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NCERT Exemplar · Class 10 Mathematics Triangles

57 questions · 57 still being checked

EXERCISE 6.4 11–18 (part 6 of 6)

  1. Exercise 11

    In ΔPQR,PD⊥QR\displaystyle \Delta \mathrm{PQR}, \mathrm{PD} \perp \mathrm{QR} such that D lies on QR . If PQ=a,PR=b,QD=c\displaystyle \mathrm{PQ}=a, \mathrm{PR}=b, \mathrm{QD}=c and DR=d\displaystyle \mathrm{DR}=d, prove that (a+b)(a−b)=(c+d)(c−d)\displaystyle (a+b)(a-b)=(c+d)(c-d).

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q11 \[PQ^2 = PD^2+QD^2 \quad \text{(Pythagoras, } \triangle PDQ, \angle PDQ=90^\circ\text{)} \] \[a^2 = PD^2+c^2 \quad \cdots(i) \] \[PR^2 = PD^2+DR^2 \quad \text{(Pythagoras, } \triangle PDR, \angle PDR=90^\circ\text{)} \] \[b^2 = PD^2+d^2 \quad \cdots(ii) \] \[(i)-(ii): \quad a^2-b^2 = c^2-d^2 \] \[(a+b)(a-b) = (c+d)(c-d) \]Answer: \(\displaystyle (a+b)(a-b)=(c+d)(c-d)\)

  2. Exercise 12

    In a quadrilateral ABCD,∠A+∠D=90∘\displaystyle \mathrm{ABCD}, \angle \mathrm{A}+\angle \mathrm{D}=90^{\circ}. Prove that AC2+BD2=AD2+BC2\displaystyle \mathrm{AC}^2+\mathrm{BD}^2=\mathrm{AD}^2+\mathrm{BC}^2 [Hint: Produce AB and DC to meet at E.]

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q12 \[\angle EAD=\angle A, \quad \angle ADE=\angle D \quad \text{(E on AB, DC produced)} \] \[\angle A+\angle D = 90^\circ \quad \text{(given)} \] \[\angle AED = 180^\circ-(\angle A+\angle D) = 90^\circ \quad \text{(angle sum, } \triangle AED\text{)} \] \[AD^2 = AE^2+ED^2 \quad \text{(Pythagoras, } \triangle AED\text{)} \] \[BC^2 = BE^2+EC^2 \quad \text{(Pythagoras, } \triangle BEC, \angle BEC=90^\circ\text{)} \] \[AC^2 = AE^2+EC^2, \quad BD^2 = BE^2+ED^2 \quad \text{(Pythagoras, } \triangle AEC, \triangle BED\text{)} \] \[AC^2+BD^2 = (AE^2+ED^2)+(BE^2+EC^2) \] \[AC^2+BD^2 = AD^2+BC^2 \]Answer: \(\displaystyle AC^2+BD^2=AD^2+BC^2\)

  3. Exercise 13

    In fig. 6.20\displaystyle 6.20, l∥m\displaystyle l \| \mathrm{m} and line segments AB, CD and EF are concurrent at point P. Prove that AEBF=ACBD=CEFD\displaystyle \frac{\mathrm{AE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{BD}}=\frac{\mathrm{CE}}{\mathrm{FD}}. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q13

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    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q13 \[\angle APC = \angle BPD \quad \text{(vert.\ opp.)} \] \[\angle PCA = \angle PDB \quad \text{(alt.\ }\angle s,\ l\|m\text{)} \] \[\triangle APC \sim \triangle BPD \implies \frac{AP}{BP}=\frac{CP}{DP}=\frac{AC}{BD} \] \[\angle APE = \angle BPF \quad \text{(vert.\ opp.)} \] \[\angle PEA = \angle PFB \quad \text{(alt.\ }\angle s,\ l\|m\text{)} \] \[\triangle APE \sim \triangle BPF \implies \frac{AP}{BP}=\frac{EP}{FP}=\frac{AE}{BF} \] \[\angle CPE = \angle DPF \quad \text{(vert.\ opp.)} \] \[\angle PEC = \angle PFD \quad \text{(alt.\ }\angle s,\ l\|m\text{)} \] \[\triangle CPE \sim \triangle DPF \implies \frac{CP}{DP}=\frac{EP}{FP}=\frac{CE}{FD} \] \[\frac{AP}{BP}=\frac{CP}{DP}=\frac{EP}{FP} \implies \frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD} \] Answer: \(\displaystyle \dfrac{AE}{BF}=\dfrac{AC}{BD}=\dfrac{CE}{FD} \)

  4. Exercise 14

    In Fig. 6.21\displaystyle 6.21, PA,QB,RC\displaystyle \mathrm{PA}, \mathrm{QB}, \mathrm{RC} and SD are all perpendiculars to a line l,AB=6 cm\displaystyle l, \mathrm{AB}=6 \mathrm{~cm}, BC=9 cm,CD=12 cm\displaystyle \mathrm{BC}=9 \mathrm{~cm}, \mathrm{CD}=12 \mathrm{~cm} and SP=36 cm\displaystyle \mathrm{SP}=36 \mathrm{~cm}. Find PQ,QR\displaystyle \mathrm{PQ}, \mathrm{QR} and RS. NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q14

