5 Marks Questions - Maths STD 7 Questions - Vidyadip

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5 Marks Questions

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

Question 15 Marks

Construct a right triangle having hypotenuse of length $5.6\ cm$ and one of whose acute angles measures $30^\circ .$

Answer

Here, $\angle\text{A}=30^\circ$ and $\angle\text{C}=90^\circ$ By angle sum property: $\angle\text{B}=60^\circ$
Steps for construction:
Step I: Draw the hypotenuse $AB$ of length $5.6\ cm.$
Step II: Draw $\angle\text{BAX}=30^\circ$ and $\angle\text{ABY}=60^\circ$
Step III: The ray $AX$ and $BY$ intersect at $C.$ Then, $ABC$ is the required triangle.

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Question 25 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 6.2\ cm$,$\angle\text{B}=60^\circ.$ and $\angle\text{C}=45^\circ.$

Answer

Steps for construction:
Step I: Draw $BC = 6.2 cm$
Step II: Draw $\angle\text{BCX}=45^\circ$
Step III: Draw $\angle\text{BCX}=60^\circ$
Step IV: The ray $CX$ an $BY$ intersect at $A$. Then, $ABC$ is the required triangle.

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Question 35 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 4.8\ cm$,$\angle\text{C}=90^\circ.$and $AB = 6.3\ cm.$

Answer

Steps for construction:
Step I: Draw $BC = 4.8\ cm$
Step II: Draw a perpendicular on $C$ such that $\angle\text{C}$ is equal to $90°.$
Step III: Draw an arc of radius $6.3\ cm$ from the centre $B.$
Step IV: Join $AB.$

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Question 45 Marks

Construct an equilateral triangle each of whose sides measures $6.2\ cm$. Measure each of its angles.

Answer

Steps for construction:
Step I: Draw $AB$ of length $6.2\ cm.$
Step II: By taking the centres as $A$ and $B$, draw equal arcs of $6.2\ cm$ on the same side of $AB$, cutting each other at $C.$
Step III: Join $AC$ and $BC$. When we will measure angles of triangle using protractor then we find that all angles are equal to $60^\circ .$

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Question 55 Marks

Construct a right-angled triangle one side of which measures $3.5\ cm$ and the length of whose hypotenuse is $6\ cm.$

Answer

Steps for construction: Step I: Draw $AB = 3.5\ cm$
Step II: Construct $\angle\text{ABX}=90^\circ$
Step III: With centre $A$, draw an arc of radius $6 \ cm$ cutting $BX$ at $C.$
Step IV: Join $AC$. Then, $ABC$ is the required triangle.

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Question 65 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 4.3\ cm$, $\angle\text{C}=45^\circ$ and $AC = 6\ cm.$

Answer

Steps for construction:
Step I: Draw $AC = 6\ cm.$
Step II: Draw $\angle\text{ACZ}=45^\circ$
Step III: With $C$ as the centre, cut ray $BZ$ at $4.3\ cm$ at point $B.$
Step IV: Join $AB$.Then, $ABC$ is the required triangle.

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Question 75 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 7\ cm$,$\angle\text{A}=45^\circ.$and $\angle\text{C}=75^\circ.$

Answer

By angle sum property:$\angle\text{B}=180^\circ-\angle\text{A}-\angle\text{C}$
$=180^\circ-45^\circ-75^\circ$
$=60^\circ$
Steps for construction:
Step I: Draw $AB = 7\ cm$
Step II: Draw $\angle\text{BAX}=45^\circ$
Step III: Draw $\angle\text{ABY}=60^\circ$
Step IV: The ray $AX$ an $BY$ intersect at $C$.
Then, $ABC$ is the required triangle.

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Question 85 Marks

Construct a $\triangle\text{ABC}$ in which $AB = 3.8cm$, $\angle\text{A}=60^\circ$ and $AC = 5cm.$

Answer

Steps for construction:
Step I: Draw $AB$ of length $3.8\ cm$.
Step II: Draw $\angle\text{BAZ}=60^\circ$.
Step III: With the centre as $A$, cut ray $AZ$ at $5\ cm$ at $C.$
Step IV: Join $BC$. Then, $ABC$ is the required triangle.

