NCERT Solutions for Class 9 Maths Chapter 7 Triangles

Ans : In ∆ ABC and ∆ ABD,
AC = AD (Given)
∠CAB = ∠ DAB (AB bisects ∠A)
AB = AB (Common)
Therefore,  ∆ ABC ≅ ∆ ABD( By SAS congruence rule)
BC = BD (By CPCT)
Therefore, BC and BD are Of equal lengths. 

Ans : In ∠ABD and ∠BAC,
AD = BC (Given)
∠DAB = ∠CBA(Given)
AB = BA (Common)
\(\begin{array}{l}{\therefore \triangle \mathrm{ABD}=\Delta \mathrm{BAC}(\mathrm{By} \text { SAS congruence rule) }} \\ {\therefore \mathrm{BD}=\mathrm{AC}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
And, ∠ABD = ∠BAC (By CPCT) 

Ans : In ∆ BOC and ∆ AOD,
\(\begin{array}{l}{\angle \mathrm{BOC}=\angle \mathrm{AOD}(\text { Vertically opposite angles })} \\ {\angle \mathrm{CBO}=\angle \mathrm{DAO}\left(\text { Each } 90^{\circ}\right)}\end{array}\)
\(\begin{array}{l}{\mathrm{BC}=\mathrm{AD}(\text { Given })} \\ {\therefore \Delta \mathrm{BOC}=\triangle \mathrm{AOD} \text { (AAS congruence rule) }} \\ {\therefore \mathrm{BO}=\mathrm{AO}(\mathrm{By} \mathrm{CPCT})} \\ {\Rightarrow \mathrm{CD} \text { bisects } \mathrm{AB}}\end{array}\) 

Ans : In ∆ ABC and  ∆ CDA, \(\begin{array}{l}{\angle B A C=\angle D C A(\text { Alternate interior angles, as } p \| q)} \\ {A C=C A(\text { Common })}\end{array}\)
\(\begin{array}{l}{\angle B C A=\angle D A C(\text { Alternate interior angles, as } l \| m)} \\ { \triangle A B C = \Delta C D A(B y \text { ASA congruence rule) }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { In } \triangle A P B \text { and } \triangle A Q B \text { . }} \\ {\angle A P B=\angle A Q B\left(E a c h 90^{\circ}\right)}\end{array}\)
\(\begin{array}{l}{\angle P A B=\angle Q A B(1 \text { is the angle bisector of } \angle A)} \\ {A B=A B(\text { Common })}\end{array}\)
\(\begin{array}{l}{\therefore \triangle \mathrm{APB} \cong \triangle \mathrm{AQB}(\mathrm{By} \text { AAS congruence rule) }} \\ {\therefore \mathrm{BP}=\mathrm{BQ}(\mathrm{By} \mathrm{CPCT})}\end{array}\)
Or, it can be said that B is equidistant from the arms of ∠ A. 

Ans : It is given that ∠ BAD = ∠ EAC
∠BAD + ∠DAC = ∠ EAC + ∠DAC
\(\begin{array}{l}{\angle B A C=\angle D A E} \\ {\text { In } \Delta B A C \text { and } \Delta D A E,} \\ {A B=A D(G \text { iven })}\end{array}\)
∠BAC = ∠DAE (Proved above)
AC = AE (Given)
Therefore, ∆BAC ≅ ∆DAE (By SAS congruence rule)
Therefore, BC = DE (By CPCT) 

Ans : It is given that ∠EPA = ∠DPB
∠EPA + ∠DPE = ∠DPB + ∠DPE
Therefore,  ∠DPA = ∠EPB
In ∠DAP and ∠EBP,
∠DAP = ∠EPB(Given)
AP = BP (P is mid-point of AB)
∠DPA = ∠EPB (From above)
Therefore, ∆ DAP ≅ ∆ EBP (ASA congruence rule)
Therefore, AD - BE (By CPCT) 

Ans : (i) In ∆ AMC and  ∆ BMD,
AM = BM (M is the mid-point Of AB)
∠AMC = ∠BMD (Vertically opposite angles)
CM = DM (Given)
Therefore, ∆ AMC ≅ ∆ BMD (By SAS congruence rule)
Therefore AC = BD (By CPCT)
And, ∠ACM = ∠BDM (By CPCT)

(ii) ∠ACM = ∠BDM
However, ∠ACM and ∠BDM are alternate interior angles.
Since alternate angles are equal,
It can be said that DB II AC
∠DBC + ∠ACB = 180° (Co-interior angles)
∠DBC + 90° = 180°
∠DBC = 90°

