NCERT Solutions Class 10th Maths Chapter – 6 Triangles Examples

Example 1. If a line intersects sides AB and AC of a AABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC (see Fig. 6.13).

Solution: DE || BC
So, AD/DB = AE/EC (Theorem 6.1)
Or, DB/AD = EC/AE
Or, DB/AD + 1 = EC/AE + 1
Or, AB/AD = AC/AE
So, 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. Show that AE/ED = BF/ FC

Solution: Let us join AC to intersect EF at G

AB || DC and EF || AB (Given)
So, EF || DC (Lines parallel to the same line are parallel to each other)
Now, in △ ADC,
EG || DC (As EF || DC)
AE/ED = AG/GC (Theorem 6.1)
Similarly, from A CAB,
CG/AG = CF/BF
i.e., AG/GC = BF/FC
Therefore, form (1) and (2)
AG/GC = BF/FC

Example 3. In Fig. 6.16, PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle.

Solution: It is given that
So, ST || QR (Theorem 6.2)
Therefore, ∠PST = ∠PQR (Corresponding angles) …(1)
Also, it is given that
∠PST = ∠PQR ….(2)
So, ∠PST = ∠PQR [From (1) and (2)]
Therefore, PQ = PR (Sides opposite the equal angles)
i.e., PQR is an isosceles triangle.

Example 4. In Fig. 6.29, if PQ || RS, prove that A POQ ~ A SOR.

Solution :
So, PQ || RS (Given)
∠P = ∠S (Alternate angles)
and ∠Q= ∠R
Also, ∠POQ = ∠SOR (Vertically opposite angles)
Therefore, △ POQ ~ △ SOR (AAA similarity criterion)

Example 5. Observe Fig. 6.30 and then find ∠P.

Solution: In △ ABC and △ POR,
AB/RQ = 3.8/7.6 = 1/2, BC/QP = 6/12 = 1/2 and CA/PR = 3√3/6√3 = 1/2
That is, AB/RQ = BC/QP = CA/PR
So, △ABC ~ △RQP (SSS similarity)
Therefore, ∠C = ∠P (Corresponding angles of similar triangles)
But ∠C = 180° – ∠A – ∠B (Angle sum property)
= 180° – 80° – 60° = 40°
So, ∠P = 40°

Example 6. In Fig. 6.31,
OA . OB OC. OD.
Show that ∠A = ∠C and ∠B = ∠D.

Solution: OA . OB = OC . OD (Given)
So, OA/OC = OD/OB (1)
Also, we have
∠AOD = ∠COB (Vertically opposite angles) (2)
Therefore, from (1) and (2), △ AOD ~ △ 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.

Solution: Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post (see Fig. 6.32). From the figure, you can see that DE is the shadow of the girl. Let DE be x metres.

Now, BD = 1.2 m x 4 = 4.8 m.
Note that in A ABE and A CDE,
∠B = ∠D (Each is of 90° because lamp-post as well as the girl are standing vertical to the ground)
and ∠E = ∠E (Same angle)
So, △ ABE ~ △ CDE (AA similarity criterion)
Therefore, BE/DE = AB/CD
i.e., 4.8 + x/x = 3.6/0.9 (90 cm = 90/100 m = 0.9)
i.e., 4.8 + x = 4x
i.e., 3x = 4.8
i.e., x = 1.6
So, the shadow of the girl after walking for 4 seconds is 1.6 m long.

Example 8. In Fig. 6.33, CM and RN are respectively the medians of A ABC and
△ PQR. If △ ABC ~ △ PQR, prove that:
(i) △ AMC ~ △ PNR
(ii) CM/RN = AB/PQ
(iii) △ CMB ~ △ RNQ

Solution: (i)
△ ABC △ PQRSo, AB/PQ = BC/QR = CA/RP ….(1)
and ∠A = ∠P, ∠B = ∠Q And ∠C = ∠R ….(2)
But AB = 2 AM and PQ = 2 PN (As CM and RN are medians)
so, from (1), 2AM/2PN = CA/RP
i.e., AM/PN = CA/RP ….(3)
Also, ∠MAC = ∠NPR [From (2)] ….(4)
So, from (3) and (4),
△AMC ~ △PNR (SSS Similarity) ….(5)

(ii) From (5), CM/RN = CA/RP …(6)
But CA/RP = AB/PQ [from 1] ….(7)
Therefore CM/RN = AB/PQ [From (6) and (7)] ….(8)

(iii) Again AB/PQ = BC/QR [From (1)]
Therefore, CM/RN = BC/QR [form (8)] ….(9)
Also, CM/RN = AB/PQ = 2 BM/QN
i.e., CM/RN = BM/QN ….(10)
i.e., CM/RN = BC/QR = BM/QN [from (9) and (10)]
Therefore △CMB ~ △RNQ (SSS similarity)
[Note: You can also prove part (iii) by following the same method as used for proving part (i).]