<\/span><\/h2>\nMark the correct alternative in each of the following.<\/strong>
Question 1.<\/strong>
The 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) 81: 16
(d) 16: 81
Solution:<\/strong>
(d)<\/strong> Triangles are similar and the ratio of their sides is 4: 9
The ratio of the areas of two similar triangles are proportional to the square oT their corresponding sides
Ratio in their areas = (4)\u00b2 : (9)\u00b2 = 16 : 81<\/p>\nRD Sharma Class 10 Solutions Chapter 4 MCQs Question 2.<\/strong>
The areas of two similar triangles are respectively 9 cm\u00b2 and 16 cm\u00b2. The ratio of their corresponding sides is
(a) 3: 4
(b) 4 : 3
(c) 2 : 3
(d) 4: 5
Solution:<\/strong>
(a)<\/strong>\u00a0Ratio in the areas of two similar triangles = 9 cm\u00b2 : 16 cm\u00b2 = 9 : 16
The areas of similar triangles are proportional to the squares of their corresponding sides
Ratio in their corresponding sides = \u221a916\u00a0=\u00a034\u00a0= 3 : 4<\/p>\nQuestion 3.<\/strong>
The areas of two similar triangles \u2206ABC and \u2206DEF are 144 cm\u00b2 and 81 cm\u00b2 respectively. If the longest side of larger \u2206ABC be 36 cm, then. The longest side of the smaller triangle \u2206DEF is :
(a) 20 cm
(b) 26 cm
(c) 27 cm
(d) 30 cm
Solution:<\/strong>
(c)<\/strong>\u00a0Area of the larger triangle ABC = 144 cm\u00b2
and area of smaller \u2206DEF = 81 cm\u00b2
Longest side of larger triangle = 36 cm
Let the longest side of smaller triangle = x cm
The ratio of the areas of two similar triangles is proportional to the squares of their corresponding sides
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-1.png\")
<\/p>\nQuestion 4.<\/strong>
\u2206ABC and \u2206BDE are two equilateral triangles such that D is the mid-point of BC. The ratio of the areas of triangles ABC and BDE is :
(a) 2: 1
(b) 1: 2
(c) 4: 1
(d) 1: 4
Solution:<\/strong>
(c)<\/strong>\u00a0\u2206ABC and \u2206BDE are equilateral triangles and D is the mid-point of PC
\u2206ABC and \u2206BDE are both equilateral triangles
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-2.png\")
<\/p>\nQuestion 5.<\/strong>
If \u2206ABC and \u2206DEF are similar such that 2AB = DE and BC = 8 cm, then EF =
(a) 16 cm
(b) 12 cm
(c) 8 cm
(d) 4 cm
Solution:<\/strong>
(a)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-3.png\")
<\/p>\nQuestion 6.<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-4.png\")
(a) 2 : 5
(b) 4 : 25
(c) 4 : 15
(d) 8 : 125
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-5.png\")
<\/p>\nQuestion 7.<\/strong>
XY is drawn parallel to the base BC of a \u2206ABC cutting AB at X and AC at Y. If AB = 4 BX and YC = 2 cm, then AY =
(a) 2 cm
(b) 4 cm
(c) 6 cm
(d) 8 cm
Solution:<\/strong>
(c)<\/strong>\u00a0In \u2206ABC, XY || BC
AB = 4BX, YC = 2 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-6.png\")
<\/p>\nQuestion 8.<\/strong>
Two poles of height 6 m and 11 m stand vertically upright on a plane ground. If the distance between their foot is 12 m, the distance between their tops is
(a) 12 m
(b) 14 m
(c) 13 m
(d) 11 m
Solution:<\/strong>
(c)<\/strong>\u00a0Let length of pole AB = 6 m
and of pole CD = 11 m
and distance between their foot = 12 m
i.e., BD = 12 m
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-7.png\")
<\/p>\nQuestion 9.<\/strong>
In \u2206ABC, D and E are points on side AB and AC respectively such that DE || BC and AD : DB = 3 : 1. If EA = 3.3 cm, then AC =
(a) 1.1 cm
(b) 4 cm
(c) 4.4 cm
(d) 5.5 cm
Solution:<\/strong>
(c)<\/strong>\u00a0In \u2206ABC, DE || BC
