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NCERT Exemplar · Class 9 Mathematics Triangles

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EXERCISE 7.4 1–10 (part 4 of 5)

  1. Exercise 1

    Find all the angles of an equilateral triangle.

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    NCERT’s answer
    $\displaystyle 60$°, $\displaystyle 60$°, $\displaystyle 60$°
    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q1 \[AB = BC = CA \quad \text{(equilateral, given)} \] \[\angle A = \angle B = \angle C \quad \text{(angles opposite equal sides)} \] \[\angle A + \angle B + \angle C = 180^\circ \] \[3\angle A = 180^\circ \; \Rightarrow \; \angle A = 60^\circ \]Answer: Each angle is \(\displaystyle 60^\circ\).

  2. Exercise 2

    The image of an object placed at a point A before a plane mirror LM is seen at the point B by an observer at D as shown in Fig. 7.12. Prove that the image is as far behind the mirror as the object is in front of the mirror. [Hint: CN is normal to the mirror. Also, angle of incidence = angle of reflection]. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-4_Q2

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    Let \(\displaystyle AB\) meet \(\displaystyle LM\) at \(\displaystyle P\); \(\displaystyle AB \perp LM\) at \(\displaystyle P\) (Fig. $\displaystyle 7.12$). NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q2 \[\angle NCP = \angle NCM = 90^\circ \quad (CN \perp LM) \] \[\angle ACP = \angle NCP - \angle ACN = 90^\circ - i \] \[\angle BCP = \angle DCM \quad \text{(vertically opposite, lines } BD, LM \text{ meet at } C) \] \[\angle DCM = \angle NCM - \angle DCN = 90^\circ - r \] \[i = r \; \Rightarrow \; \angle ACP = \angle BCP \] In \(\displaystyle \triangle APC\) and \(\displaystyle \triangle BPC\): \[\angle APC = \angle BPC = 90^\circ \quad (AB \perp LM \text{ at } P) \] \[PC = PC \quad \text{(common)}, \quad \angle ACP = \angle BCP \] \[\triangle APC \cong \triangle BPC \quad \text{(ASA)} \] \[AP = BP \quad \text{(CPCT)} \]Answer: \(\displaystyle AP = BP\) — the image is as far behind the mirror as the object is in front.

  3. Exercise 3

    ABC is an isosceles triangle with AB=AC\displaystyle \mathrm{AB}=\mathrm{AC} and D is a point on BC such that AD⊥BC\displaystyle \mathrm{AD} \perp \mathrm{BC} (Fig. 7.13\displaystyle 7.13). To prove that ∠BAD=∠CAD\displaystyle \angle \mathrm{BAD}=\angle \mathrm{CAD}, a student proceeded as follows: In ΔABD\displaystyle \Delta \mathrm{ABD} and ΔACD\displaystyle \Delta \mathrm{ACD}, AB=AC( Given )∠B=∠C( because AB=AC)\begin{array}{ll} A B=A C & (\text { Given }) \\ \angle B=\angle C & (\text { because } A B=A C) \end{array} and ∠ADB=∠ADC\displaystyle \angle \mathrm{ADB}=\angle \mathrm{ADC} Therefore, ΔABD≅ΔACD(AAS)\displaystyle \Delta \mathrm{ABD} \cong \Delta \mathrm{ACD}(\mathrm{AAS}) So, ∠BAD=∠CAD\displaystyle \angle \mathrm{BAD}=\angle \mathrm{CAD} (CPCT) What is the defect in the above arguments? [Hint: Recall how ∠B=∠C\displaystyle \angle \mathrm{B}=\angle \mathrm{C} is proved when AB=AC\displaystyle \mathrm{AB}=\mathrm{AC} ]. NCERT_Question_Class9_Maths_Exemplar_Ch7_Ex7-4_Q3

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    NCERT’s answer
    It is defective to use $\displaystyle \angle \mathrm{ABD}=\angle \mathrm{ACD}$ for proving this result.
    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q3 The step \(\displaystyle \angle B=\angle C\) is not established independently: NCERT proves \(\displaystyle AB=AC \Rightarrow \angle B=\angle C\) by drawing the bisector of \(\displaystyle \angle A\) and showing \(\displaystyle \triangle BAD \cong \triangle CAD\) (SAS), which already needs \(\displaystyle \angle BAD=\angle CAD\) — the very conclusion this argument claims to reach. So the argument presupposes \(\displaystyle \angle BAD=\angle CAD\).Answer: circular reasoning — \(\displaystyle \angle B=\angle C\) rests on \(\displaystyle \angle BAD=\angle CAD\), so AAS does not prove it.

  4. Exercise 4

    P is a point on the bisector of ∠ABC\displaystyle \angle \mathrm{ABC}. If the line through P , parallel to BA meet BC at Q, prove that BPQ is an isosceles triangle.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q4 \[\angle ABP = \angle PBC \quad (BP \text{ bisects } \angle ABC) \] \[\angle BPQ = \angle ABP \quad \text{(alternate angles, } PQ \parallel BA) \] \[\Rightarrow \angle BPQ = \angle PBC = \angle PBQ \] \[\Rightarrow BQ = PQ \quad \text{(sides opposite equal angles)} \]Answer: \(\displaystyle \triangle BPQ\) is isosceles, with \(\displaystyle BQ = PQ\).

  5. Exercise 5

    ABCD is a quadrilateral in which AB=BC\displaystyle \mathrm{AB}=\mathrm{BC} and AD=CD\displaystyle \mathrm{AD}=\mathrm{CD}. Show that BD bisects both the angles ABC and ADC .