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Question 95 Marks

In the adjoining figure, $ABC$ is a triangle in which $AD$ is the bisector of $\angle\text{A}.$ If $\text{AD}\bot\text{BC}$ show that $\triangle\text{ABC}$ is isosceles.

Answer

Given:
$\angle\text{BAD}=\angle\text{DAC}\ \dots(\text{i})$
To show that $\triangle\text{ABC}$ is isoceles, we should show that $\angle\text{B}=\angle\text{C}.$
$\therefore\text{AD}\bot\text{BC},\angle\text{ADB}=\angle\text{ADC}=90^\circ$
$\angle\text{ADC}=\angle\text{ADB}$
$\angle\text{BAD}+\angle\text{ABD}=\angle\text{DAC}+\angle\text{ACD}$ (exterior angle property)
$\angle\text{DAC}+\angle\text{ABD}=\angle\text{DAC}+\angle\text{ACD}$ [from equation (i)]
$\angle\text{ABD}=\angle\text{ACD}$
This is because opposite angles of a triangle $\triangle\text{ABC}$ are equal.
Hence, $ABS$ is an isosceles triangle.

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Question 105 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 5.8\ cm$,$\angle\text{B}=\angle\text{C}=30^\circ.$ Measure $AB$ and $AC$. What do you observe?

Answer

Steps for construction:
Step I: Draw $BC = 5.8cm$
Step II: Draw $\angle\text{BCY}=30^\circ$
Step III: Draw $\angle\text{CBX}=30^\circ$
Step IV: The ray $BX$ an $CY$ intersect at $A$. Then, $ABC$ is the required triangle. On measuring $AB$ and $AC: AB = AC = 3.4cm.$

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Question 115 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 5.3cm$, $\angle\text{B}=60^\circ$ and $AB = 4.2cm$. Also, draw the perpendicular bisector of $AC.$

Answer

Steps of construction:
Step 1: Draw $BC = 5.3cm$
Step 2: Construct $\angle\text{CBX}=60^\circ$
Step 3: With $B$ as the centre and radius $4.2cm$, cut the ray $BX$ at point $A.$
Step 4: Join $A$ and $C.$
Step 5: With $A$ as the centre and radius more than half of $AC$, draw an arc on either side of $AC.$
Step 6: With $C$ as the centre and the same radius, draw another arc cutting the previously drawn arc at $M$ and $N.$
Step 7: Join $M$ and $N.$

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Question 125 Marks

Construct a $\triangle\text{ABC}$ in which $AB = AC = 4.8\ cm$ and $BC = 5.3\ cm$. Measure $\angle\text{B}$ and $\angle\text{C}$. Draw $\text{AD}\bot\text{BC}$.

Answer

Steps for construction:
Step I: Draw $BC = 5.3cm$
Step II: Draw an arc of radius $4.8cm$ from the centre, $B.$
Step III: Draw another arc of radius $4.8cm$ from the centre, $C.$
Step IV: Both of these arcs intersect at $A$.
Step V: Join $AB$ and $AC.$
Step VI: With $A$ as the centre and any radius, draw an arc cutting $BC$ at $M$ and $N.$
Step VII: With $M$ as the centre and the radius more than half of $MN$, draw an arc.
Step VIII: With $N$ as the centre and the same radius, draw another arc cutting the previously drawn arc at $P$.
Step IX: Join $AP$, cutting $BC$ at $D$.Then, $\text{AD}\bot\text{BC}$

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Question 135 Marks

Construct a $\triangle\text{ABC}$ in which $BC = 3.6\ cm, AB = 5\ cm$ and $AC = 5.4\ cm.$ Draw the perpendicular bisector of the side $BC.$

Answer

Steps for construction:
Step I: Draw a line segment $(AB)$ of length $5\ cm.$
Step II: Draw an arc of radius $5.4\ cm$ from the centre $(A).$
Step III: With $B$ as the centre, draw another arc of radius $3.6\ cm,$ cutting the previous arc at $C.$
Step IV: Join $AC$ and $BC.$
Step V: Taking $B$ as the centre and the radius more than half of $BC,$ draw two arcs on both the sides of $BC.$
Step VI: Similarly, taking $C$ as the centre and the same radius, draw arcs on both the sides of $BC,$ cutting the previous arcs at $P$ and $Q.$ Then, $PQ$ is the required perpendicuar bisector of $BC,$ meeting $BC$ at $D.$