AD : DB = 3 : 1, EA = 3.3 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-8.png\")
<\/p>\nQuestion 10.<\/strong>
In triangles ABC and DEF, \u2220A = \u2220E = 40\u00b0, AB : ED = AC : EF and \u2220F = 65\u00b0, then \u2220B =
(a) 35\u00b0
(b) 65\u00b0
(c) 75\u00b0
(d) 85\u00b0
Solution:<\/strong>
(c)<\/strong>\u00a0In \u2206ABC and \u2206DEF,
\u2220A = \u2220E = 40\u00b0
AB : ED = AC : EF, \u2220F = 65\u00b0
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-9.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-10.png\")
<\/p>\nQuestion 11.<\/strong>
If ABC and DEF are similar triangles such that \u2220A = 47\u00b0 and \u2220E = 83\u00b0, then \u2220C =
(a) 50\u00b0
(b) 60\u00b0
(c) 70\u00b0
(d) 80\u00b0
Solution:<\/strong>
(a)<\/strong>\u00a0\u2206ABC ~ \u2206DEF
\u2220A = 47\u00b0, \u2220E = 83\u00b0
\u2206ABC and \u2206DEF are similar
\u2220A = \u2220D, \u2220B = \u2220E and \u2220C = \u2220F
\u2220A = 47\u00b0
\u2220B = \u2220E = 83\u00b0
But \u2220A + \u2220B + \u2220C = 180\u00b0 (Sum of angles of a triangle)
47\u00b0 + 83\u00b0 + \u2220C = 180\u00b0
=> 130\u00b0 + \u2220C = 180\u00b0
=> \u2220C = 180\u00b0 \u2013 130\u00b0
=> \u2220C = 50\u00b0<\/p>\nQuestion 12.<\/strong>
If D, E, F are the mid-points of sides BC, CA, and AB respectively of \u2206ABC, then the ratio of the areas of triangles DEF and ABC is
(a) 1: 4
(b) 1: 2
(c) 2 : 3
(d) 4: 5
Solution:<\/strong>
(a)<\/strong> D, E, and F are the midpoints of the sides. BC, CA, and AB respectively of \u2206ABC
DE, EF, and FD are joined
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-11.png\")
<\/p>\nQuestion 13.<\/strong>
In an equilateral triangle ABC, if AD \u22a5 BC, then
(a) 2AB\u00b2 = 3AD\u00b2
(b) 4AB\u00b2 = 3AD\u00b2
(c) 3AB\u00b2 = 4AD\u00b2
(d) 3AB\u00b2 = 2AD\u00b2
Solution:<\/strong>
(c)<\/strong>\u00a0In equilateral \u2206ABC,
AD \u22a5 BC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-12.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-13.png\")
<\/p>\nQuestion 14.<\/strong>
If \u2206ABC is an equilateral triangle such that AD \u22a5 BC, then AD\u00b2 =
(a)\u00a032\u00a0DC\u00b2
(b) 2 DC\u00b2
(c) 3 CD\u00b2
(d) 4 DC\u00b2
Solution:<\/strong>
(c)<\/strong>\u00a0In equilateral \u2206ABC, AD \u22a5 BC
AD bisects BC at D
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-14.png\")
<\/p>\nQuestion 15.<\/strong>
In a \u2206ABC, AD is the bisector of \u2220BAC. If AB = 6 cm, AC = 5 cm and BD = 3 cm, then DC =
(a) 11.3 cm
(b) 2.5 cm
(c) 3.5 cm
(d) None of these
Solution:<\/strong>
(b)<\/strong>\u00a0In \u2206ABC, AD is the bisector of \u2220BAC
AB = 6 cm, AC = 5 cm, BD = 3 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-15.png\")
<\/p>\nQuestion 16.<\/strong>
In a \u2206ABC, AD is the bisector of \u2220BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC
(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 8 cm
Solution:<\/strong>
(a)<\/strong>\u00a0In \u2206ABC, AD is the bisector of \u2220BAC
AB = 8 cm, BD = 6 cm and DC = 3 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-16.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-17.png\")
<\/p>\nQuestion 17.<\/strong>
ABCD is a trapezium such that BC || AD and AB = 4 cm. If the diagonals AC and BD intersect at O such that\u00a0AOOC\u00a0=\u00a0DOOB\u00a0=\u00a012\u00a0, then BC =
(a) 7 cm
(b) 8 cm
(c) 9 cm
(d) 6 cm
Solution:<\/strong>
(b)<\/strong>\u00a0In trapezium ABCD, BC || AD
AD = 4 cm, diagonals AC and BD intersect
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-18.png\")
=> x = 8
