R.d Sharma 2022 _mcqs for Class 9 Maths Chapter 11 - Triangles And Its Angles

Answer:

In a given ΔABC we are given that the three angles are equal. So,

According to the angle sum property of a triangle, in ΔABC

Therefore, all the three angles of the triangle are equal to

So, the correct option is (c).

Answer:

In the given problem, we have a right angled triangle and the other two angles are equal.

So, In ΔABC

Now, using the angle sum property of the triangle, in ΔABC, we get,

()

Therefore, the correct option is (b).

Answer:

In the ΔABC, CD is the ray extended from the vertex C of ΔABC. It is given that the exterior angle of the triangle is and two of the interior opposite angles are equal.

So, and_.

So, now using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get.

In ΔABC

Therefore, each of the two opposite interior angles is

So, the correct option is (d).

Answer:

In the given problem, one angle of a triangle is equal to the sum of the other two angles.

Thus, 

     ..........(1)

Now, according to the angle sum property of the triangle

In ΔABC

                  .........(2)

Further, using (2) in (1),

Thus,

Therefore, the correct option is (d).

Answer:

In the given problem, side BC of ΔABC has been produced to a point D. Such that and. Here, we need to find

Given

We get,

Now, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

In ΔABC

Also, (Using 1)

Thus,

Therefore, the correct option is (a).

Answer:

In the given ΔABC, . D is the ray extended from point A. AX bisectsand

Here, we need to find

As ray AX bisects

Thus,

Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

Thus,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, and

Now, according to the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get,

So,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, all the three sides of the triangle are produced. We need to find the sum of the three exterior angles so produced.

Now, according to the angle sum property of the triangle

        .......(1)

Further, using the property, “an exterior angle of the triangle is equal to the sum of two opposite interior angles”, we get,

            ......(2)

Similarly,

           .......(3)

Also, 

           .......(4)

Adding (2) (3) and (4)

We get,

Thus, 

Therefore, the correct option is (d).

Answer:

In the given ΔABC,, AD bisects and .

Here, we need to find.

As, AD bisects,

We get, 

Now, according to angle sum property of the triangle

In ΔABD

Hence,

Therefore, the correct option is (c).

Answer:

In the given ΔABC, an exterior angle and its interior opposite angles are in the ratio 4:5.

Let us take,

Now using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”

We get,

Thus, 

Also, using angle sum property in ΔABC

Thus, 

Therefore, the correct option is (a).

Answer:

In the given ΔABC,and . Bisectors of and meet at O. 

We need to find

Since, OB is the bisector of.

Thus,

Now, using the angle sum property of the triangle

In ΔABC, we get,

Similarly, in ΔBOC

Hence,

Therefore, the correct option is (b).

Answer:

In the given problem, line segment AB and CD intersect at O, such that,and .

We need to find

As

Applying the property, “alternate interior angles are equal”, we get,

    .......(1)

Now, using the angle sum property of the triangle

In ΔODB, we get,

(using 1)

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure,,and . We need to find.

Here, and CD is the transversal, so using the property, “corresponding angles are equal”, we get

Also, using the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in ΔOBD, we get, 

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, we need to find

Here, AB and CD are straight lines intersecting at point O, so using the property, “vertically opposite angles are equal”, we get,

Further, applying the property, “an exterior angle of a triangle is equal to the sum of the two opposite interior angles”, in ΔAOC, we get,

Similarly, in ΔBOD

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, measures of the angles of ΔABC are in the ratio. We need to find the measure of the smallest angle of the triangle.

Let us take,

Now, applying angle sum property of the triangle in ΔABC, we get,

Substituting the value of x in,and

Since, the measure of is the smallest

Thus, the measure of the smallest angle of the triangle is

Therefore, the correct option is (c).

Answer:

In the given problem, we need to find the value of x if

Here, if , then using the property, “if the two lines are parallel, then the alternate interior angles are equal”, we get,

Further, applying angle sum property of the triangle

In ΔABC

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure, we need to find the value of x

Here, according to the angle sum property of the triangle

In ΔABD

Also, ABC is a straight line. So, using the property, “angles forming a linear pair are supplementary”, we get,

Further, using the property, “exterior angle of a triangle is equal to the sum of two opposite interior angles”, we get

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure,and

We need to find the measure of x

Here, we draw a line RS parallel to BP, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

Now, and BA is the transversal, so using the property, “alternate interior angles are equal”

Similarly, and AC is the transversal

                              ........(2)

Adding (1) and (2), we get

Also, as

Using the property,”angles opposite to equal sides are equal”, we get

Further, using the property, “an exterior angle is equal to the sum of the two opposite interior angles”

In ΔABC

Thus,

Therefore, the correct option is (c).