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Question 145 Marks

Draw a line $AB$ and take a point $P$ outside it. Draw a line $CD$ parallel to $AB$ and passing through the point $P.$

Answer

Steps of construction:
Step I: Draw a line $AB.$
Step II: Take a point $Q$ on $AB$ and a point $P$ outside $AB$, and join $PQ.$
Step III: With $Q$ as the centre an any radius, draw an arc to cut $AB$ at $X$ and $PQ$ at $Z.$
Step IV: With $P$ as the centre and the same radius, draw an arc cutting $QP$ at $Y.$
Step V: With $y$ as the centre and the radius equal to $XZ,$ draw an arc to cut the pevious arc at $E.$
Step VI: Join $PE$ and produce it on both the sides to get the required line.

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Question 155 Marks

Construct a $\triangle\text{PQR}$ in which $QR = 6\ cm, PQ = 4.4\ cm$ and $PR = 5.3\ cm.$ Draw the bisector of $\angle\text{P.}$

Answer

Steps for construction:
Step I: Draw a line segment $QR$ of length $6\ cm.$
Step II: Draw arcs of $4.4\ cm$ and $5.3\ cm$ from $Q$ and $R,$ respectively. They intersect at $P.$
Step III: Draw an arc of any radius from the centre $(P),$ cutting $PQ$ and $PR$ at $S$ and $T,$ respectively.
Step IV: With $S$ as the centre and the radius more than half of $ST,$ draw an arc.
Step V: With $T$ as the centre and the same radius, draw another arc cutting the previously drawn arc at $X.$
Step VI: Join $P$ and $X.$ Then, $PX$ is the bisector of $\angle\text{P}$.

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Question 165 Marks

Draw a line l and draw another line m parallel to l at a distance of $4.3\ cm$ from it.

Answer

Steps for construction:
Step I: Let $l$ be the given line.
Step II: Take any two points $A$ and $B$ on line $l.$
Step III: Construct $\angle\text{BAE}=90^\circ$ and $\angle\text{ABF}=90^\circ$.
Step IV: With $A$ as the centre and the radius equal to $4.3\ cm,$ cut $AE$ at $C.$
Step V: With $B$ as the centre and the radius equal to $4.3\ cm$, cut $BF$ at $D.$
Step VI: Join $CD$ and produce it on either side to get the required line $m,$ parallel to l and at a distance of $4.3\ cm$ from it.

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Question 175 Marks

Draw a line $AB$ and draw another line $CD$ parallel to $AB$ at a distance of $3.5\ cm$ from it.

Answer

Steps for construction:
Step I: Let $AB$ be the given line.
Step II: Take any two points $P$ and $Q$ on $AB.$
Step III: Construct $\angle\text{BPE}=90^\circ$ and $\angle\text{BQF}=90^\circ$.
Step IV: With $P$ as the centre and the radius equal to $3.5\ cm,$ cut $PE$ at $R$.
Step V: With $Q$ as the centre and the radius equal to $3.5\ cm,$ cut $QF$ at $S.$
Step VI: Join $RS$ and produce it on both the sides to get the required line, parallel to $AB$ and at a distance of $3.5\ cm$ from it.

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Question 185 Marks

Construct a $\triangle\text{ABC}$ in which $AB = AC = 5.2\ cm$ and $\angle\text{A}=120^\circ.$ Draw $\text{AD}\bot\text{BC}.$

Answer

Steps for construction:
Step I: Draw $AB = 5.2\ cm$
Step II: Draw $\angle\text{BAX}=120^\circ$
Step III: With $A$ as the centre, cut the ray $AX$ at $5.3\ cm$ at point $C.$
Step IV: Join $BC.$
Step V: $A$ as the centre and any radius, draw an arc cutting $BC$ at $M$ and $N.$
Step VI: With $M$ as the centre and the radius more than half of $MN,$ draw an arc.
Step VII: With $N$ as the centre and the same radius as before, draw another arc cutting the previously drawn arc at $P.$
Step VIII: Join $AP$ meeting $BC$ at $D.$
$\therefore\text{AD}=\text{BC}$

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