Problem Set 1 Geometry 10th Std Maths Part 2 Answers Chapter 1 Similarity
Question 1
Maharashtra Board Solution
Select the appropriate alternative. i. In ∆ABC and ∆PQR, in a one to one correspondence = = , then (A) ∆PQR – ∆ABC (B) ∆PQR – ∆CAB (C) ∆CBA – ∆PQR (D) ∆BCA – ∆PQR
Solution & Step-by-Step Answer:
(B)
ii. If in ∆DEF and ∆PQR, ∠D ≅ ∠Q, ∠R ≅ ∠E, then which of the following statements is false?
(A) =
(B) =
(C) =
(D) =
Answer:
∆DEF ~ ∆QRP … [AA test of similarity]
∴ = = …[Corresponding sides of similar triangles]
(B)
iii. In ∆ABC and ∆DEF, ∠B = ∠E, ∠F = ∠C and AB = 3 DE, then which of the statements regarding the two triangles is true?
(A) The triangles are not congruent and not similar.
(B) The triangles are similar but not congruent.
(C) The triangles are congruent and similar.
(D) None of the statements above is true.
Answer:
(B)
iv. ∆ABC and ∆DEF are equilateral triangles, A(∆ABC) : A(∆DEF) = 1 : 2. If AB = 4, then what is length of DE?
(A) 2√2
(B) 4
(C) 8
(D) 4√2
Answer:
Refer Q. 6 Practice Set 1.4
(D)
v. In the adjoining figure, seg XY || seg BC, then which of the following statements is true?
(A) =
(B) =
(C) =
(D) =
Answer:
∆ABC ~ ∆AXY … [AA test of similarity]
∴ = …[Corresponding sides of similar triangles]
∴ = …[Altemendo]
(A)
Question 2
Maharashtra Board Solution
In ∆ABC, B-D-C and BD = 7, BC = 20, then find following ratios.
Solution & Step-by-Step Answer:
Draw AE ⊥ BC, B – E – C. BC = BD + DC [B – D – C] ∴ 20 = 7 + DC ∴ DC = 20 – 7 = 13
i. ∆ABD and ∆ADC have same height AE.
[Triangles having equal height]
∴
ii. ∆ABD and ∆ABC have same height AE.
[Triangles having equal height]
∴
iii. ∆ADC and ∆ABC have same height AE.
[Triangles having equal height]
∴
Question 3
Maharashtra Board Solution
Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm, then what is the corresponding base of the bigger triangle?
Solution & Step-by-Step Answer:
Let A1 and A2 be the areas of two triangles. Let b1 and b2 be their corresponding bases. A1 : A2 = 2 : 3
∴ The corresponding base of the bigger triangle is 9 cm.
Question 4
Maharashtra Board Solution
In the adjoining figure, ∠ABC = ∠DCB = 90°, AB = 6, DC = 8, then
Solution & Step-by-Step Answer:
∆ABC and ∆DCB have same base BC.
Question 5
Maharashtra Board Solution
In the adjoining figure, PM = 10 cm, A(∆PQS) = 100 sq. cm, A(∆QRS) = 110 sq. cm, then find NR.
Solution & Step-by-Step Answer:
∴ NR = 11 cm
Question 6
Maharashtra Board Solution
∆MNT ~ ∆QRS. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio
Solution & Step-by-Step Answer:
∆MNT- ∆QRS [Given] ∴ ∠M ≅ ∠Q (i) [Corresponding angles of similar triangles] In ∆MLT and ∆QPS, ∠M ≅ ∠Q [From (i)] ∠MLT ≅ ∠QPS [Each angle is of measure 90°] ∴ ∆MLT ~ ∆QPS [AA test of similarity]
Question 7
Maharashtra Board Solution
In the adjoining figure, A – D – C and B – E – C. seg DE || side AB. If AD = 5, DC = 3, BC = 6.4, then find BE.
Solution & Step-by-Step Answer:
In ∆ABC, seg DE || side AB [Given] ∴ = [Basic proportionality theorem] ∴ = ∴ 3x = 5 (6.4 – x) ∴ 3x = 32 – 5x ∴ 8x = 32 ∴ x = =4 ∴ BE = 4 units
Question 8
Maharashtra Board Solution
In the adjoining figure, seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. AB = 60, BC = 70, CD = 80, PS = 280, then find PQ, QR and RS.
