## Chapter 1 – Similarity

## Practice Sums

**1 Mark Questions:**

(1) In ∆ ABC, AB = 3 cm, BC = 2 cm and AC = 2.5 cm. ∆ DEF ~ ∆ ABC, EF = 4 cm. What is the perimeter of ∆ DEF?

(2) The sides of two similar triangles are 4 : 9. What is the ratio of their area?

(3) The areas of two similar triangles are 18 cm² and 32 cm² respectively. What is the ratio of their corresponding sides?

(4) ∆ ABC ~ ∆ PQR, AB = 6 cm, BC = 8 cm, CA = 10 cm and QR = 6 cm. What is the length of side PR?

(5) In ∆ ABC, AB = 6 cm, BC = 8 cm and AC = 10 cm. ∆ ABC is enlarged to ∆ PQR such that the largest side is 12.5 cm. What is the length of the smallest side of ∆ PQR?

(6) In ∆ ABC, B – D – C and BD = 6 cm, DC = 4 cm. What is the ratio of A (∆ ABC) to A (∆ ACD)?

(7) In ∆ XYZ, PQ YZ, X – P – Y and X – Q – Z. If \(\large \frac {XP}{PY}\)= \(\large \frac {4}{13}\) and XQ = 4.8 cm. What is XZ?

(8) In ∆ ABC, P is a point on side BC such that BP = 4 cm and PC = 7 cm. Then what is the ratio of A (∆ APC) : A (∆ ABC)?

(9) In ∆ PQR, seg RS is the bisector of ∠PRQ, PS = 8, SQ = 6, PR = 20 then what is the value of QR?

(10) In ∆ ABC, line PQ || side BC, AP = 3, BP = 6, AQ = 5, then what is the value of CQ?

(11) In the adjoining figure, seg BE ⊥ seg AB and seg BA ⊥ seg AD. If BE = 6, AD = 9, then find A (∆ ABE) : A (∆ BAD)

(12) The ratio of the areas of two triangles with the common base is 6 : 5. Height of the larger triangles is 9 cm. Then find the corresponding height of the smaller triangle.

(1) 15 cm

(2) 16 : 81

(3) 3 : 4

(4) 7.5 cm

(5) 7.5 cm

(6) 5 : 2

(7) 20.4 cm

(8) 7 : 11

(9) 15 units

(10) 10

(11) 2 : 3

(12) 7.5 cm

**2 Marks Questions:**

(1) The ratio of the areas of two triangles with equal height is 3 : 2. The base of the larger triangle is 18 cm. Find the corresponding base of the smaller triangle.

(2) In ∆ DEF, line PQ || side EF. DQ = 1.8, QF = 5.4, PE = 7.2. Find DE.

(3) In ∆ PQR, seg RS is bisector of ∠PRQ. PS = 6, SQ = 8, PR = 15. Find QR.

(4) In ∆ XYZ, XY = YZ, Ray YM bisects ∠XYZ. X – M – Z. Prove that M is the midpoint of seg XZ.

(5) A vertical pole of a length 6 m casts a shadow of 4 m long on the ground. At the same time a tower casts a shadow 28 m long. Find the height of the tower.

(6) Let X be any point on side BC of ∆ ABC. Seg XM || side AB and seg XN || side CA. M – N – T, T – B – X. Prove that, TX² = TB × TC.

(7) ∆ DEF ~ ∆ MNK, If DE = 5 and MN = 6, then find the value of A (∆ DEF) : A (∆ MNK)

(8) If ∆ ABC ~ ∆ DEF such that the area of ∆ ABC is 9 cm² and the area of ∆ DEF is 16 cm². If BC = 2.1 cm. Find length of EF.