Answer:

In the given problem, the exterior angles obtained on producing the base of a triangle both ways areand. So, let us draw ΔABC and extend the base BC, such that:

Here, we need to find

Now, since BCD is a straight line, using the property, “angles forming a linear pair are supplementary”, we get

Similarly, EBS is a straight line, so we get,

Further, using angle sum property in ΔABC

Thus,

Therefore, the correct option is (c).

Answer:

In the given figure,

We need to find the value of x.

Now, since AB and CD are straight lines intersecting at point O, using the property, “vertically opposite angles are equal”, we get,

Further, applying angle sum property of the triangle 

In ΔBOC 

Then, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get,

In ΔEOC

Further solving for x, we get,

Thus,

Therefore, the correct option is (b).

Answer:

In the given ΔABC, we need to convert z in terms of x and y 

Now, BC is a straight line, so using the property, “angles forming a linear pair are supplementary”

Similarly,

Also, using the property, “vertically opposite angles are equal”, we get,

Further, using angle sum property of the triangle

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, we need to find y in terms of x

Now, using the property, “an exterior angle of the triangle is equal to the sum of the two opposite interior angles”, we get

In ΔABC

                          ..........(1)

Similarly, in ΔOCD

(using 1)

Thus,

Therefore, the correct option is (a).

Answer:

In the given figure, we need to find the value of x.

Here, DBA is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Now, applying the value of y inand

Also,

Further, applying angle sum property of the triangle

In ΔABC

Thus,

Therefore, the correct option is (d).

Answer:

In the given figure,,,and

We need to find the value of x and y

Here, we draw a line ST parallel to AB, i.e

Also, using the property, “two lines parallel to the same line are parallel to each other”

As,

Thus,

Now, and EF is the transversal, so using the property, ”alternate interior angles are equal”, we get,

Similarly,and EF is the transversal

                          .......(2)

Adding (1) and (2), we get

Further,FPE is a straight line

Applying the property, angles forming a linear pair are supplementary

Also, applying angle sum property of the triangle

In ΔPRQ

Thus,

Therefore, the correct option is (a).

Answer:

In the given problem, bisectors of the acute angles of a right angled triangle meet at O. We need to find .

Now, using the angle sum property of a triangle

In ΔABC

Now, further multiplying each of the term by in (1)

Also, applying angle sum property of a triangle

In ΔAOC

Thus,

Therefore, the correct option is (c).

Answer:

In the given figure, bisects of exterior anglesand meet at O and

We need to find

Now, according to the theorem, “if the sides AB and AC of a ΔABC are produced to P and Q respectively and the bisectors of and intersect at O, therefore, we get,

Hence, in ΔABC

Thus,

Therefore, the correct option is (b).

Answer:

In the given figure, bisectors of and meet at E and

We need to find

Here, using the property, “an exterior angle of the triangle is equal to the sum of the opposite interior angles”, we get,

In ΔABC with as its exterior angle

         ........(1)

Similarly, in with as its exterior angle

(CE and BE are the bisectors of and)

    .......(2)

Now, multiplying both sides of (1) by

We get, 

   ......(3)

From (2) and (3) we get,

Thus,

Therefore, the correct option is (a).

Answer:

In the given problem, BC of ΔABC is produced to point D. bisectors of meet side BC at L, and

Here, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,

In ΔABC

Now, as AL is the bisector of

Also, is the exterior angle of ΔALC

Thus,

Again, using the property, “exterior angle of a triangle is equal to the sum of the two opposite interior angles”, we get,

In

Thus,

Therefore the correct option is (b).

Answer:

In the given problem,

We need to find the value of x

Here, as, using the property, “consecutive interior angles are supplementary”, we get

               ..........(1)

Further, applying angle sum property of the triangle 

In ΔABC

(using 1)

Now, AB is a straight line, so using the property, “angles forming a linear pair are supplementary”, we get,

Thus,

Therefore, the correct option is (c).

Answer:

In the given problem, we need to find the value of x.

Here, according to the corollary, “if bisectors of and of a ΔABC meet at a point O, then

In ΔRST

Further solving for x, we get,

Thus,

Therefore, the correct option is (d).

