TRIANGLE EXERCISE 6.4

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1 TRIANGLE EXERCISE 6.4. Let Δ ABC ~ Δ DEF and their areas be, respectively, 64 cm2 and 2 cm2. If EF =5.4 cm, find BC. Ans; According to question ABC~ DEF ar( DEF) = AB2 DE 2 = BC2 EF 2 = AC2 DF 2 64cm 2 2cm 2 = 64 2 = BC 5.4 8cm cm = BC 5.4 BC = BC =.2 cm (Bc)2 (5.4) 2 2. Diagonals of a trapezium ABCD with AB DC intersect each other at the point O.If AB = 2 CD, find the ratio of the areas of triangles AOB and COD. Ans:2 According to question AB DC OAB = OCD and OBA = ODC (Alternate interior angle) AOB = COD (Vertical opposite angle) (ii) By equation and (ii) AOB~ COD

2 ar( COD) = AB2 CD 2 = OA2 OC 2 = OB2 OD 2 ar( COD) = AB2 CD 2 AB = 2 CD ar( COD) = (2CD)2 CD 2 ar( COD) = 4CD2 CD 2 ar( COD) = 4: 3. In Fig. 6.44, ABC and DBC are two triangles on thesame base BC. If AD intersects BC at O, show that ar (ABC) ar (DBC = AO DO Ans:3 Draw a perpendicular from A to BC is AL and from D to DM on BC. ar ( ABC) = 2 ar ( ABC) = 2 ar ( DBC) = 2 ar ( ABC) = 2 base height BC AL base height BC DM (ii) In ALO and DMO AOL = DOM (Vertical opposite angle) ALO = DMO = 90 ALO ~ DMO (By AA similarity rule)

3 AL DM = AO DO By equation and (ii) ar( DBC) = 2 2 ar( DBC) = AL DM AL Bc DM BC By using in above ar( DBC) = AL DM = OA OD ar( DBC) = OA OD 4. If the areas of two similar triangles are equal, provethat they are congruent. Ans:4 Let we have two triangles PQR and LMN both are similar and equal. When triangles are similar then the ratio of their areas will be equal to the square of the ratio of their corresponding sides. PQR ~ LMN ar( LMN) = PQ2 LN 2 = ar( LMN) (given) (ii) By using equation (ii) ar( LMN) = By using in equation ar( LMN) = PQ2 LN 2 =

4 PQ 2 LN 2 = 2 = LM 2 ; PR 2 = LN 2 ; QR 2 = MN 2 = LM; PR = LN; QR = MN 5. D, E and F are respectively the mid-points of sides AB, BC and CA of Δ ABC. Find theratio of the areas of Δ DEF and Δ ABC. Ans:5 According to question D, E, F are the mid-points of sides AB, BC, CA of ABC. DE ACand DE = AC, DF BC and DF = BC, EF 2 2 AB and EF = 2 AB DF BC and BE = 2 BC DF BE and DF = BE DE AC and AF = 2 AC DE AF and DE = AF (ii) From equation and (ii) it is clear that BDFC is a parallelogram. ar ( BDE)=ar( DEF) Similarly, DECF is a parallelogram.(de CF, CE DF) ar ( EFC)=ar( DEF) (iv) Similarly, ADEF is a parallelogram.(ad EF, DE AF)

5 ar ( ADF)=ar( DEF) (v) By equation, (iv)and (v) ar ( ADF)=ar( DEF)= ar (BDE)=ar( CEF) (vi) and we know, ar ( ABC)= ar( DEF) + ar (BDE) + ar( CEF) + ar ( ADF) by equation (vi) ar ( ABC)=4ar( DEF) ar( DEF) ar ( ABC) = 4 6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratioof their corresponding medians. Ans:6 Let the two triangles are PQR and LMN and their medians are PD and LS. PQR ~ LMN therefore the ratio of their corresponding sides are equal. LM = QR MN = PR LN And their corresponding angles are also equal P = L, Q = M, R = N (ii) QD = 2 QR and MS = 2 MN By using equation, we get

6 = 2 QD PR = LM MN LN 2 LM = QD MN = PR LN In PQD and LMS Q = M PQ = QD LM MN PQD ~ LMS (By equation ii) (By equation iii) (By SAS similarity rule) LM = QD MN = PD LS (iv) ar( LMN) = PQ2 LN 2 By using equation and (iv), we get PQ = QD = PR = PD LM MN LN LS ar( LMN) = PD2 LS 2 7. Prove that the area of an equilateral triangle described on one side of a square is equalto half the area of the equilateral triangle described on one of its diagonals. Ans:7 Let the sides of be x. By using Pythagorean Theorem The length of diagonal is 2 x Side of an equilateral triangle formed by the diagonal of square is 2 x.

7 And, Side of an equilateral triangle formed by the side of square is x. All equilateral triangles are similar by AAA similarity rule because each angle of equilateral triangle is 60. ar (PQL) ar(rsn) = PQ2 RS 2 ar (PQL) = x 2 = ar(rsn) ( 2x) 2 2 Tick the correct answer and justify : 8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio ofthe areas of triangles ABC and BDE is (A) 2 : (B) : 2 (C) 4 : (D) : 4 Ans: 8 Equilateral triangles are similar by AAA similarity rule because each angle of equilateral triangle is 60. ( ABC) ~( BDE) Let the side of triangle ABC = a Therefore side of triangle BDE = a/2 ar (ABC) = 2 2 = 4 ar(bde) (a/2) 2 9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 8 : 6 (D) 6 : 8 Ans:9 when two triangles are similar then the area of two similar triangles is equal to the square of the ratio of their corresponding sides. Area of the triangles are in the ratio of (6/8)

Mathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2.

Mathematics. Exercise 6.4. (Chapter 6) (Triangles) (Class X) Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. () Exercise 6.4 Question 1: Let and their areas be, respectively, 64 cm 2 and 121 cm 2. If EF = 15.4 cm, find BC. Answer 1: 1 () Question 2: Diagonals of a trapezium ABCD with AB DC intersect each other