TOPICS

Unit-6(Examples)

Triangles

**Example-1 :-** If a line intersects sides AB and AC of a Δ ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC (see figure).

Given that : DE || BC. Prove that : AD/AB = AE/EC Proof : We have DE || BC (Given) so, AD/DB = AE/EC (Theorem-1) DB/AD = EC/AE Both sides adding by 1, we get DB/AD + 1 = EC/AE + 1 (DB + AD)/AD = (EC + AE)/AE AB/AD = AC/AE or AD/AB = AE/AC

**Example-2 :-** ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB (see Figure). Show that AE/ED = BF/FC.

Given that : ABCD is a trapezium with AB || DC and EF || AB. Prove that : AE/ED = BF/FC. Construction : Let us join AC to intersect EF at G. Proof : We have AB || DC and EF || AB (Given) so, EF || DC (Lines parallel to the same line are parallel to each other) Now, in Δ ADC, EG || DC (As EF || DC) So, AE/ED = AG/GC ......(i) (Theorem-1) Similarly, from Δ CAB, CG/AG = CF/BF or AG/GC = BF/FC .......(ii) Therefore, from (i) and (ii) AE/ED = BF/FC

**Example-3 :-** In figure, PS/SQ = PT/TR and ∠ PST = ∠ PRT. Prove that PQR is an isosceles triangle.

Given that : PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that : PQR is an isosceles triangle. Proof : We have PS/SQ = PT/TR (Given) so, ST || QR (Theorem-2) Therefore, ∠ PST = ∠ PQR....(i) (Corresponding angles) Also, we have ∠ PST = ∠ PRQ ......(ii) (Given) So, ∠ PRQ = ∠ PQR [From (i) and (ii)] Therefore, PQ = PR (Sides opposite the equal angles) i.e., PQR is an isosceles triangle.

**Example-4 :-** In Figure, if PQ || RS, prove that Δ POQ ~ Δ SOR.

Given that : PQ || RS Prove that : Δ POQ ~ Δ SOR Proof : We have PQ || RS (Given) So, ∠ P = ∠ S (Alternate angles) and ∠ Q = ∠ R Also, ∠ POQ = ∠ SOR (Vertically opposite angles) Therefore, Δ POQ ~ Δ SOR (AAA similarity criterion)

**Example-5 :-** Observe Figure and then find ∠ P.

In Δ ABC and Δ PQR, AB/RQ = 3.8/7.6 = 1/2 BC/QP = 6/12 = 1/2 CA/PR = 3√3/6√3 = 1/2 i.e. AB/RQ = BC/QP = CA/PR So, Δ ABC ~ Δ RQP (SSS similarity) Therefore, ∠ C = ∠ P (Corresponding angles of similar triangles) But ∠ C = 180^{o}– ∠ A – ∠ B (Angle sum property) = 180^{o}– 80^{o}– 60^{o}= 40^{o}So, ∠ P = 40^{o}

**Example-6 :-** In Figure, OA . OB = OC . OD. Show that ∠ A = ∠ C and ∠ B = ∠ D.

Given that : OA . OB = OC . OD Prove that : ∠ A = ∠ C and ∠ B = ∠ D Proof : We have OA . OB = OC . OD (Given) So, OA/OC = OD/OB ......(i) (Alternate angles) Also, we have ∠ AOD = ∠ COB .....(ii) (Vertically opposite angles) Therefore, from (i) and (ii), Δ AOD ~ Δ COB (SAS similarity criterion) So, ∠ A = ∠ C and ∠ D = ∠ B (Corresponding angles of similar triangles)

**Example-7 :-** A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post. In figure, DE is the shadow of the girl. Let DE be x metres. Now, BD = 1.2 m × 4 = 4.8 m. In Δ ABE and Δ CDE, ∠ B = ∠ D (Each is of 90^{o}because lamp-post as well as the girl are standing vertical to the ground) and ∠ E = ∠ E (Same angle) So, Δ ABE ~ Δ CDE (AA similarity criterion) Therefore, (4.8 + x)/x = 3.6/0.9 4.8 + x = 4x 4.8 = 4x - x 4.8 = 3x 4.8/3 = x x = 1.6 So, the shadow of the girl after walking for 4 seconds is 1.6 m long.

