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NCERT Exemplar · Class 9 Mathematics Triangles

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EXERCISE 7.4 11–21 (part 5 of 5)

  1. Exercise 11

    Show that in a quadrilateral ABCD,AB+BC+CD+DA<2(BD+AC)\displaystyle \mathrm{ABCD}, \mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{DA}<2(\mathrm{BD}+\mathrm{AC})

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    Let diagonals AC and BD meet at O. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q11 \[OA + OB > AB, \quad OB + OC > BC \quad \text{(triangle inequality)} \] \[OC + OD > CD, \quad OD + OA > DA \quad \text{(triangle inequality)} \] \[2(OA+OB+OC+OD) > AB+BC+CD+DA \] \[OA+OC = AC, \quad OB+OD = BD \] \[2(AC+BD) > AB+BC+CD+DA \]Answer: \(\displaystyle AB+BC+CD+DA < 2(BD+AC)\).

  2. Exercise 12

    Show that in a quadrilateral ABCD, AB+BC+CD+DA>AC+BD\displaystyle A B+B C+C D+D A>A C+B D

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q12 \[\triangle ABC:\ AB+BC>AC, \qquad \triangle ACD:\ CD+DA>AC \quad \text{(triangle inequality)} \] \[\triangle ABD:\ AB+DA>BD, \qquad \triangle BCD:\ BC+CD>BD \quad \text{(triangle inequality)} \] \[2(AB+BC+CD+DA) > 2AC+2BD \] \[AB+BC+CD+DA > AC+BD \]Answer: \(\displaystyle AB+BC+CD+DA > AC+BD\).

  3. Exercise 13

    In a triangle ABC,D\displaystyle \mathrm{ABC}, \mathrm{D} is the mid-point of side AC such that BD=12AC\displaystyle \mathrm{BD}=\frac{1}{2} \mathrm{AC}. Show that ∠ABC\displaystyle \angle \mathrm{ABC} is a right angle.

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    \[AD = DC = \frac{1}{2}AC \quad \text{(D is the mid-point of }AC\text{)} \] \[BD = \frac{1}{2}AC \quad \text{(given)} \] \[AD = BD = DC \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q13 \[\angle DAB = \angle DBA \quad \text{(}AD=BD\text{, angles opposite equal sides)} \] \[\angle DBC = \angle DCB \quad \text{(}BD=DC\text{, angles opposite equal sides)} \] \[\angle DAB + \angle ABC + \angle DCB = 180^\circ \quad \text{(angle sum of }\triangle ABC\text{)} \] \[\angle DAB + (\angle DBA+\angle DBC) + \angle DCB = 180^\circ \] \[2(\angle DBA+\angle DBC) = 180^\circ \] \[\angle ABC = \angle DBA+\angle DBC = 90^\circ \]Answer: \(\displaystyle \angle ABC = 90^\circ\).

  4. Exercise 14

    In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

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    Produce \(\displaystyle BM\) to \(\displaystyle E\) such that \(\displaystyle ME = BM\). NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q14 \[AM = CM, \quad BM = EM \quad \text{(M is the mid-point of }AC\text{; construction)} \] \[\angle AMB = \angle CME \quad \text{(vertically opposite angles)} \] \[\triangle AMB \cong \triangle CME \quad \text{(SAS)} \] \[AB = CE, \quad \angle ABM = \angle CEM \quad \text{(CPCT)} \] \[AB \parallel CE \quad \text{(alternate angles, transversal }BE\text{)} \] \[\angle ABC = 90^\circ \; \Rightarrow \; \angle ECB = 90^\circ \quad \text{(co-interior angles, }AB\parallel CE\text{)} \] In \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle ECB\): \[AB = EC, \quad \angle ABC = \angle ECB = 90^\circ, \quad BC = CB \] \[\triangle ABC \cong \triangle ECB \quad \text{(SAS)} \] \[AC = EB = 2BM \quad \text{(CPCT; }BE = BM+ME\text{)} \]Answer: \(\displaystyle BM = \frac{1}{2}AC\).