(iii) In ∆ DBC and  ∆ ACB,
DB = AC (Already proved)
∠DBC = ∠ACB(Each 90°)
BC = CB (Common)
Therefore, ∆ DBC ≅ ∆ ACB (SAS congruence rule)

(iv) ∆DBC ≅ ∆ACB
AB = DC (By CPCT)
AB = 2 CM
\(\therefore C M=\frac{1}{2} A B\) 

Ans : 
(i) It is given that in triangle ABC, AB = AC
∠ACB = ∠ABC (Angles opposite to equal sides Of a triangle are equal)
\(\frac{1}{2} \angle A C B=\frac{1}{2} \angle A B C\)
∠OCB =∠OBC
Therefore, OB = OC (Sides opposite to equal angles of a triangle are also equal)

(ii) In ∆OAB and ∆OAC,
AO = AO(Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ∆OAB ≅ ∆OAC (By SSS congruence rule)
∠BAO = ∠CAO (CPCT)
Therefore, AO bisects ∠A. 

Ans : In ∆ADC and ∆ADB,
AD = AD (Common)
∠ADC = ∠ADB (Each 90°)
CD = BD (AD is the perpendicular bisector of BC)
Therefore, ∆ADC ≅ ∆ADB (By SAS congruence rule)
AB = AC (By CPCT)
Therefore, ABC is an isosceles triangle in which AB = AC 

Ans : In ∆AEB and ∆AFC,
∠AEB and ∠AFC (Each 90°)
∠A = ∠A (Common angle)
AB = AC (Given)
Therefore, ∆AEB ≅ ∆AFC (By AAS congruence rule)
BE = CF (By CPCT) 

Ans : (i) ∆ ABE and ∆ ACF,
\(\begin{array}{l}{\angle A B E \text { and } \angle A C F\left(E a c h 90^{\circ}\right)} \\ {\angle A=\angle A(\text { Common angle })} \\ {B E=C F(G \text { iven })}\end{array}\)
Therefore, ∆ ABE ≅ ∆ ACF(By AAS congruence rule)

(ii) It has already been proved that
∆ ABE ≅ ∆ ACF
Therefore, AB = AC(By CPCT) 

Ans : 
Let us join AD.
In ∆ABD and ∆ACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common side)
Therefore, ∆ABD ≅ ∆ACD (By SSS congruence rule)
∠ABD = ∠ACD (By CPCT) 

Ans : In ∆ABC,
\(\begin{array}{l}{\mathrm{AB}=\mathrm{AC}(\text { Given) }} \\ {\therefore \angle \mathrm{ACB}=\angle \mathrm{ABC} \text { (Angles opposite to equal sides of a triangle are also equal) }}\end{array}\)
In ∆ACD,
\(\begin{array}{l}{\mathrm{AC}=\mathrm{AD}(\text { Given) }} \\ {\therefore \angle \mathrm{ADC}=\angle \mathrm{ACD} \text { (Angles opposite to equal sides of a triangle are also equal) }}\end{array}\)
In ∆BCD,
∠ABC + ∠BCD + ∠ADC = 1800 (Angle sum property of a triangle)
∠ACB + ∠ACB +∠ACD + ∠ACD = 180°
2(∠ACB + ∠ACD) = 180°
2(∠BCD) = 180°
Therefore, ∠BCD = 90° 

Ans : 
It is given that
AB = AC\(\begin{array}{l}{\therefore \angle C=\angle B(\text { Angles opposite to equal sides are also equal) }} \\ {\text { In } \Delta A B C \text { . }} \\ {\angle A+\angle B+\angle C=180^{\circ} \text { (Angle sum property of a triangle) }} \\ {90^{\circ}+\angle B+\angle C=180^{\circ}} \\ {90^{\circ}+\angle B+\angle B=180^{\circ}}\end{array}\)
\(\begin{array}{l}{2 \angle B=90^{\circ}} \\ {\angle B=45^{\circ}} \\ {\therefore \angle B=\angle C=45^{\circ}}\end{array}\) 

Ans : 
Let us consider that ABC is an equilateral triangle.
Therefore, AB = BC + AC
AB = AC
Therefore, ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)
Also,
AB = AC
∠B = ∠A (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
∠A = ∠B = ∠C
\(\begin{array}{l}{\text { In } \triangle A B C,} \\ {\angle A+\angle B+\angle C=180^{\circ}} \\ {\angle A+\angle A+\angle A=180^{\circ}} \\ {\angle A=180^{\circ}} \\ {\angle A=60^{\circ}} \\ {\therefore \angle A=\angle B=\angle C=60^{\circ}}\end{array}\)
Hence, in a equilateral triangle, all interior angles are of measure 60°. 