BC = 8 cm<\/p>\nQuestion 18.<\/strong>
If ABC is a right triangle right-angled at B and M, N are the mid-points of AB and BC respectively, then 4 (AN\u00b2 + CM\u00b2) =
(a) 4 AC\u00b2
(b) 5 AC\u00b2
(c)\u00a054\u00a0AC\u00b2
(d) 6 AC\u00b2
Solution:<\/strong>
(b)<\/strong>\u00a0In right \u2206ABC, \u2220B = 90\u00b0
M and N are the midpoints of AB and BC respectively
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-19.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-20.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-21.png\")
<\/p>\nQuestion 19.<\/strong>
If in \u2206ABC and \u2206DEF,\u00a0ABDE\u00a0=\u00a0BCFD, then \u2206ABC ~ \u2206DEF when
(a) \u2220A = \u2220F
(b) \u2220A = \u2220D
(c) \u2220B = \u2220D
(d) \u2220B = \u2220E
Solution:<\/strong>
(c)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-22.png\")
<\/p>\nQuestion 20.<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-23.png\")
Solution:
<\/strong>(a)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-24.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-25.png\")
<\/p>\nQuestion 21.<\/strong>
\u2206ABC ~ \u2206DEF, ar (\u2206ABC) = 9 cm\u00b2, ar (\u2206DEF) = 16 cm\u00b2. If BC = 2.1 cm, then the measure of EF is
(a) 2.8 cm
(b) 4.2 cm
(c) 2.5 cm
(d) 4.1 cm
Solution:<\/strong>
(a)<\/strong>\u00a0\u2206ABC ~ \u2206DEF
ar (\u2206ABC) = 9 cm\u00b2, ar (\u2206DEF) =16 cm\u00b2,
BC = 2.1 cm
\u2206ABC ~ \u2206DEF
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-26.png\")
<\/p>\nQuestion 22.<\/strong>
The length of the hypotenuse of an isosceles right triangle whose one side is 4\u221a2 cm is
(a) 12 cm
(b) 8 cm
(c) 8\u221a2 cm
(d) 12\u221a2 cm
Solution:<\/strong>
(b)<\/strong>\u00a0In isosceles right \u2206ABC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-27.png\")
\u2220B = 90\u00b0, AB = BC = 4\u221a2
AC = \u221a2
equal side = \u221a2 x 4\u221a2 = 8 cm<\/p>\nQuestion 23.<\/strong>
A man goes 24 m due west and then 7 m due north. How far is he from the starting point?
(a) 31 m
(b) 17 m
(c) 25 m
(d) 26 m
Solution:<\/strong>
(c)<\/strong>\u00a0In the figure, O is starting point OA = 24 m and AB = 7 m
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-28.png\")
By Pythagoras Theorem
OB\u00b2 = OA\u00b2 + AB\u00b2
= (24)\u00b2 + (7)\u00b2 = 576 + 49 = 625 = (25)\u00b2
OB = 25 m<\/p>\nQuestion 24.<\/strong>
\u2206ABC ~ \u2206DEF. If BC = 3 cm, EF = 4 cm and ar (\u2206ABC) = 54 cm\u00b2, then ar (\u2206DEF)
(a) 108 cm\u00b2
(b) 96 cm\u00b2
(c) 48 cm\u00b2
(d) 100 cm\u00b2
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-29.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-30.png\")
<\/p>\nQuestion 25.<\/strong>
\u2206ABC ~ \u2206PQR such that ar (\u2206ABC) = 4 ar (\u2206PQR). If BC = 12 cm, then QR =
(a) 9 cm
(b) 10 cm
(c) 6 cm
(d) 8 cm
Solution:<\/strong>
(c)<\/strong>\u00a0\u2206ABC ~ \u2206PQR
ar (\u2206ABC) = 4ar (\u2206PQR), BC = 12 cm
\u2206ABC ~ \u2206PQR
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-31.png\")
<\/p>\nQuestion 26.<\/strong>
The areas of two similar triangles are 121 cm\u00b2 and 64 cm\u00b2 respectively. If the median of the first triangle is 12.1 cm, then the corresponding median of the other triangles is
(a) 11 cm
(b) 8.8 cm
(c) 11.1 cm
(d) 8.1 cm
Solution:<\/strong>
(b)<\/strong>\u00a0Areas of two similar triangles are 121 cm\u00b2 and 64 cm\u00b2
Median of first triangle = 12.1 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-32.png\")
In \u2206ABC, area \u2206DEF
AD and PS are their corresponding median
\u2206ABC ~ \u2206DEF
