3 Marks Question - MATHS STD 7 Questions - Vidyadip

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3 Marks Question

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

Question 13 Marks

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of $RHS$ congruence condition which are congruent. State each result in symbolic form.

Answer

In figure, $AD = AD$ (common) hypoteuse $AC$
= hypoteuse $AB$ $\angle\text{ADB}+\angle\text{ADC}=180^\circ$ (Linear pair)
$\Rightarrow\angle\text{ADB}+90^\circ=180^\circ$
$\Rightarrow\angle\text{ADB}=180^\circ-90^\circ$
$\angle\text{ADB}=90^\circ$
$\angle\text{ADB}=\angle\text{ADC}=90^\circ$
Therefore, by $RHS$, $\triangle\text{ADB}\cong\triangle\text{ADC}$.

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Question 23 Marks

In figure, $a = b = c$, name the angle which is congruent to $\angle\text{AOC}$

Answer

We have, $\angle\text{AOB}=\angle\text{BOC}=\angle\text{COD}$
$\therefore\angle\text{AOB}=\angle\text{COD}$
Also, $\angle\text{AOB}+\angle\text{BOC}=\angle\text{BOC}+\angle\text{COD}$
$\angle\text{AOC}=\angle\text{BOD}$
Hence, $\angle\text{BOD}$ is congruent to $\angle\text{AOC}.$

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Question 33 Marks

In figure, $\angle\text{AOC}\cong\angle\text{PYR}$ and $\angle\text{AOC}\cong\angle\text{PYR}$. Name the angle which is congruent to $\angle\text{AOB}.$

Answer

$\angle\text{AOC}\cong\angle\text{PYR}\ ...(\text{i})$
Also, $\angle\text{BOC}\cong\angle\text{QYR}$
Now, $\angle\text{AOC}=\angle\text{AOB}+\angle\text{BOC}$ and $\angle\text{PYR}=\angle\text{PYQ}+\angle\text{QYR}$
By putting the value of $\angle\text{AOC}$ and $\angle\text{PYR}$ in equation $(i)$,
we get: $\angle\text{AOB}+\angle\text{BOC}\cong\angle\text{PYQ}+\angle\text{QYR}$$\angle\text{AOB}\cong\angle\text{PYQ}$
$(\because \angle\text{BOC}\cong\angle\text{QYR})$
Hence, $\angle\text{AOB}\cong\angle\text{PYQ}$

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Question 43 Marks

In Figure, $AB = AD$ and $\angle\text{BAC}=\angle\text{DAC}$.

State in symbolic form the congruence of two triangles $ABC$ and $ADC$ that is true.
Complete each of the following, so as to make it true:

$\angle\text{ABC}$ =
$\angle\text{ACD}$ =
Line segment $AC$ bisects and .

Answer

$AB = AD$ (given)

$\angle\text{BAC}=\angle\text{DAC}$ (given)
$AC = CA$ (common)
Therefore by $SAS$ condition of congruency, $\triangle\text{ABC}\cong\triangle\text{ADC}$

$\angle\text{ABC}=\angle\text{ADC}$ $(c.p.c.t)$
$\angle\text{ACD}=\angle\text{ACB}$ $(c.p.c.t)$
Line segment $AC$ bisects $AD$ and $AB$.

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Question 53 Marks

Draw a right triangle $ABC$. Use $RHS$ condition to construct another triangle congruent to it.

Answer

Consider $\triangle\text{ABC}$ with $\angle\text{B}$ as right angle.
We know construct another right angle triangle on base $BC$, such that
$\angle\text{C}$ is a right angle and $AB = DC$
Also, $BC = CB$
Therefore, by $RHS$, $\triangle\text{ABC}\cong\triangle\text{DCB}$

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Question 63 Marks

Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.

Answer

Two right angles are congruent to each other because they both measure $90$ degrees.
We know that two angles are congruent if they have the same measure.

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Question 73 Marks

In each of the following pairs of right triangles, the measures of some parts are indicated along side. State by the application of $RHS$ congruence condition which are congruent. State each result in symbolic form.

