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Maths 4 Mark Question

Constructions · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 16 questions.

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4 Mark Question

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

Question 14 Marks
Draw $\triangle\text{ABC}$ in which BC = 8cm, $\angle\text{B}=50^\circ$ and $\angle\text{A}=50^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment BC of length 8cm.
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 24 Marks
Construct a right triangle ABC in which AB = 5.8cm, BC = 4.5cm and $\angle\text{C}=90^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment BC = 4.5cm.
Step II: Draw $\angle\text{BCX}$ of measure 90°.
Step III: With centre B and radius AB = 5.8cm, 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 34 Marks
Draw an equilateral triangle one of whose sides is of length 7cm.
Answer

Steps of construction:
Step I: Draw a line segment AB of length 7cm.
Step II: With centre A, draw an arc of radius 7cm.
Step III: With centre B, draw an arc of radius 7cm intersecting the previously drawn arc at C.
Step IV: Join AC and BC to get the required triangle.
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Question 44 Marks
Draw a right triangle having hypotenuse of length 5.4cm, and one of the acute angles of measure 30°.
Answer
Let ABC be the right triangle at A such that hypotenuse BC = 5.4cm. Let C = 30°.
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 = 5.4cm. Step II: Draw angle CBY = 60°. Step III: Draw angle BCX of measure 30° 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 54 Marks
Draw $\triangle\text{ABC}$ in which AB = 3cm, BC = 5cm and $\angle\text{Q}=70^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment AB of length 3cm.
Step II: Draw $\angle\text{XBA}=70^\circ.$
Step III:Cut an arc on BX at a distance of 5cm at C.
Step IV: Join AC to get the required triangle.
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Question 64 Marks
Draw a right triangle with hypotenuse of length 5cm and one side of length 4cm.
Answer

Steps of construction:
Step I: Draw a line segment QR = 4cm.
Step II: Draw $\angle\text{QRX}$ of measure 90°.
Step III: With centre Q and radius PQ = 5cm, 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 74 Marks
Construct a right triangle, right angled at C in which AB = 5.2cm and BC= 4.6cm.
Answer

Steps of construction:
Step I: Draw a line segment BC = 4.6cm.
Step II: Draw $\angle\text{BCX}$ of measure 90°.
Step III: With centre B and radius AB = 5.2cm, 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 84 Marks
Draw $\triangle\text{ABC}$ in which$\angle\text{C}=90^\circ$ and AC = BC = 4cm.
Answer

Steps of construction:
Step I: Draw a line segment BC of length 4cm.
Step II: At C, draw $\angle\text{BCY}=90^\circ.$
Step III: Cut an arc on CY at a distance of 4cm at A.
Step IV: Join AB. ABC is the required triangle.
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Question 94 Marks
Draw a right triangle whose hypotenuse is of length 4cm and one side is of length 2.5cm.
Answer

Steps of construction:
Step I: Draw a line segment QR = 2.5cm.
Step II: Draw $\angle\text{QRX}$ of measure 90°.
Step III: With centre Q and radius PQ = 4cm, 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 104 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 = 6cm.
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 114 Marks
Construct $\triangle\text{ABC}$ in which BC = 4cm, $\angle\text{B}=50^\circ$ and $\angle\text{C}=70^\circ.$
Answer
Steps of construction:
Step I: Draw a line segment BC of length 4cm. 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 124 Marks
Construct $\triangle\text{ABC}$ in which AB = 6.4cm, $\angle\text{A}=45^\circ$ and $\angle\text{B}=60^\circ.$
Answer

Steps of construction:
Step I: Draw a line segment AB = 6.4cm.
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 134 Marks
Draw $\triangle\text{ABC}$ in which $\angle\text{A}=70^\circ,$ AB = 4cm and AC= 6cm. Measure BC.
Answer

Steps of construction:
Step I: Draw a line segment AC of length 6cm.
Step II: Draw $\angle\text{XAC}=70^\circ.$
Step III: Cut an arc on AX at a distance of 4cm at B.
Step IV: Join BC to get the desired triangle.
Step V: We see that BC = 6cm.
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Question 144 Marks
Draw an isosceles triangle in which each of the equal sides is of length 3cm and the angle between them is 45°.
Answer

Steps of construction:
Step I: Draw a line segment PQ of length 3cm.
Step II: Draw $\angle\text{QPX}=45^\circ.$
Step II: Cut an arc on PX at a distance of 3cm at R.
Step IV: Join QR to get the required triangle.
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Question 154 Marks
Draw $\triangle\text{PQR}$ in which $\angle\text{Q}=80^\circ,$ $\angle\text{R}=55^\circ$ and QR = 4.5cm. Draw the perpendicular bisector of side QR.
Answer
Steps of construction:
Step I: Draw a line segment QR = 4.5cm. 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 2.25cm, draw arcs on either sides of QR. Step V: With R as centre and radius more than 2.25cm, 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 164 Marks
Draw $\triangle\text{ABC}$ in which $\angle\text{A}=120^\circ,$ AB = AC = 3cm. Measure $\angle\text{B}$ and $\angle\text{C}.$
Answer

Steps of construction:
Step I: Draw a line segment AC of length 3cm.
Step II: Draw $\angle\text{XAC}=120^\circ.$
Step III: Cut an arc on AX at a distance of 3cm 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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