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    NCERT’s answer
    $\displaystyle 8$ cm, $\displaystyle 12$ cm, $\displaystyle 16$ cm
    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q14 Let \(\displaystyle \theta\) be the angle line \(\displaystyle PQRS\) makes with \(\displaystyle l\). \[PQ\cos\theta = AB,\quad QR\cos\theta = BC,\quad RS\cos\theta = CD \] \[PQ:QR:RS = AB:BC:CD = 6:9:12 = 2:3:4 \] Let \(\displaystyle PQ=2k,\ QR=3k,\ RS=4k\). \[PQ+QR+RS = SP \] \[2k+3k+4k = 36 \] \[9k = 36 \implies k = 4 \] \[PQ = 8\text{ cm},\quad QR = 12\text{ cm},\quad RS = 16\text{ cm} \] Answer: \(\displaystyle PQ=8\text{ cm},\ QR=12\text{ cm},\ RS=16\text{ cm}\)

  5. Exercise 15

    O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB∥DC\displaystyle \mathrm{AB} \| \mathrm{DC}. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO=QO\displaystyle \mathrm{PO}=\mathrm{QO}.

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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_Class10_Maths_Exemplar_Ch6_Ex6-4_Q15 \[\angle AOB=\angle COD \quad \text{(vert.\ opp.)} \] \[\angle OAB=\angle OCD \quad \text{(alt.\ }\angle s,\ AB\|DC\text{)} \] \[\triangle AOB \sim \triangle COD \implies \frac{AO}{CO}=\frac{BO}{DO} \] \[\implies \frac{AO}{AC}=\frac{BO}{BD} \quad \text{(componendo)} \] In \(\displaystyle \triangle ADC\), \(\displaystyle PO\|DC\): \[\triangle APO \sim \triangle ADC \implies \frac{PO}{DC}=\frac{AO}{AC} \] In \(\displaystyle \triangle BDC\), \(\displaystyle QO\|DC\): \[\triangle BQO \sim \triangle BDC \implies \frac{QO}{DC}=\frac{BO}{BD} \] \[\frac{PO}{DC}=\frac{AO}{AC}=\frac{BO}{BD}=\frac{QO}{DC} \implies PO=QO \] Answer: \(\displaystyle PO=QO\)

  6. Exercise 16

    In Fig. 6.22\displaystyle 6.22, line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and ∠AEF=∠AFE\displaystyle \angle \mathrm{AEF}=\angle \mathrm{AFE}. Prove that BDCD=BFCE\displaystyle \frac{\mathrm{BD}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{CE}}. [Hint: Take point G on AB such that CG∥DF\displaystyle \mathrm{CG} \| \mathrm{DF}.] NCERT_Question_Class10_Maths_Exemplar_Ch6_Ex6-4_Q16

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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_Class10_Maths_Exemplar_Ch6_Ex6-4_Q16 Construction: \(\displaystyle G\) on \(\displaystyle AB\) with \(\displaystyle CG\|DF\). In \(\displaystyle \triangle AGC\), \(\displaystyle FE\|GC\): \[\frac{AF}{FG}=\frac{AE}{EC}=1 \quad (E\text{ midpt.\ of }AC) \implies AF=FG \] \[AE=AF \quad (\angle AEF=\angle AFE) \implies FG=AE=EC \] In \(\displaystyle \triangle BFD\), \(\displaystyle CG\|FD\): \[\frac{BC}{CD}=\frac{BG}{GF} \implies \frac{BC+CD}{CD}=\frac{BG+GF}{GF} \] \[\frac{BD}{CD}=\frac{BF}{FG}=\frac{BF}{CE} \] Answer: \(\displaystyle \dfrac{BD}{CD}=\dfrac{BF}{CE}\)

  7. Exercise 17

    Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the 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_Class10_Maths_Exemplar_Ch6_Ex6-4_Q17 Area of a semicircle of diameter \(\displaystyle d\) is \(\displaystyle \dfrac{\pi d^2}{8}\). \[AC^2 = AB^2+BC^2 \quad \text{(Pythagoras, }\angle B=90^\circ\text{)} \] \[\frac{\pi\, AC^2}{8} = \frac{\pi\, AB^2}{8}+\frac{\pi\, BC^2}{8} \] Answer: area on \(\displaystyle AC\) = area on \(\displaystyle AB\) + area on \(\displaystyle BC\)

  8. Exercise 18

    Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

    Not cross-checked

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

    NCERT_Solution_Class10_Maths_Exemplar_Ch6_Ex6-4_Q18 Area of an equilateral triangle of side \(\displaystyle a\) is \(\displaystyle \dfrac{\sqrt3}{4}a^2\). \[AC^2 = AB^2+BC^2 \quad \text{(Pythagoras, }\angle B=90^\circ\text{)} \] \[\frac{\sqrt3}{4}AC^2 = \frac{\sqrt3}{4}AB^2+\frac{\sqrt3}{4}BC^2 \] Answer: area on \(\displaystyle AC\) = area on \(\displaystyle AB\) + area on \(\displaystyle BC\)