  5. Exercise 15

    Two lines l\displaystyle l and m\displaystyle m intersect at the point O and P is a point on a line n\displaystyle n passing through the point O such that P is equidistant from l\displaystyle l and m\displaystyle m. Prove that n\displaystyle n is the bisector of the angle formed by l\displaystyle l and m\displaystyle m.

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    Let Q, R be the feet of the perpendiculars from P to l, m. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q15 \[PQ \perp l, \quad PR \perp m \] \[PQ = PR \quad \text{(P is equidistant from }l, m\text{)} \] In \(\displaystyle \triangle OQP\) and \(\displaystyle \triangle ORP\): \[\angle OQP = \angle ORP = 90^\circ, \quad OP = OP, \quad PQ = PR \] \[\triangle OQP \cong \triangle ORP \quad \text{(RHS)} \] \[\angle POQ = \angle POR \quad \text{(CPCT)} \]Answer: \(\displaystyle n\) bisects the angle between \(\displaystyle l\) and \(\displaystyle m\), since \(\displaystyle \angle POQ = \angle POR\).

  6. Exercise 16

    Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD=BC\displaystyle \mathrm{AD}=\mathrm{BC}.

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    Join AN and BN. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q16 \[AM = BM \quad \text{(M is the mid-point of }AB\text{)} \] \[\angle AMN = \angle BMN = 90^\circ \quad \text{(}MN \perp AB\text{)} \] \[MN = MN \quad \text{(common)} \] \[\triangle AMN \cong \triangle BMN \quad \text{(SAS)} \] \[AN = BN, \quad \angle ANM = \angle BNM \quad \text{(CPCT)} \] \[\angle DNA = 90^\circ - \angle ANM = 90^\circ - \angle BNM = \angle CNB \quad \text{(}MN \perp DC\text{)} \] In \(\displaystyle \triangle ADN\) and \(\displaystyle \triangle BCN\): \[AN = BN, \quad \angle DNA = \angle CNB, \quad DN = CN \quad \text{(N is the mid-point of }DC\text{)} \] \[\triangle ADN \cong \triangle BCN \quad \text{(SAS)} \] \[AD = BC \quad \text{(CPCT)} \]Answer: \(\displaystyle AD = BC\).

  7. Exercise 17

    ABCD is a quadrilateral such that diagonal AC bisects the angles A and C. Prove that AB=AD\displaystyle \mathrm{AB}=\mathrm{AD} and CB=CD\displaystyle \mathrm{CB}=\mathrm{CD}.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q17 In \(\displaystyle \triangle ABC\) and \(\displaystyle \triangle ADC\): \[\angle BAC = \angle DAC \quad \text{(AC bisects }\angle A\text{)} \] \[AC = AC \quad \text{(common)} \] \[\angle BCA = \angle DCA \quad \text{(AC bisects }\angle C\text{)} \] \[\triangle ABC \cong \triangle ADC \quad \text{(ASA)} \] \[AB = AD, \quad CB = CD \quad \text{(CPCT)} \]Answer: \(\displaystyle AB = AD\) and \(\displaystyle CB = CD\).

  8. Exercise 18

    ABC is a right triangle such that AB=AC\displaystyle \mathrm{AB}=\mathrm{AC} and bisector of angle C intersects the side AB at D . Prove that AC+AD=BC\displaystyle \mathrm{AC}+\mathrm{AD}=\mathrm{BC}.