Answer

In figure,
$BD = DB$
hypoteuse $AB =$ hypoteuse $BC$
$\Rightarrow\angle\text{BDA}+\angle\text{BDC}=180^\circ$
$\Rightarrow\angle\text{BDA}+90^\circ=180^\circ$
$\Rightarrow\angle\text{BDA}=180^\circ-90^\circ$
$\angle\text{BDA}=90^\circ$
$\angle\text{BDA}=\angle\text{BDC}=90^\circ$
Therefore, by $RHS$, $\triangle\text{ABD}\cong\triangle\text{CBD}$.

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Question 83 Marks

In Figure, $\angle\text{POQ}\cong\angle\text{ROS}$, can we say that $\angle\text{POR}\cong\angle\text{QOS}$

Answer

We have, $\angle\text{POQ}\cong\angle\text{ROS}\ ...(\text{i})$
Also, $\angle\text{ROQ}\cong\angle\text{ROQ}$ (same angle)
Therefore, adding $\angle\text{ROQ}$ to both sides of $(i)$,
we get: $\angle\text{POQ}+\angle\text{ROQ}\cong\angle\text{ROQ}+\angle\text{ROS}$
$\therefore\angle\text{ROQ}\cong\angle\text{ROQ}$ Hence proved.

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Question 93 Marks

$\triangle\text{ABC}$ is isoseles with $AB = AC$. Also, $\text{AD}\perp\text{BC}$ meeting $BC$ in $D$. Are the two triangles $ABD$ and $ACD$ congruent? State in symbolic form. Which congruence condtion do you use? Which side of $\triangle\text{ADC}$ equals $BD$? Which angle of $\triangle\text{ADC}$ equals $\angle\text{B}?$

Answer

We have $AB = AC ...(i)$
$AD = DA$ (common) $...(ii)$
and $\angle\text{ADC}=\angle\text{ADB}$ $($$\text{AD}\perp\text{BC}$ at point $D) ...(iii)$
Therefore from $(i), (ii)$ and $(iii)$ by $RHS$ congruence condition,
$\triangle\text{ABD}\cong\triangle\text{ACD}$
Now, the triangles are congurent.
Therefore, $BD = CD$. and $\angle\text{ABD}=\angle\text{ACD}$ $(c.p.c.t.)$

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Question 103 Marks

Triangles $ABC$ and $PQR$ are both isosceles with $AB = AC$ and $PO = PR$ respectively. If also, $AB = PQ$ and $BC = QR,$ are the two triangles congruent? Which condition do you use?It $\angle\text{B}=50^\circ,$ what is the measure of $\angle\text{R}?$

Answer

We have $AB = AC$ in isosceles $\triangle\text{ABC}$ And $PQ = PR$ in isosceles $\triangle\text{PQR}.$
Also, we are given that $AB = PQ$ and $QR = BC$.
Therefore, $AC = PR$ $(AB = AC, PQ = PR$ and $AB = PQ)$
Hence, $\triangle\text{ABC}\cong\triangle\text{PQR}$
Now, $\angle\text{ABC}=\angle\text{PQR}$ (Since triangles are congruent)
However, $\triangle\text{PQR}$ is isosceles.
Therefore, $\angle\text{PRQ}=\angle\text{PQR}=\angle\text{ABC}=50^\circ$

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Question 113 Marks

Which of the following pairs of triangle are congruent by $ASA$ condition?

Answer

In $\triangle\text{ABC},$
Now $AB =AC$ (Given)
$\angle\text{ABD}=\angle\text{ACD}=40^\circ$ (Angles opposite to equal sides)
$\Rightarrow\angle\text{ABD}+\angle\text{ACD}+\angle\text{BAC}=180^\circ$ (Angle sum property)
$\Rightarrow40^\circ+40^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-80^\circ$
$\angle\text{BAC}=100^\circ$
Now,
$\angle\text{BAD}+\angle\text{DAC}=\angle\text{BAC}$
$\Rightarrow\angle\text{BAD}=\angle\text{BAC}-\angle\text{DAC}$
$\Rightarrow\angle\text{BAD}=100^\circ-50^\circ$
$\angle\text{BAD}=50^\circ$
$\therefore\angle\text{BAD}=\angle\text{CAD}=50^\circ$
Therefore, by $ASA$, $\triangle\text{ABD}\cong\triangle\text{ADC}$.