Ans : (i) In \( \triangle A B D \text { and } \triangle A C D, \)
\( \begin{array}{l}{A B=A C(G \text { wen })} \\ {B D=C D(\text { Given })} \\ {A D=A D \text { (Common) }}\end{array} \)
\( \triangle A B D \cong \triangle A C D(B y \text { SSS congruence rule) } \)
\( \angle B A D=\angle C A D(B y  C P C T) \)
\( \angle B A P=\angle C A P \) ….(1)
(ii) In \( \triangle \mathrm{ABP} \text { and } \triangle \mathrm{ACP} , \)
\( A B=A C(\text { Given }) \)
\( \begin{array}{l}{\angle B A P=\angle C A P[\text { From equation }(1)]} \\ {A P=A P(\text { Common })}\end{array} \)
\( \therefore \triangle \mathrm{ABP} \cong \triangle \mathrm{ACP} \) ( By SAS congruence rule)
\( \therefore B P=C P(B y  C P C T) \) ….(2)
(iii) From equation (1),
\( \angle B A P=\angle C A P \)
Hence, AP bisects \( \angle A \),
In \( \Delta B D P \text { and } \Delta C D P \)
\( \begin{array}{l}{\mathrm{BD}=\mathrm{CD}(\text { Given })} \\ {\mathrm{DP}=\mathrm{DP}(\text { Common })} \\ {\mathrm{BP}=\mathrm{CP}[\text { From equation }(2)]}\end{array} \)
\( \therefore \triangle B D P \cong \Delta C D P \) (By SSS congruence rule)
\( \triangle \angle B D P=\angle C D P(B y C P C T) \) …(3)
Hence, AP bisects \( \angle \mathrm{D} \)
(iv) \( \triangle B D P \cong \Delta C D P \)
\( \therefore \angle B P D=\angle C P D(B y C P C T) \) ….(4)
\( \angle B P D+\angle C P D=180^{\circ}(\text { Linear pair angles) } \)
\( \angle B P D+\angle B P D=180^{\circ} \)
\( 2 \angle 8 P D=180^{\circ}[\text { From Equation }(4)] \)
\( \angle B P D=90^{\circ} \quad \dots(5) \)
FRom equations (2) and (5), it can be said that AP is the perpendicular bisector of BC. 

Ans : 
(i) In \( \Delta B A D \text { and } \triangle C A D ,\)
\( \angle \mathrm{ADB}=\angle \mathrm{ADC}\left(\text { Each } 90^{\circ} \text { as } \mathrm{AD} \text { is an altitude) }\right. \)
\( \begin{array}{l}{A B=A C(G \text { iven })} \\ {A D=A D(\text { Common })}\end{array} \)
\( \therefore \triangle B A D \cong \triangle C A D \) (By RHS Congruence rule)
\( \triangle B D=C D \) (By CPCT)
Hence, AD bisects BC.
(ii) Also, by CPCT,
\( \angle B A D=\angle C A D \)
Hence, AD bisects \( \angle \mathrm{A} \) 

Ans : (i) In \( \triangle A B C, A M \) is the median to BC.
\( \therefore \) \( \mathrm{BM}=\frac{1}{2} \mathrm{BC} \)
In \( \Delta \mathrm{PQR} \), PN is the median to QR.
\( \therefore Q N=\frac{1}{2} Q R \)
However, BC = QR
\( \therefore \frac{1}{2} B C=\frac{1}{2} Q R \)
\( \therefore \mathrm{BM}=\mathrm{QN} \quad \ldots(1) \)
In \( \triangle A B M \text { and } \Delta P Q N, \)
AB = PQ(Given)
\( \begin{array}{l}{\mathrm{BM}=\mathrm{QN}[\text { From Equation }(1)]} \\ {\mathrm{AM}=\mathrm{PN}(\text { Given })} \\ {\text { } \mathrm{ABM} \cong \Delta \mathrm{PQN} \text { (By SSS congruence rule }} \\ {\angle \mathrm{ABM}=\angle \mathrm{PQN}(\mathrm{By} \mathrm{CPCT})} \\ {\angle \mathrm{ABC}=\angle \mathrm{PQR}}\end{array} \) …(2)
(iii) In \( \triangle A B C \text { and } \Delta P Q R \),
\( \begin{array}{l}{\mathrm{AB}=\mathrm{PQ}(\text { Given })} \\ {\angle \mathrm{ABC}=\angle \mathrm{PQR}[\mathrm{From} \text { Equation }(2)]} \\ {\mathrm{BC}=\mathrm{QR}(\text { Given })}\end{array} \)
\( \therefore \triangle A B C \cong \Delta P Q R \) (By SAS congruence rule) 