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-33.png\")
<\/p>\nQuestion 27.<\/strong>
In an equilateral triangle ABC if AD \u22a5 BC, then AD\u00b2 =
(a) CD\u00b2
(b) 2 CD\u00b2
(c) 3 CD\u00b2
(d) 4 CD\u00b2
Solution:<\/strong>
(c)<\/strong>\u00a0In equilateral \u2206ABC, AD \u22a5 BC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-34.png\")
<\/p>\nQuestion 28.<\/strong>
In an equilateral triangle ABC if AD \u22a5 BC, then
(a) 5 AB\u00b2 = 4 AD\u00b2
(b) 3 AB\u00b2 = 4 AD\u00b2
(c) 4 AB\u00b2 = 3 AD\u00b2
(d) 2 AB\u00b2 = 3 AD\u00b2
Solution:<\/strong>
(b)<\/strong>\u00a0In equilateral \u2206ABC, AD \u22a5 BC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-35.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-36.png\")
<\/p>\nQuestion 29.<\/strong>
In an isosceles triangle ABC if AC = BC and AB\u00b2 = 2 AC\u00b2, then \u2220C =
(a) 30\u00b0
(b) 45\u00b0
(c) 90\u00b0
(d) 60\u00b0
Solution:<\/strong>
(c)<\/strong>\u00a0In isosceles \u2206ABC, AC = BC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-37.png\")
and AB2 = 2 AC\u00b2 = AC\u00b2 + AC\u00b2
= AC\u00b2 + BC\u00b2 ( AC = BC)
By converse of Pythagoras Theorem,
\u2220C = 90\u00b0<\/p>\nQuestion 30.<\/strong>
\u2206ABC is an isosceles triangle in which \u2220C = 90\u00b0. If AC = 6 cm, then AB =
(a) 6\u221a2 cm
(b) 6 cm
(c) 2\u221a6 cm
(d) 4\u221a2 cm
Solution:<\/strong>
(a)<\/strong>\u00a0\u2206ABC is an isosceles with \u2220C= 90\u00b0
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-38.png\")
AC = BC
AC = 6 cm
AB\u00b2 = AC\u00b2 + BC\u00b2 (Pythagoras Theorem)
(6)\u00b2 + (6)\u00b2 = 36 + 36 = 72 (AC = BC)
AB = \u221a72 = \u221a(36 x 2) = 6\u221a2 cm<\/p>\nQuestion 31.<\/strong>
If in two triangles ABC and DEF, \u2220A = \u2220E, \u2220B = \u2220F, then which of the following is not true?
(a)\u00a0BCDF\u00a0=\u00a0ACDE
(b)\u00a0ABDE\u00a0=\u00a0BCDF
(c)\u00a0ABEF\u00a0=\u00a0ACDE
(d)\u00a0BCDF\u00a0=\u00a0ABEF
Solution:<\/strong>
(b)<\/strong>\u00a0In two triangles ABC and DEF
\u2220A = \u2220E, \u2220B = \u2220F
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-39.png\")
<\/p>\nQuestion 32.<\/strong>
In the figure, the measures of \u2220D and \u2220F are respectively
(a) 50\u00b0, 40\u00b0
(b) 20\u00b0, 30\u00b0
(c) 40\u00b0, 50\u00b0
(d) 30\u00b0, 20\u00b0
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-40.png\")
Solution:<\/strong>
(b)<\/strong>\u00a0In \u2206ABC
\u2220A = 180\u00b0 \u2013 (\u2220B + \u2220C)
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-41.png\")
<\/p>\nQuestion 33.<\/strong>
In the figure, the value of x for which DE || AB is
(a) 4
(b) 1
(c) 3
(d) 2
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-42.png\")
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-43.png\")
<\/p>\nQuestion 34.<\/strong>
In the figure, if \u2220ADE = \u2220ABC, then CE =
(a) 2
(b) 5
(c)\u00a092
(c) 3
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-44.png\")
Solution:<\/strong>
(c)<\/strong>\u00a0In the figure \u2220ADE = \u2220ABC
AB = 2, DB = 3, AE = 3
Let EC = x
\u2220ADE = \u2220ABC
But these are corresponding angles DE || BC
\u2206ADE ~ \u2206ABC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-45.png\")
<\/p>\nQuestion 35.<\/strong>
In the figure, RS || DB || PQ. If CP = PD = 11 cm and DR = RA = 3 cm. Then the values of x and y are respectively
(a) 12, 10
(b) 14, 6
(c) 10, 7
(d) 16, 8
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-46.png\")
Solution:<\/strong>
(d)<\/strong>\u00a0In the figure RS || DB || PQ