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Question 123 Marks

Which of the following pairs of triangle are congruent by $ASA$ condition?

Answer

In $\triangle\text{ABC},$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ (Angle sum property)
$\Rightarrow\angle\text{C}=180^\circ-\angle\text{A}-\angle\text{B}$
$\Rightarrow\angle\text{C}=180^\circ-30^\circ-90^\circ$
$\angle\text{C}=60^\circ$
In $\triangle\text{PQR},$
$\Rightarrow\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ$ (Angle sum property)
$\Rightarrow\angle\text{P}=180^\circ-\angle\text{Q}-\angle\text{R}$
$\Rightarrow\angle\text{P}=180^\circ-60^\circ-90^\circ$
$\angle\text{P}=30^\circ$
Now,
$\angle\text{BAC}=\angle\text{QPR}=30^\circ$
$\angle\text{BCA}=\angle\text{PRQ}=60^\circ$
and $AC = PR$ (Given)
Therefore, by $ASA$, $\triangle\text{ABC}\cong\triangle\text{PQR}$.

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Question 133 Marks

In figure, $BD$ and $CE$ are altitudes of $\triangle\text{ABC}$ and $BD = CE.$
$i.$ Is $\triangle\text{BCD}\cong\triangle\text{CBE}?$
$ii.$ State the three pairs of matching parts you have used to answer $(i)$.

Answer

$i.$ Yes, $\triangle\text{BCD}\cong\triangle\text{CBE}$ by $\ce{RHS}$ congruence condition.
$ii.$ We have used, hypotenuse $BC =$ hypotenuse $CB$
$BD = CE ($given in question$)$
and $\angle\text{BDC}=\angle\text{CEB}=90^\circ$

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Question 143 Marks

Explain the concept of congruence of figures with the help of certain examples.

Answer

Congruent objects or figures are exact copies of each other or we can say mirror images of each other.
The relation of two objects being congruent is called congruence.Consider Ball $A$ and Ball $B$. These two balls are congruent.

Now consider the two stars below. Star $A$ and Star $B$ are exactly the same in size, colour and shape. These are congruent stars.

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Question 153 Marks

$ABC$ and $DBC$ are both isosceles triangles on a common base $BC$ such that $A$ and $D$ lie on the same side of $BC$. Are triangles $ADB$ and $ADC$ congruent? Which condition do you use? If $\angle\text{BAC}=40^\circ$ and $\angle\text{BDC}=100^\circ,$then find $\angle\text{ADB}.$

Answer

$YES$ $\triangle\text{ADB}\cong\triangle\text{ADC}$ (By $SSS$)
$AB = AC , DB = DC$ and $AD= DA$
$\angle\text{BAD}=\angle\text{CAD}$ $(C.P.C.T)$
$\angle\text{BAD}+\angle\text{CAD}=40^\circ$
$\Rightarrow2\angle\text{BAD}=40^\circ$
$\Rightarrow\angle\text{BAD}=\frac{40^\circ}{2}$
$\angle\text{BAD}=20^\circ$
$\angle\text{ABC}+\angle\text{BCA}+\angle\text{BAC}=180^\circ$ (Angle sum property)
Since $\triangle\text{ABC}$ is an isosceles triangle,
$\angle\text{ABC}=\angle\text{BCA}$
$\angle\text{ABC}+\angle\text{ABC}+40^\circ=180^\circ$
$\Rightarrow2\angle\text{ABC}=180^\circ-40^\circ$
$\Rightarrow2\angle\text{ABC}=140^\circ$
$\Rightarrow\angle\text{ABC}=\frac{140^\circ}{2}$
$\angle\text{ABC}=70^\circ$
$\angle\text{DBC}+\angle\text{BCD}+\angle\text{BDC}=180^\circ$ (Angle sum property)
Since $\triangle\text{ABC}$ is an isosceles triangle,
$\angle\text{DBC}=\angle\text{BCD}$
$\angle\text{DBC}+\angle\text{DBC}+100^\circ=180^\circ$
$\Rightarrow2\angle\text{DBC}=180^\circ-100^\circ$
$\Rightarrow2\angle\text{DBC}=80^\circ$
$\Rightarrow\angle\text{DBC}=\frac{80^\circ}{2}$

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