Ans : 
In \( \triangle B E C \text { and } \Delta C F B \) ,
\( \angle B E C=\angle C F B\left(E a c h 90^{\circ}\right) \)
\( \begin{array}{l}{\mathrm{BC}=\mathrm{CB}(\mathrm{Common})} \\ {\mathrm{BE}=\mathrm{CF}(\mathrm{Given})}\end{array} \)
\( \therefore \triangle B E C \cong \Delta C F B \) (By RHS congruency)
\( \therefore \angle B C E=\angle C B F(B y C P C T) \)
\( \therefore A B=A C \) (Sides opposite to equal angles of a triangle are equal)
Hence, \( \triangle \mathrm{ABC} \) is isosceles. 

Ans : 
In \( \triangle A P B \text { and } \triangle A P C \) ,
\( \angle A P B=\angle A P C\left(E a c h 90^{\circ}\right) \)
\( \begin{aligned} A B &=A C(G i v e n) \\ A P &=A P(C o m m o n) \end{aligned} \)
\( \therefore \triangle A P B \cong \triangle A P C \) (Using RHS congruence rule)
\( \therefore \angle B=\angle C \) (By using CPCT) 

Ans : 
Let us consider a right-angled triangle ABC, right-angled at B.
In \( \triangle A B C \),
\( \begin{array}{l}{\angle A+\angle B+\angle C=180^{\circ} \text { (Angle sum property of a triangle) }} \\ {\angle A+90^{\circ}+\angle C=180^{\circ}} \\ {\angle A+\angle C=90^{\circ}}\end{array} \)
Hence, the other two angles have to be acute(i.e., less than \( 90^{\circ} \) ).
\( \angle B \text { is the largest angle in } \triangle A B C . \)
\( \begin{array}{l}{\angle B>\angle A \text { and } \angle B>\angle C} \\ {A C>B C \text { and } A C>A B}\end{array} \)
[In any triangle, the side opposite to the larger (greater) angle is longer.]
Therefore, AC is the largest side in \( \Delta A B C \)
However, AC is the hypotenuse of \( \triangle \mathrm{ABC} \). Therefore, hypotenuse is the longest side in a right-angled triangle. 

Ans : In the given figure,
\( \angle A B C+\angle P B C=180^{\circ}(\text { Linear pair) } \)
\( \angle A B C=180^{\circ}-\angle P B C \quad \ldots(1) \)
Also,
\( \angle A C B+\angle Q C B=180^{\circ} \)
\( \angle A C B=180^{\circ}-\angle Q C B \) …(2)
As \( \angle P B C<\angle Q C B \)
\( 180^{\circ}-\angle P B C>180^{\circ}-\angle Q C B \)
\( \begin{array}{l}{\angle A B C>\angle A C B[\text { From Equations }(1) \text { and }(2)]} \\ {A C>A B(\text { Side opposite to the larger angle is larger.) }}\end{array} \)
Hence proved AC > AB 

Ans : In ΔAOB,
∠B < ∠A
⇒ AO < BO (Side opposite to smaller angle is smaller) … (1)
In ΔCOD,
∠C < ∠D
⇒ OD < OC (Side opposite to smaller angle is smaller) … (2)
On adding equations (1) and (2), we obtain
AO + OD < BO + OC
AD < BC 

Ans : 

In ΔABC,
AB < BC (AB is the smallest side of quadrilateral ABCD)
∴ ∠2 < ∠1 (Angle opposite to the smaller side is smaller) … (1)
In ΔADC,
AD < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠4 < ∠3 (Angle opposite to the smaller side is smaller) … (2)
On adding equations (1) and (2), we obtain
∠2 + ∠4 < ∠1 + ∠3
⇒ ∠C < ∠A
⇒ ∠A > ∠C
Let us join BD.



