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Maths 3 Marks Question

Constructions · Jharkhand Board (JAC) STD 7 (Jharkhand - English Medium). 16 questions.

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Draw $\triangle\text{ABC}$ in which $BC = 8\ cm$, $\angle\text{B}=50^\circ$ and $\angle\text{A}=50^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $BC$ of length $8\ cm$.
Step II: Draw $\angle\text{CBX}$ such that $\angle\text{CBX}=50^\circ.$
Step III: Draw $\angle\text{BCY}$ with $Y$ on the same side of $BC$ as $X$ such that $\angle\text{BCY}=80^\circ.$
Step IV: Let $CY$ and $BX$ intersect at $A$.
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Question 23 Marks
Construct a right triangle $ABC$ in which $AB = 5.8\ cm, BC = 4.5\ cm$ and $\angle\text{C}=90^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $BC = 4.5\ cm.$
Step II: Draw $\angle\text{BCX}$ of measure $90^\circ .$​​​​​​​
Step III: With centre $B$ and radius $AB = 5.8\ cm$, draw an arc of the circle to intersect ray $BX$ at $A$.
Step IV: Join $AB$ to obtain the desired triangle $ABC$.
Step V: $ABC$ is the required triangle.
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Question 33 Marks
Draw an equilateral triangle one of whose sides is of length $7\ cm$.
Answer

Steps of construction:
Step I: Draw a line segment $AB$ of length $7\ cm$.
Step II: With centre $A$, draw an arc of radius $7\ cm$.
Step III: With centre $B$, draw an arc of radius $7\ cm$ intersecting the previously drawn arc at $C$.
Step IV: Join $AC$ and $BC$ to get the required triangle.
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Question 43 Marks
Draw a right triangle having hypotenuse of length $5.4\ cm$, and one of the acute angles of measure $30^\circ $.
Answer

Let $ABC$ be the right triangle at $A$ such that hypotenuse $BC = 5.4\ cm$. Let $C = 30^\circ .$
Therefore $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ\angle\text{B}$ $=180^\circ-30^\circ-90^\circ=60^\circ.$
Steps of construction:
Step I: Draw a line segment $BC = 5.4\ cm.$
Step II: Draw angle $CBY = 60^\circ .$​​​​​​​
Step III: Draw angle $BCX$ of measure $30^\circ $ with $X$ on the same side of $BC$ as $Y$.
Step IV: Let $BY$ and $CX$ intersect at $A$.
Step V: Then $ABC$ is the required triangle.
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Question 53 Marks
Draw $\triangle\text{ABC}$ in which $AB = 3\ cm, BC = 5\ cm$ and $\angle\text{Q}=70^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $AB$ of length $3\ cm.$
Step II: Draw $\angle\text{XBA}=70^\circ.$
Step III:Cut an arc on $BX$ at a distance of $5\ cm$ at $C$.
Step IV: Join $AC$ to get the required triangle.
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Question 63 Marks
Draw a right triangle with hypotenuse of length $5\ cm$ and one side of length $4\ cm$.
Answer

Steps of construction:
Step I: Draw a line segment $QR = 4\ cm$.
Step II: Draw $\angle\text{QRX}$ of measure $90^\circ $.
Step III: With centre $Q$ and radius $PQ = 5\ cm$, draw an arc of the circle to intersect ray $RX$ at $P$.
Step IV: Join $PQ$ to obtain the desired triangle $PQR$.
Step V: $PQR$ is the required triangle.
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Question 73 Marks
Construct a right triangle, right angled at $C$ in which $AB = 5.2\ cm$ and $BC= 4.6\ cm.$
Answer

Steps of construction:
Step I: Draw a line segment $BC = 4.6\ cm.$
Step II: Draw $\angle\text{BCX}$ of measure $90^\circ $.
Step III: With centre $B$ and radius $AB = 5.2\ cm$, draw an arc of the circle to intersect ray $CX$ at $A$.
Step IV: Join $AB$ to obtain the desired triangle $ABC$.
Step V: $ABC$ is the required triangle.
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Question 83 Marks
Draw $\triangle\text{ABC}$ in which$\angle\text{C}=90^\circ$ and $AC = BC = 4\ cm.$
Answer

Steps of construction:
Step I: Draw a line segment $BC$ of length $4\ cm$.
Step II: At $C,$ draw $\angle\text{BCY}=90^\circ.$
Step III: Cut an arc on $CY$ at a distance of $4\ cm$ at $A$.
Step IV: Join $AB. ABC$ is the required triangle.
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Question 93 Marks
Draw a right triangle whose hypotenuse is of length $4\ cm$ and one side is of length $2.5\ cm$.
Answer