**Example-8 :-** In Figure, CM and RN are respectively the medians of Δ ABC and Δ PQR. If Δ ABC ~ Δ PQR, prove that :

(i) Δ AMC ~ Δ PNR

(ii) CM/RN = AB/PQ

(iii) Δ CMB ~ Δ RNQ

Given that : CM and RN are respectively the medians of Δ ABC and Δ PQR. Δ ABC ~ Δ PQR Prove that :(i) Δ AMC ~ Δ PNR (ii) CM/RN = AB/PQ (iii) Δ CMB ~ Δ RNQ Proof : (i) We have Δ ABC ~ Δ PRQ (Given) So, AB/PQ = BC/QR = CA/RP ......(i) and ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R .....(ii) AB = 2 AM and PQ = 2 PN (As CM and RN are medians) So, from (i), 2AM/2PN = CA/RP i.e., AM/PN = CA/RP .......(iii) Also, ∠ MAC = ∠ NPR [From (ii)] ......(iv) So, from (iii) and (iv), Δ AMC ~ Δ PNR .......(v) (SAS similarity)

(ii) From (v) CM/RN = CA/RP ....(vi) CA/RP = AB/PQ [from (i)] ....(vii) Therefore, CM/RN = AB/PQ [from (vi) and (vii)] .....(viii)

(iii) Again, AB/PQ = BC/QR [frrom (i)] Therefore, CM/RN = BC/QR [from (viii)] .....(ix) Also, CM/RN = AB/PQ = 2BM/2QN CM/RN = BM/QN .......(x) CM/RN = BC/QR = BM/QN [from (ix) and (x)] Therefore, Δ CMB ~ Δ RNQ (SSS similarity)

**Example-9 :-** In Figure, the line segment XY is parallel to side AC of Δ ABC and it divides the triangle into two parts of equal areas. Find the ratio AX/AB.

Given that The line segment XY is parallel to side AC of Δ ABC. i.e., XY || AC So, ∠ BXY = ∠ A and ∠ BYX = ∠ C (Corresponding angles) Therefore, Δ ABC ~ Δ XBY (By AA similarity Rule) So, ar(ABC)/ar(XBY) = (AB/XB)^{2}....(i) (Therorem-6) Also given It divides the triangle into two parts of equal areas. i.e., ar (ABC) = 2 ar (XBY) So, ar(ABC)/ar(XBY) = 2/1 .........(ii) Therefore, from (i) and (ii), (AB/XB)^{2}= 2/1 AB/XB = √2/1 XB/AB = 1/√2 1 - XB/AB = 1 - 1/√2 (AB - XB)/AB = (√2 - 1)/√2 AX/AB = (2 - √2)/2

**Example-10 :-** In Figure, ∠ ACB = 90^{o} and CD ⊥ AB. Prove that BC^{2}/AC^{2} = BD/AD.

Given that : ∠ ACB = 90^{o}and CD ⊥ AB. Prove that : BC^{2}/AC^{2}= BD/AD Proof : We have ∠ ACB = 90^{o}and CD ⊥ AB (Given) Δ ACD ~ Δ ABC (Theorem-7) So, AC/AB = AD/AC AC^{2}= AB . AD ......(i) Similarly, Δ BCD ~ Δ BAC (Theorem-7) So, BC/BA = BD/BC BC^{2}= BA . BD ......(ii) Therefore, from (i) and (ii), BC^{2}/AC^{2}= (BA . BD)/(AB . AD) = BD/AD

**Example-11 :-** A ladder is placed against a wall such that its foot is at a distance of 2.5 m from the wall and its top reaches a window 6 m above the ground. Find the length of the ladder.