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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_Q5 \[AB = CB,\quad AD = CD,\quad BD = BD \quad \text{(given, common)} \] \[\triangle ABD \cong \triangle CBD \quad \text{(SSS)} \] \[\angle ABD = \angle CBD, \quad \angle ADB = \angle CDB \quad \text{(CPCT)} \]Answer: \(\displaystyle BD\) bisects both \(\displaystyle \angle ABC\) and \(\displaystyle \angle ADC\).

  6. Exercise 6

    ABC is a right triangle with AB=AC\displaystyle \mathrm{AB}=\mathrm{AC}. Bisector of ∠A meets BC at D. Prove that BC=2AD\displaystyle \mathrm{BC}=2 \mathrm{AD}.

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    NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q6 \[\angle A = 90^\circ,\ AB = AC \; \Rightarrow \; \angle B = \angle C = 45^\circ \] \[\angle BAD = \angle CAD = 45^\circ \quad (AD \text{ bisects } \angle A) \] \[\triangle ABD \cong \triangle ACD \quad \text{(SAS: } AB{=}AC,\ \angle BAD{=}\angle CAD,\ AD{=}AD) \] \[\Rightarrow BD = DC \quad \text{(CPCT)} \] \[\text{In } \triangle ABD:\ \angle ABD = \angle BAD = 45^\circ \; \Rightarrow \; AD = BD \quad \text{(sides opposite equal angles)} \] \[BC = BD + DC = 2\,BD = 2\,AD \]Answer: \(\displaystyle BC = 2\,AD\).

  7. Exercise 7

    O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ΔOCD\displaystyle \Delta \mathrm{OCD} is an isosceles triangle.

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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_Q7 \[OA = OB = AB \quad \text{(}\triangle OAB\text{ equilateral, given)} \] \[\angle OAB = \angle OBA = 60^\circ \] \[\angle DAB = \angle ABC = 90^\circ \quad \text{(angles of square }ABCD\text{)} \] \[\angle OAD = \angle DAB - \angle OAB = 90^\circ - 60^\circ = 30^\circ \] \[\angle OBC = \angle ABC - \angle OBA = 90^\circ - 60^\circ = 30^\circ \] \[\angle OAD = \angle OBC, \quad AD = BC \quad \text{(sides of square)} \] \[\triangle OAD \cong \triangle OBC \quad \text{(SAS)} \] \[OD = OC \quad \text{(CPCT)} \]Answer: \(\displaystyle OD = OC\), so \(\displaystyle \triangle OCD\) is isosceles.

  8. Exercise 8

    ABC and DBC are two triangles on the same base BC such that A and D lie on the opposite sides of BC,AB=AC\displaystyle \mathrm{BC}, \mathrm{AB}=\mathrm{AC} and DB=DC\displaystyle \mathrm{DB}=\mathrm{DC}. Show that AD is the perpendicular bisector of BC.

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

    Let AD meet BC at O. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q8 \[AB = AC, \quad DB = DC \quad \text{(given)}, \quad AD = AD \quad \text{(common)} \] \[\triangle ABD \cong \triangle ACD \quad \text{(SSS)} \] \[\angle BAO = \angle CAO \quad \text{(CPCT)} \] \[AB = AC, \quad \angle BAO = \angle CAO, \quad AO = AO \] \[\triangle ABO \cong \triangle ACO \quad \text{(SAS)} \] \[BO = CO, \quad \angle AOB = \angle AOC \quad \text{(CPCT)} \] \[\angle AOB + \angle AOC = 180^\circ \quad \text{(}B, O, C\text{ collinear)} \] \[\angle AOB = \angle AOC = 90^\circ \]Answer: \(\displaystyle BO = CO\) and \(\displaystyle AO \perp BC\), so AD is the perpendicular bisector of BC.

  9. Exercise 9

    ABC is an isosceles triangle in which AC=BC.AD\displaystyle \mathrm{AC}=\mathrm{BC} . \mathrm{AD} and BE are respectively two altitudes to sides BC and AC . Prove that AE=BD\displaystyle \mathrm{AE}=\mathrm{BD}.

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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_Q9 \[AC = BC \quad \text{(given)} \; \Rightarrow \; \angle CAB = \angle CBA \quad \text{(angles opposite equal sides)} \] \[\angle ADB = \angle BEA = 90^\circ \quad \text{(}AD \perp BC,\ BE \perp AC\text{)} \] \[\angle ABD = \angle BAE \quad (\angle CBA = \angle CAB) \] \[AB = BA \quad \text{(common)} \] \[\triangle ADB \cong \triangle BEA \quad \text{(AAS)} \] \[BD = AE \quad \text{(CPCT)} \]Answer: \(\displaystyle AE = BD\).

  10. Exercise 10

    Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side.

    Not cross-checked

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

    Produce AD to E with \(\displaystyle DE = AD\); join CE. NCERT_Solution_Class9_Maths_Exemplar_Ch7_Ex7-4_Q10 \[BD = DC \quad \text{(D is midpoint of BC)}, \quad AD = DE \quad \text{(construction)} \] \[\angle ADB = \angle EDC \quad \text{(vertically opposite angles)} \] \[\triangle ADB \cong \triangle EDC \quad \text{(SAS)} \] \[AB = EC \quad \text{(CPCT)} \] \[AC + CE > AE \quad \text{(triangle inequality, }\triangle ACE\text{)} \] \[AE = AD + DE = 2\,AD, \quad CE = AB \] \[AC + AB > 2\,AD \]Answer: \(\displaystyle AB + AC > 2\,AD\).