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Question 13 Marks
Construct a quadrilateral $ABCD$ given that $AB = 4\ cm, BC = 3\ cm,$
$\angle\text{A}=75^\circ,$
$\angle\text{B}=80^\circ$ and $\angle\text{C}=120^\circ.$
Answer


Steps of construction:
Step I: Draw $AB = 4\ cm.$
Step II: Construct $\angle\text{XAB}=75^\circ$ at $A$ and $\angle\text{ABY}=80^\circ$ at $B.$​​​​​​​
Step III: With $B$ as the centre and radius $3\ cm,$ cut off $BC = 3\ cm.$​​​​​​​
Step IV: At $C,$ draw $\angle\text{BCZ}=120^\circ$ such that it meets $AX$ at $D.$
The quadrilateral so obtained is the required quadrilateral.
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Question 23 Marks
Construct, if possible, a quadrilateral $ABCD$ in which $AB = 6\ cm, BC = 7\ cm, CD = 3\ cm, AD = 5.5\ cm$ and $AC = 11\ cm.$ Give reasons for not being able to construct, if you cannot. $($Not possible, because in triangle $ACD, AD + CD < AC).$
Answer
Such a quadrilateral cannot be constructed because in a triangle, the sum of the length of its two sides must be greater than the that of the third side But here in triangle $ACD. AD + CD = 5.5 + 3 = 8.5\ cm$ and $AC = 11\ cm.$
i.e., $AD + CD < AC,$ which is not possible. So, the construction is not possible.
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Question 33 Marks
Construct a quadrilateral $ABCD,$ when $AB = 3\ cm, CD = 3\ cm, DA = 7.5\ cm, AC = 8\ cm$ and $BD = 4\ cm.$
Answer
If we consider a triangle $ABD$ from the given data, then $AB = 3\ cm ,BD = 4\ cm, AD = 7.5\ cm, AB + BD = 3 + 4 = 7\ cm$ However, we know that the sum of the lengths of two sides of a triangle is always greater than the third side. Therefore, construction is not possible from the given data.
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3 Marks Question - MATHS STD 8 Questions - Vidyadip