Answer:

Statement-2 (Reason): In a ∆ABC, if the bisectors of angles ∠B and ∠C meet at a point O, then ∠BOC = $90º+\frac{\angle \mathrm{A}}{2}.$
Given that, In an ∆ABC, bisectors of angles ∠B and ∠C meet at a point O.

In ∆ BOC, we have
$\Rightarrow \angle 1+\angle 2+\angle \mathrm{BOC}=180° .....\left(1\right)$

Also, ∆ ABC
$$\begin{gathered} \Rightarrow \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{A}+2\angle 1+2\angle 2=180° \left(\because \angle \mathrm{B}=2\angle 1 \mathrm{and} \angle \mathrm{C}=2\angle 2\right) \\ \Rightarrow \frac{\angle \mathrm{A}}{2}+\angle 1+\angle 2=90° \left(\mathrm{Divide} \mathrm{whole} \mathrm{equation} \mathrm{by} 2\right) \\ \Rightarrow \angle 1+\angle 2=90°-\frac{\angle \mathrm{A}}{2} .....\left(2\right) \end{gathered}$$

From (1) and (2), we have
$$\begin{gathered} \Rightarrow 90°-\frac{\angle \mathrm{A}}{2}+\angle \mathrm{BOC}=180° \\ \Rightarrow \angle \mathrm{BOC}=90°+\frac{\angle \mathrm{A}}{2} \end{gathered}$$

Thus, Statement-2 is true.

Statement-1 (Assertion): In a ∆ ABC, if the bisectors of angles ∠B and ∠C meet at a point O, then ∠BOC is always an obtuse angle.

According to Statement-2: In a ∆ABC, if the bisectors of angles ∠B and ∠C meet at a point O, then ∠BOC = $90º+\frac{\angle \mathrm{A}}{2}.$
⇒ ∠BOC is greater than 90∘.
⇒ ∠BOC is always an obtuse angle.
Thus,  Statement-2 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. 

Hence the correct answer is option (a).



 

Answer:

Statement-2 (Reason): In a ∆ABC, sides AB and AC are produced to P and Q respectively. If the bisectors of ∠PBC and ∠QCB intersect at O, then $\angle \mathrm{BOC}=90º–\frac{\mathrm{A}}{2}.$
Given that, in a ∆ABC, sides AB and AC are produced to P and Q respectively. The bisectors of ∠PBC and ∠QCB intersect at O.
∴ $\angle 2=\frac{1}{2}\angle \mathrm{QCB} \mathrm{and} \angle 1=\frac{1}{2}\angle \mathrm{CBP}$           .....(1)

Since $\angle \mathrm{ABC} \mathrm{and} \angle \mathrm{CBP}$ forms a linear pair.
$$\begin{gathered} \therefore \angle \mathrm{ABC}+\angle \mathrm{CBP}=180° \\ \Rightarrow \angle \mathrm{B}+2\angle 1=180° \left[\mathrm{From}\left(1\right)\right] \\ \Rightarrow 2\angle 1=180°-\angle \mathrm{B} \\ \Rightarrow \angle 1=90°-\frac{\angle \mathrm{B}}{2} .....\left(2\right) \end{gathered}$$

Again, $\angle \mathrm{ACB} \mathrm{and} \angle \mathrm{QCB}$ form a linear pair.
$$\begin{gathered} \therefore \angle \mathrm{ACB}+\angle \mathrm{QCB}=180° \\ \Rightarrow \angle \mathrm{C}+2\angle 2=180° \left[\mathrm{From} \left(1\right)\right] \\ \Rightarrow 2\angle 2=180°-\angle \mathrm{C} \\ \Rightarrow \angle 2=90°-\frac{\angle \mathrm{C}}{2} .....\left(3\right) \end{gathered}$$

In ∆ABC, by Angle Sum Property of Triangle
$$\begin{gathered} \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180° \\ \Rightarrow \angle \mathrm{B}+\angle \mathrm{C}=180°-\angle \mathrm{A} .....\left(4\right) \end{gathered}$$

In ∆BOC, we have
$$\begin{gathered} \angle 1+\angle 2+\angle \mathrm{BOC}=180° \\ \Rightarrow 90°-\frac{1}{2}\angle \mathrm{B}+90°-\frac{1}{2}\angle \mathrm{C}+\angle \mathrm{BOC}=180° \left[\mathrm{From} \left(2\right) \mathrm{and} \left(3\right)\right] \\ \Rightarrow 180°-\frac{1}{2}\left(\angle \mathrm{B}+\angle \mathrm{C}\right)+\angle \mathrm{BOC}=180° \\ \Rightarrow \angle \mathrm{BOC}=\frac{1}{2}\left(\angle \mathrm{B}+\angle \mathrm{C}\right) \\ \Rightarrow \angle \mathrm{BOC}=\frac{1}{2}\left(180°-\angle \mathrm{A}\right) \left[\mathrm{From} \left(4\right)\right] \\ \Rightarrow \angle \mathrm{BOC}=90°-\frac{\angle \mathrm{A}}{2} \end{gathered}$$
Thus, Statement-2 is true.