Solution & Step-by-Step Answer:
seg PA, seg QB, seg RC and seg SD are perpendicular to line AD. [Given] ∴ seg PA || seg QB || seg RC || seg SD (i) [Lines perpendicular to the same line are parallel to each other] Let the value of PQ be x and that of QR be y. PS = PQ + QS [P – Q – S] ∴ 280 – x + QS ∴ QS = 280 – x (ii) Now, seg PA || seg QB || seg SD [From (i)] ∴ = [Property of three parallel lines and their transversals] ∴ = [B – C – D] ∴ = ∴ = ∴ = ∴ 5x = 2 (280 – x) ∴ 5x = 560 – 2x ∴ 7x = 560 ∴ x = = 80 ∴ PQ = 80 units QS = 280 – x [From (ii)] = 280 – 80 = 200 units But, QS = QR + RS [Q – R – S] ∴ 200 = y + RS ∴ RS = 200 – y (ii) Now, seg QB || seg RC || seg SD [From (i)] ∴ = [Property of three parallel lines and their transversals] ∴ = ∴ = ∴ 8y = 7(200 – y) ∴ 8y = 1400 – 7y ∴ 15y = 1400 ∴ y = = ∴ QR = units RS = 200 – 7 [From (iii)] = 200 – = = ∴ RS = units
Question 9
Maharashtra Board Solution
In ∆PQR, seg PM is a median. Angle bisectors of ∠PMQ and ∠PMR intersect side PQ and side PR in points X and Y respectively. Prove that XY || QR Complete the proof by filling in the boxes.
Solution & Step-by-Step Answer:
Proof: In ∆PMQ, ray MX is bisector of ∠PMQ.
Question 10
Maharashtra Board Solution
In the adjoining figure, bisectors of ∠B and ∠C of ∆ABC intersect each other in point X. Line AX intersects side BC in point Y. AB = 5, AC = 4, BC = 6, then find .
Solution & Step-by-Step Answer:
Let the value of BY be x. BC = BY + YC [B – Y – C] ∴ 6 = x + YC ∴ YC = 6 – x in ∆BAY, ray BX bisects ∠B. [Given] ∴ = (i) [Property of angle bisector of a triangle] Also, in ∆CAY, ray CX bisects ∠C. [Given]
Question 11
Maharashtra Board Solution
In ꠸ABCD, seg AD || seg BC. Diagonal AC and diagonal BD intersect each other in point P. Then show that =
Solution & Step-by-Step Answer:
proof: seg AD || seg BC and BD is their transversal. [Given] ∴ ∠DBC ≅ ∠BDA [Alternate angles] ∴ ∠PBC ≅ ∠PDA (i) [D – P – B] In ∆PBC and ∆PDA, ∠PBC ≅ ∠PDA [From (i)] ∠BPC ≅ ∠DPA [Vertically opposite angles] ∴ ∆PBC ~ ∆PDA [AA test of similarity] ∴ = [Corresponding sides of similar triangles] ∴ = [By altemendo]
Question 12
Maharashtra Board Solution
In the adjoining figure, XY || seg AC. If 2 AX = 3 BX and XY = 9, complete the activity to find the value of AC.
Solution & Step-by-Step Answer:
2 AX = 3 BX [Given]
Question 13
Maharashtra Board Solution
In the adjoining figure, the vertices of square DEFG are on the sides of ∆ABC. If ∠A = 90°, then prove that DE2 = BD × EC. (Hint: Show that ∆GBD is similar to ∆ CFE. Use GD = FE = DE.)
Solution & Step-by-Step Answer:
proof: ꠸DEFG is a square. ∴ DE = EF = GF = GD (i) [Sides of a square] ∠GDE = ∠DEF = 90° [Angles of a square] ∴ seg GD ⊥ side BC, seg FE ⊥ side BC (ii) In ∆BAC and ∆BDG, ∠BAC ≅ ∠BDG [From (ii), each angle is of measure 90°] ∠ABC ≅ ∠DBG [Common angle] ∴ ∆BAC – ∆BDG (iii) [AA test of similarity] In ∆BAC and ∆FEC, ∠BAC ≅ ∠FEC [From (ii), each angle is measure 90°] ∠ACB ≅ ∠ECF [Common angle] ∴ ∆BAC – ∆FEC (iv) [AA test of similarity] ∴ ∆BDG – ∆FEC [From (iii) and (iv)] ∴ = (v) [Corresponding sides of similar triangles] ∴ = [From (i) and (v)] ∴ DE2 = BD × EC