Steps of construction:
Step I: Draw a line segment $QR = 2.5\ cm$.
Step II: Draw $\angle\text{QRX}$ of measure $90^\circ $.
Step III: With centre $Q$ and radius $PQ = 4\ cm$, draw an arc of the circle to intersect ray $RX$ at $P$.
Step IV: Join $PQ$ to obtain the desired triangle $PQR$.
Step V: $PQR$ is the required triangle.
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Question 103 Marks
Draw $\triangle\text{ABC}$ in which $AC = 6cm,$ $\angle\text{A}=90^\circ$ and $\angle\text{B}=60^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $AC = 6\ cm.$
Step II: Draw $\angle\text{ACX}=30^\circ.$
Step III: Draw $\angle\text{CAY}$ with $Y$ on the same side of $AC$ as $X$ such that $\angle\text{CAY}=90^\circ.$
Step IV: Join $CX$ and $AY$. Let these intersect at $B$.
Step V: $ABC$ is the required triangle where angle $\angle\text{ABC}=60^\circ.$
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Question 113 Marks
Construct $\triangle\text{ABC}$ in which $BC = 4\ cm$, $\angle\text{B}=50^\circ$ and $\angle\text{C}=70^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $BC$ of length $4\ cm$.
Step II: Draw $\angle\text{CBX}$ such that $\angle\text{CBX}=50^\circ.$
Step III: Draw $\angle\text{BCY}$ with $Y$ on the same side of $BC$ as $X$ such that $\angle\text{BCY}=70^\circ.$
Step IV: Let $CY$ and $BX$ intersect at $A$.
Step V: $ABC$ is the required triangle.
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Question 123 Marks
Construct $\triangle\text{ABC}$ in which $AB = 6.4\ cm$, $\angle\text{A}=45^\circ$ and $\angle\text{B}=60^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment $AB = 6.4\ cm$.
Step II: Draw $\angle\text{BAX}=45^\circ.$
Step III: Draw $\angle\text{ABY}$ with $Y$ on the same side of $AB$ as $X$ such that $\angle\text{ABY}=60^\circ.$
Step IV: Let $AX$ and $BY$ intersect at $C$.
Step V: $ABC$ is the required triangle.
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Question 133 Marks
Draw $\triangle\text{ABC}$ in which $\angle\text{A}=70^\circ,$ $AB = 4\ cm$ and $AC= 6\ cm$. Measure $BC$.
Answer

Steps of construction:
Step I: Draw a line segment $AC$ of length $6\ cm$.
Step II: Draw $\angle\text{XAC}=70^\circ.$
Step III: Cut an arc on $AX$ at a distance of $4\ cm$ at $B$.
Step IV: Join $BC$ to get the desired triangle.
Step V: We see that $BC = 6\ cm$.
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Question 143 Marks
Draw an isosceles triangle in which each of the equal sides is of length $3\ cm$ and the angle between them is $45^\circ $.
Answer

Steps of construction:
Step I: Draw a line segment $PQ$ of length $3\ cm.$
Step II: Draw $\angle\text{QPX}=45^\circ.$
Step II: Cut an arc on $PX$ at a distance of $3\ cm$ at $R$.
Step IV: Join $QR$ to get the required triangle.
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Question 153 Marks
Draw $\triangle\text{PQR}$ in which $\angle\text{Q}=80^\circ,$ $\angle\text{R}=55^\circ$ and $QR = 4.5\ cm$. Draw the perpendicular bisector of side $QR$.
Answer

Steps of construction:
Step I: Draw a line segment $QR = 4.5\ cm.$
Step II: Draw $\angle\text{RQX}=80^\circ$ and $\angle\text{QRY}=55^\circ.$
Step III: Let $QX$ and $RY$ intersect at $P$ so that $PQR$ is the required triangle.
Step IV: With $Q$ as centre and radius more that $2.25\ cm$, draw arcs on either sides of $QR$.
Step V: With $R$ as centre and radius more than $2.25\ cm$, draw arcs intersecting the previous arcs at $M$ and $N$.
Step VI: Join $MN$ is the required perpendicular bisector of $QR$.
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Question 163 Marks
Draw $\triangle\text{ABC}$ in which $\angle\text{A}=120^\circ,$ $AB = AC = 3\ cm$. Measure $\angle\text{B}$ and $\angle\text{C}.$
Answer

Steps of construction:
Step I: Draw a line segment $AC$ of length $3\ cm$.
Step II: Draw $\angle\text{XAC}=120^\circ.$
Step III: Cut an arc on $AX$ at a distance of $3\ cm$ at $B$.
Step IV: Join $BC$ to get the required triangle.
Step V: By measuring, we get $\angle\text{B}=\angle\text{C}=30^\circ.$
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