Statement-1 (Assertion): In a ∆ABC, sides AB and AC are produced to P and Q respectively. If the bisectors of ∠PBC and ∠QCB intersect at O, then ∠BOC is an acute angle.

According to Statement-2: In a ∆ABC, sides AB and AC are produced to P and Q respectively. If the bisectors of ∠PBC and ∠QCB intersect at O, then $\angle \mathrm{BOC}=90º–\frac{\mathrm{A}}{2}.$
⇒ ∠BOC is less than 90∘.
⇒ ∠BOC is always an acute angle.

Thus,  Statement-2 is true.
So, Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. 

Hence the correct answer is option (a).

Answer:

Statement-2 (Reason): The angle between the internal bisector of one base angle and the external bisector of the other angle of a triangle is equal to one-half of the vertical angle.

Using the Exterior Angle Theorem in $△ABC$, we have
$$\begin{gathered} \mathrm{ext}.\angle \mathrm{ACD}=\angle \mathrm{A}+\angle \mathrm{B} \\ \Rightarrow \frac{1}{2}\mathrm{ext}.\angle \mathrm{ACD}=\frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{B} \\ \Rightarrow \angle \mathrm{ECD}=\frac{1}{2}\angle \mathrm{A}+\angle \mathrm{EBC} .....\left(1\right) \left[\because \mathrm{BE} \mathrm{and} \mathrm{CE} \mathrm{are} \mathrm{bisectors} \mathrm{of} \angle \mathrm{B} \mathrm{and} \angle \mathrm{ACD}\right] \end{gathered}$$

Using the Exterior Angle Theorem in $△\mathrm{BCE}$, we obtain
$\angle \mathrm{ECD}=\angle \mathrm{EBC}+\angle \mathrm{E}$                   .....(2)

From (1) and (2), we get
$$\begin{gathered} \Rightarrow \frac{1}{2}\angle \mathrm{A}+\angle \mathrm{EBC}=\angle \mathrm{EBC}+\angle \mathrm{E} \\ \Rightarrow \angle \mathrm{E}=\frac{1}{2}\angle \mathrm{A} \end{gathered}$$
Thus, Statement-2 is true.

Statement-1 (Assertion): In the given figure, side BC of ∆ABC, is produced to a point D such that the bisectors of ∠ABC and ∠ACD meet at a point E. If ∠BAC = 80º, then ∠BEC = 50º.

Given that, ∠BAC = 80º.

According to Statement-2: The angle between the internal bisector of one base angle and the external bisector of the other angle of a triangle is equal to one-half of the vertical angle.
$$\begin{gathered} \Rightarrow \angle \mathrm{E}=\frac{1}{2}\angle \mathrm{A} \\ \Rightarrow \angle \mathrm{E}=\frac{1}{2}\times 80° \\ \Rightarrow \angle \mathrm{E}=40° \\ \Rightarrow \angle \mathrm{BEC}=40° \end{gathered}$$
Thus, Statement-1 is false.
So, Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).

 

Answer:

Disclaimer: In figure ∠OAC = 60º.

Statement-2 (Reason): If two parallel lines are intersected by a transversal, then the corresponding angles are equal.
Thus, Statement-2 is True.

Statement-1 (Assertion): In the given Figure, if CP ∥ BQ, then ∠ACP = 140º.

Given that, CP ∥ BQ
$\Rightarrow \angle \mathrm{QBO}=\angle \mathrm{AOC}=110° \left(\mathrm{Corresponding} \mathrm{angles}\right)$
Now,
$\begin{array}{rcl}\angle \mathrm{ACP}& =& \angle \mathrm{AOC}+\angle \mathrm{OAC} \left(\mathrm{Exterier} \mathrm{angle} \mathrm{property}\right)\\ & =& 110°+60°\\ & =& 170°\end{array}$

Thus, Statement-1 is false.
So, ​Statement-1 is false, Statement-2 is true.

Hence, the correct answer is option (d).