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    Draw \(\displaystyle DE \perp BC\), E on BC. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q18 \[\angle DAC = \angle DEC = 90^\circ \quad \text{(given; construction)} \] \[\angle ACD = \angle ECD \quad \text{(CD bisects }\angle C\text{)} \] \[DC = DC \quad \text{(common)} \] \[\triangle ADC \cong \triangle EDC \quad \text{(AAS)} \] \[AD = ED, \quad AC = EC \quad \text{(CPCT)} \] \[\angle B = \angle ACB = 45^\circ \quad \text{(}AB=AC,\ \angle A = 90^\circ\text{)} \] \[\angle BDE = 180^\circ - 90^\circ - \angle B = 45^\circ = \angle B \] \[DE = BE \quad \text{(sides opposite equal angles in }\triangle BDE\text{)} \] \[BC = BE + EC = DE + AC = AD + AC \]Answer: \(\displaystyle AC + AD = BC\).

  9. Exercise 19

    AB and CD are the smallest and largest sides of a quadrilateral ABCD . Out of ∠B\displaystyle \angle \mathrm{B} and ∠D\displaystyle \angle \mathrm{D} decide which is greater.

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    NCERT’s answer
    $\displaystyle \angle \mathrm{B}$ will be greater.
    Join \(\displaystyle BD\). \[AD > AB \quad \text{(AB is the smallest side)} \] \[\Rightarrow \angle ABD > \angle ADB \quad \text{(angle opposite the longer side)} \] \[CD > CB \quad \text{(CD is the largest side)} \] \[\Rightarrow \angle CBD > \angle CDB \quad \text{(angle opposite the longer side)} \] Adding, \[\angle ABD+\angle CBD > \angle ADB+\angle CDB \] \[\angle ABC > \angle ADC \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q19 Answer: \(\displaystyle \angle B > \angle D\).

  10. Exercise 20

    Prove that in a triangle, other than an equilateral triangle, angle opposite the longest side is greater than 23\displaystyle \frac{2}{3} of a right angle.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Let \(\displaystyle BC\) be the longest side of \(\displaystyle \Delta ABC\), opposite \(\displaystyle \angle A\). \[BC \ge AB,\ BC \ge AC \] \[\Rightarrow \angle A \ge \angle B,\ \angle A \ge \angle C \quad \text{(angle opposite the longer side)} \] \[\angle A+\angle B+\angle C=180^\circ \] \[\Rightarrow 3\angle A \ge 180^\circ \Rightarrow \angle A \ge 60^\circ \] Equality forces \(\displaystyle \angle A=\angle B=\angle C=60^\circ\), an equilateral \(\displaystyle \Delta\), excluded. \[\angle A > 60^\circ = \tfrac{2}{3}\times 90^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q20 Answer: \(\displaystyle \angle A > 60^\circ\).

  11. Exercise 21

    ABCD is quadrilateral such that AB=AD\displaystyle \mathrm{AB}=\mathrm{AD} and CB=CD\displaystyle \mathrm{CB}=\mathrm{CD}. Prove that AC is the perpendicular bisector of BD.

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    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Join \(\displaystyle AC\); let it meet \(\displaystyle BD\) at \(\displaystyle O\). In \(\displaystyle \Delta ABC\) and \(\displaystyle \Delta ADC\): \[AB=AD,\ CB=CD,\ AC=AC \quad \text{(common)} \] \[\Rightarrow \Delta ABC \cong \Delta ADC \quad \text{(SSS)} \] \[\Rightarrow \angle BAC=\angle DAC \quad \text{(CPCT)} \] In \(\displaystyle \Delta ABO\) and \(\displaystyle \Delta ADO\): \[AB=AD,\ \angle BAO=\angle DAO,\ AO=AO \quad \text{(common)} \] \[\Rightarrow \Delta ABO \cong \Delta ADO \quad \text{(SAS)} \] \[\Rightarrow BO=DO,\ \angle AOB=\angle AOD \quad \text{(CPCT)} \] \[\angle AOB+\angle AOD=180^\circ \quad \text{(linear pair)} \] \[\Rightarrow \angle AOB=\angle AOD=90^\circ \] NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q21 Answer: \(\displaystyle BO=DO\) and \(\displaystyle AC \perp BD\).