BC= 9\ cm$. Then the length of $AC$ (in $cm.$) is:
The construction of $\Delta LMN$ when $MN=7$ $cm$ and $m\angle M=45^\circ$ is not possible when difference of $LM$ and $LN$ is equal to:
Which of the following could be the value of $AC-BC$ in the construction of a triangle $ABC$ in which base $AB = 5 cm, \angle A = 30^{\circ}$?
The construction of $\Delta LMN$ when $MN=6$ $cm$ and $m\angle M=45^\circ$ is not possible when difference between $LM$ and $LN$ is equal to:
To construct a triangle similar to a given triangle ABC with its sides 6/5th of the corresponding sides of $\Delta$ABC. Correct order of steps of construction -
(a) Draw a ray AX inclined at certain angle with AB on opposite side of C.
(b) Starting from A, cut off six equal line segments AX$ _1$, X$ _1$X$ _2$, X$ _2$X$ _3$, X$ _3$X$ _4$, X$ _4$X$ _5$ and X$ _5$X$ _6$ on AX.
(c) Draw a line B'C' parallel to BC to intersect AC produced at C'
(d) Join X$ _5$B and draw a line X6B' parallel to X5B, to intersect AB produced at B'.
Construct a $\Delta ABC$, whose perimeter is $10.5 cm$ and base angles are $60^o$ and $45^o$. Find the third angle.