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——— St, Hsu's, , The Learning App, CBSE Board Class 10 Maths Chapter 6- Triangles, Objective Questions, , Areas of Similar Triangles, , 1. If A ABC~ A DEF such that AB = 12 cm and DE = 14 cm. Find the ratio of areas, of A ABC and A DEF., , (A) 49/9, , (B) 36/49, (c) 49/16, (D) 25/49, , Answer: (B) 36/49, , Solution: We know that the ratio of areas of two similar triangles is equal, to the ratio of the squares., Of any two corresponding sides,, area of A ABC / area of A DEF = (AB/DE) 2= (12/14) 2= 36/49, , 2. Dand E are points on the sides AB and AC respectively of a AABC such that DE | |, BC. Which of the following statement is true?, , (i) A ADE ~ A ABC, (ii) (area of A ADE/ area of A ABC) = (AD?/AB?), (iii) (area of A ADE/ area of A ABC)= (AB2/ AD?), , (A) only (iii), (B) only (i), (C) only (i) and (ii), (D) all (i) , (ii) and (iii), Answer: (i) A ADE ~ A ABC and (ii) (area of A ADE/ area of A ABC) = (AD?/AB2), , Solution:, , https://byjus.com
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Hsu's, , The Learning App, , & Cc, In A ADE and A ABC, we have, ZADE=2B, , [Since DE || BC 2 ADE = Z B (Corresponding angles)], , and, 2 A= A A [Common], A ADE ~ A ABC, , Therefore, (area of A ADE / area of A ABC) = (AD2/AB?), , In the figure, PB and QA are perpendicular to segment AB. If OA = 5 cm, PO = 7cm, and area (AQOA) = 150 cm?, find the area of APOB., , (A) 233 cm?, (B) 294 cm?, (C) 300 cm?, (D) 420 cm?, , Answer: (B) 294 cm?, , https://byjus.com
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——— St, Hsu's, , The Learning App, Solution: Consider AYQOA and A POB, QA || PB,, Therefore, 2 AQO = Z PBO [Alternate angles], Z QAO = Z BPO [Alternate angles], and, , ZQOA = Z BOP [Vertically opposite angles], , As QOA ~ BOP [by AAA similarity], , Therefore, (OQ/ OB) = (OA/OP), , Now, area (POB)/ area (QOA) = (OP) 2/ (OA) 2= 72/ 52, , Since area (QOA) =150cm?, , =area (POB) =294cm?, , 4. Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The, , ratio of corresponding heights is:, (A) 4:5, (B) 5:4, (C) 3:2, (D) 5:7, Answer: (A) 4:5, Solution: For similar isosceles triangles,, , Area (A1) / Area (Az) = (hi)? / (hz)?, , (hi / ha) = 4/5, , Basic Proportionality Theorem, , 5. In AABC, AB = 3 and, AC = 4 cm and AD is the bisector of ZA. Then, BD : DC is —, , https://byjus.com
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fy SYvu's, , The Learning App, , (A) 9: 16, (B) 4:3, (Cc) 3:4, (D) 16:9, , Answer: (C) 3:4, , Solution:, , , , The Angle-Bisector theorem states that if a ray bisects an angle of a triangle, then, it divides the opposite side into segments that are proportional to the other two, sides (It may be similar or may not depending on type of triangle it divides), , In AABC, , as per the statement AB/ AC= BD/DC i.e. a/b= c/d, , So, BD/ DC= AB/AC= %, , So, BD: DC = 3:4, , 6. ABCD is a parallelogram with diagonal AC If a line XY is drawn such that XY || AB., , BX/XC=?, , https://byjus.com