(1) 12 cm

(2) 9.6 units

(3) 20 units

(5) 42 m

(7) 25 : 36

(8) 2.8

**3 Marks Questions:**

(1) In the adjoining figure, RP : PK = 3 : 2, then find the value of

(i) A (∆ TRP) : A (∆ TPK)

(ii) A (∆ TRK) : A (∆ TPK)

(iii) A (∆ TRP) : A (∆ TRK)

(2) In the adjoining figure, seg DH ⊥ seg EF, seg GK ⊥ seg EF. If DH = 12 cm, GK = 20 cm and A(∆ DEF) = 300 cm², then find,

(i) EF

(ii) A(∆ GEF)

(3) In the adjoining figure, seg ML || seg BC, seg NL || seg DC. Prove that AM : AB = AN : AD.

(4) □ ABCD is a trapezium in which AB || DC and its diagonals intersect each other at point O. Show that AO : BO = CO : DO.

(5) Point D and E are the points on sides AB and AC such that AB = 5.6, AD = 1.4, AC = 7.2 and AE = 1.8. Show that DE || BC.

(6) In ∆ PQR, ray QS bisects of ∠PQR. P – S – R. Show that \(\large \frac {A\, ∆ PQS}{A\, ∆ QRS}\) = \(\large \frac {PQ}{QR}\)

(7) In the adjoining figure, seg PQ || seg AB. Seg PR || seg BD. Prove that QR || AD.

(8) In the adjoining figure, seg PA, seg QB, seg RC and seg SD are ⊥ to line l AB = 6, BC = 9, CD = 12 and PS = 36, then find PQ, QR and RS.

(9) In ∆ ABC, AB = 5, BC = 6, AC = 7. ∆ PQR ~ ∆ ABC. Perimeter of ∆PQR is 360. Find PQ, QR and PR.

(10) In ∆ ABC, ∠B = 90⁰, seg DE ⊥ side AC. AD = 6, AB = 12, AC = 18, then find AE.

(11) E is a point on side CB, C – B – E, In ∆ ABC, AB = AC. If seg AD ⊥ BC, B – D – C and seg EF ⊥ side AC, A – F – C. Prove that ∆ ABD ~ ∆ ECF.

(12) D is a point on side BC of ∆ ABC such that, ∠ADC = ∠BAC. Show that AC² = BC × DC.

(13) In ∆ RES, RE = 15, SE = 10. In ∆ PEA, PE = 8, AE = 12. Prove that ∆ RES ~ ∆ AEP.

(14) In the adjoining figure, seg DE || side BC. If DE : BC = 3 : 5, then find A(∆ ADE) : A(□DBCE).

(15) In ∆ ABC, PQ is a line segment that intersects AB at point P and AC at point Q. PQ || BC. If PQ divides ∆ ABC into two equal parts equal in area, find BP : AB.

(1) (i) 3 : 2

(ii) 5 : 2

(iii) 3 : 5

(2) (i) 50 cm

(ii) 500 cm²

(9) 100 units, 120 units, 140 units

(10) 4 units

(14) 9 : 16

(15) \(\large \frac {\sqrt{2} \,– \,1}{\sqrt{2}}\)

**4 Marks Questions:**

(1) In the adjoining figure, seg CE ⊥ side AB, seg AD ⊥ side BC. Prove that,

(i) ∆ AEP ~ ∆ CDP

(ii) ∆ AEP ~ ∆ ADB

(2) In the adjoining figure, if ∆ ABN ≅ ∆ ACM, show that ∆ AMN ~ ∆ ABC.

(3) In the adjoining figure, seg AB || side DC, OD = x OB = x – 3, OC = x – 5, OA = 3x – 19. Find the value of x.

(4) In ∆ ABC, ∠ABC = 90⁰. ∆ PAB, ∆ QAC and ∆ RBC are the equilateral triangles constructed on sides AB, AC and BC respectively. Prove that : A(∆ PAB) + A(∆ RBC) = A(∆ QAC).

(5) In ∆ABC, seg DE side BC. If 2 A(∆ ADE) = A(□ DBCE). Show that BC = \(\sqrt{3}\) × DE.

(3) x = 8, x = 9