In ΔABD,
AB < AD (AB is the smallest side of quadrilateral ABCD)
∴ ∠8 < ∠5 (Angle opposite to the smaller side is smaller) … (3)
In ΔBDC,
BC < CD (CD is the largest side of quadrilateral ABCD)
∴ ∠7 < ∠6 (Angle opposite to the smaller side is smaller) … (4)
On adding equations (3) and (4), we obtain
∠8 + ∠7 < ∠5 + ∠6
⇒ ∠D < ∠B
⇒ ∠B > ∠D 

Ans : As \( P R>P Q \)
\( \angle P Q R>\angle P R Q(\text { Angle opposite to larger side is larger) } \) ...(1)
PS is the bisector of \( \angle Q P R \).
\( \angle Q P S=\angle R P S \quad \ldots(2) \)
\( \angle P S R \text { is the exterior angle of } \Delta P Q S \)
\( \angle P S R=\angle P Q R+\angle Q P S \quad-(3) \)
\( \begin{array}{l}{\angle P S Q \text { is the exterior angle of } \Delta P R S \text { . }} \\ {\angle P S Q=\angle P R Q+\angle R P S}\end{array} \) …(4)
\( \begin{array}{l}{\text { Adding Equations }(1) \text { and }(2), \text { we obtain }} \\ {\angle P Q R+\angle Q P S>\angle P R Q+\angle R P S}\end{array} \)
\( \angle \mathrm{PSR}>\angle \mathrm{PSQ}[\text { Using the values of Equations }(3) \text { and }(4)] \) 

Ans : 
Let us take a line l and from point P (i.e., not on line l), draw two line segments PN and PM. Let PN be perpendicular to line L and PM is drawn at some other angle.
In \( \triangle \mathrm{PNM} \)
\( \begin{array}{l}{\angle N=90^{\circ}} \\ {\angle P+\angle N+\angle M=180^{\circ}(\text { Angle sum property of a triangle) }} \\ {\angle P+\angle M=90^{\circ}}\end{array} \)

Clearly, \( \angle M \) is an acute angle

\( \begin{array}{l}{\angle M < A N} \\ {P N < P M (\text { side opposite to the smaller angle is smaller}) }\end{array} \)

Similarly, by drawing different line segments from P to l, it can be proved that PN is smaller in comparison to them.
Therefore, it can be observed that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest. 

Ans : Circumference of a triangle is always equidistant from all the vertices of that triangle. Circumference is the point where perpendicular bisectors of all the sides of the triangle meet together.

In \( \triangle \mathrm{ABC} \), we can find the circumference by drawing the perpendicular bisectors of sides AB, BC, AND CA of this triangle. O is the point where these bisectors are meeting together. Therefore, O is the point which is equidistant from all the vertices of \( \triangle \mathrm{ABC} \). 

Ans : The point which is equidistant from all the sides of a triangle is called the incentre of the triangle. Incentre of a triangle is the intersection point of the angle bisectors of the interior angles of that triangle.

Here, in \( \triangle \mathrm{ABC} \), we can find the incentre of this triangle by drawing the angle bisectors of the interior angles of this triangle. I is the point where these angle bisectors are intersecting each other. Therefore, I is the point equidistant from all the sides of \( \triangle \mathrm{ABC} \) 

Ans : Maximum number of persons can approach the ice-cream parlour if it is equidistant from A, B and C from a triangle. In a triangle, the circumcentre is the only point that is equidistant from its vertices. So, the ice-cream parlour should be set up at the circumcentre O of \( \triangle \mathrm{ABC} \)

In this situation, maximum number of persons can approach it. We can find circumcentre O of this triangle by drawing perpendicular bisectors of the ised of this triangle. 

Ans : It can be observed that hexagonal-shaped rangoli has 6 equilateral triangles in it.

\( \begin{array}{l}{\text { Area of } \Delta \mathrm{OAB}=\frac{\sqrt{3}}{4}(\text { side })^{2}=\frac{\sqrt{3}}{4}(5)^{2}} \\ {=\frac{\sqrt{3}}{4}(25)=\frac{25 \sqrt{3}}{4} \mathrm{cm}^{2}}\end{array} \)
Area of hexagonal-shaped rangoli \( =6 \times \frac{25 \sqrt{3}}{4}=\frac{75 \sqrt{3}}{2} \mathrm{cm}^{2} \)
Area of equilateral triangle having its sides as 1 cm \( =\frac{\sqrt{3}}{4}(1)^{2}=\frac{\sqrt{3}}{4} \mathrm{cm}^{2} \)
Number of equilateral triangles of 1 cm side that can be filled In this hexagonal-shaped rangoli \( =\frac{\frac{75 \sqrt{3}}{2}}{\frac{\sqrt{3}}{4}}=150 \)
Star-shaped rangoli has 12 equilateral triangles of side 5 cm in it.

Area of star-shaped rangoli = \( 12 \times \frac{\sqrt{3}}{4} \times(5)^{2}=75 \sqrt{3} \)
Number of equilateral triangles of 1 cm side that can be filled in this star-shaped rangoli = \( \frac{75 \sqrt{3}}{\frac{\sqrt{3}}{4}}=300 \)
Therefore, star-shaped rangoli has more equilateral triangles in it.