Let AB be the ladder and CA be the wall with the window at A. Also, BC = 2.5 m and CA = 6 m From Pythagoras Theorem, we have: AB^{2}= BC^{2}+ CA^{2}AB^{2}= (2.5)^{2}+ (6)^{2}AB^{2}= 42.25 AB = 6.5 Thus, length of the ladder is 6.5 m.

**Example-12 :-** In Figure, if AD ⊥ BC, prove that AB^{2} + CD^{2} = BD^{2} + AC^{2}.

Given that : AD ⊥ BC Prove that : AB^{2}+ CD^{2}= BD^{2}+ AC^{2}Proof : In Δ ADC, we have AC^{2}= AD^{2}+ CD^{2}....(i) (By Pythagoras Theorem) In Δ ADB, we have AB^{2}= AD^{2}+ BD^{2}....(ii)(By Pythagoras Theorem) By Subtracting (i) from (ii), we get AB^{2}– AC^{2}= BD^{2}– CD^{2}AB^{2}+ CD^{2}= BD^{2}+ AC^{2}

**Example-13 :-** BL and CM are medians of a triangle ABC right angled at A. Prove that 4(BL^{2} + CM^{2}) = 5BC^{2}.

Given that : BL and CM are medians of the Δ ABC in which ∠ A = 90^{o}Prove that : 4(BL^{2}+ CM^{2}) = 5BC^{2}Proof : In Δ ABC, we have BC^{2}= AB^{2}+ AC^{2}....(i) (By Pythagoras Theorem) In Δ ABL, we have BL^{2}= AL^{2}+ AB^{2}(By Pythagoras Theorem) BL^{2}= (AC/2)^{2}+ AB^{2}(L is the mid point of AC) BL^{2}= AC^{2}/4 + AB^{2}4BL^{2}= AC^{2}+ 4AB^{2}.......(ii) In Δ CMA, we have CM^{2}= AC^{2}+ AM^{2}(By Pythagoras Theorem) CM^{2}= AC^{2}+ (AB/2)^{2}(M is the mid point of AB) CM^{2}= AC^{2}+ AB^{2}/4 4CM^{2}= 4AC^{2}+ AB^{2}.......(iii) By Adding (ii) and (iii), we get 4(BL^{2}+ CM^{2}) = 5(AC^{2}+ AB^{2}) 4(BL^{2}+ CM^{2}) = 5BC^{2}[From (i)]

**Example-14 :-** O is any point inside a rectangle ABCD (see Figure). Prove that OB^{2} + OD^{2} = OA^{2} + OC^{2}.

Given that : O is any point inside a rectangle ABCD. Prove that : OB^{2}+ OD^{2}= OA^{2}+ OC^{2}Construction : Draw PQ || BC Proof : Through O, draw PQ || BC so that P lies on AB and Q lies on DC. Now, PQ || BC Therefore, PQ ⊥ AB and PQ ⊥ DC(∠ B = 90^{o}and ∠ C = 90^{o}) So, ∠ BPQ = 90^{o}and ∠ CQP = 90^{o}Therefore, BPQC and APQD are both rectangles. Now, in Δ OPB, OB^{2}= BP^{2}+ OP^{2}...... (i) Similarly, in Δ OQD, OD^{2}= OQ^{2}+ DQ^{2}...... (ii) Similarly, in Δ OQC, OC^{2}= OQ^{2}+ CQ^{2}...... (iii) Similarly, in Δ OAP, OA^{2}= AP^{2}+ OP^{2}...... (iv) By Adding (i) and (ii), OB^{2}+ OD^{2}= BP^{2}+ OP^{2}+ OQ^{2}+ DQ^{2}OB^{2}+ OD^{2}= CQ^{2}+ OP^{2}+ OQ^{2}+ AP^{2}(As BP = CQ and DQ = AP) OB^{2}+ OD^{2}= CQ^{2}+ OQ^{2}+ OP^{2}+ AP^{2}OB^{2}+ OD^{2}= OC^{2}+ OA^{2}[From (iii) and (iv)]

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