CP = PD = 11 cm DR = RA = 3 cm
In \u2206ABD
RS || BD and AR = RD
RS =\u00a012\u00a0BD
y =\u00a012\u00a0x or x = 2y
Only 16, 8 is possible<\/p>\nQuestion 36.<\/strong>
In the figure, if PB || CF and DP || EF, then\u00a0ADDE\u00a0=
(a)\u00a034
(b)\u00a013
(c)\u00a014
(d)\u00a023
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-47.png\")
Solution:<\/strong>
(b)<\/strong>\u00a0In the figure, PB || CF, DP || EF
AB = 2 cm, AC = 8 cm
BC = AC \u2013 AB = 8 \u2013 2 = 6 cm
In \u2206ACF, BP || CF
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-48.png\")
ADDE\u00a0=\u00a013<\/p>\nQuestion 37.<\/strong>
A chord of a circle of radius 10 cm subtends a right angle at the center. The length of the chord (in cm) is
(a) 5\u221a2
(b) 10\u221a2
(c)\u00a05\u221a2
(d) 10\u221a3\u00a0[ICSE 2014]<\/strong>
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-49.png\")
AB\u00b2 = OA\u00b2 + OB\u00b2 (Pythagoras Theorem)
AB\u00b2 = 10\u00b2 + 10\u00b2
AB\u00b2 = 2 (10)\u00b2
AB = 10\u221a2<\/p>\nQuestion 38.<\/strong>
A vertical stick 20 m long casts a shadow 10 m long on the ground. At the same time, a tower casts a shadow 50 m long on the ground. The height of the tower is
(a) 100 m
(b) 120 m
(c) 25 m
(d) 200 m
Solution:<\/strong>
(a)<\/strong>\u00a0Height of a stick = 20 m
and length of its shadow = 10 m
At the same time
Let height of tower = x m
and its shadow = 50 m
20 : x = 10 : 50
x x 10 = 20 x 50
=> x =\u00a020×5010\u00a0= 100
Height of tower = 100 m<\/p>\nQuestion 39.<\/strong>
Two isosceles triangles have equal angles and their areas are in the ratio 16: 25. The ratio of their corresponding heights is :
(a) 4: 5
(b) 5: 4
(c) 3: 2
(d) 5: 7
Solution:<\/strong>
(a)<\/strong> The corresponding angles of two isosceles triangles are equal These are similar Ratio in their areas = 16: 25
The ratio of areas of similar triangles is in proportion to the squares of their corresponding altitudes (heights)
Ratio in their altitudes =\u00a0\u221a1625=45
= i.e., 4 : 5<\/p>\nQuestion 40.<\/strong>
\u2206ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. If \u2206DEF ~ \u2206ABC and EF = 4 cm, then perimeter of \u2206DEF is
(a) 7.5 cm
(b) 15 cm
(c) 22.5 cm
(d) 30 cm
Solution:<\/strong>
(b)<\/strong>\u00a0\u2206DEF ~ \u2206ABC
AB = 3 cm, BC = 2 cm, CA = 2.5 cm, EF = 4 cm
\u2206s are similar
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-50.png\")
<\/p>\nQuestion 41.<\/strong>
In \u2206ABC, a line XY parallel to BC cuts AB at X and AC at Y. If BY bisects \u2220XYC, then :
(a) BC = CY
(b) BC = BY
(c) BC \u2260 CY
(d) BC \u2260 BY
Solution:<\/strong>
(a)<\/strong>\u00a0In \u2206ABC, XY || BC
BY is the bisector of \u2220XYC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-51.png\")
\u2220XYB = \u2220CXB \u2026.(i)
XY || BC
\u2220XYB = \u2220XBC (Alternate angles) \u2026\u2026\u2026.(ii)
From (i) and (ii)
\u2220CYB = \u2220YBC
BC = CY<\/p>\nQuestion 42.<\/strong>
In a \u2206ABC, \u2220A = 90\u00b0, AB = 5 cm and AC = 12 cm. If AD \u22a5 BC, then AD =
(a)\u00a0132\u00a0cm
(b)\u00a06013\u00a0cm
(c)\u00a01360\u00a0cm
(d)\u00a02\u221a1513\u00a0cm
Solution:<\/strong>
(b)<\/strong>\u00a0In \u2206ABC,
\u2220A = 90\u00b0, AB = 5 cm, AC = 12 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-52.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-53.png\")
<\/p>\nQuestion 43.<\/strong>
In a \u2206ABC, perpendicular AD from A on BC meets BC at D. If BD = 8 cm, DC = 2 cm and AD = 4 cm, then
(a) \u2206ABC is isosceles
(b) \u2206ABC is equilateral
(c) AC = 2 AB
(d) \u2206ABC is right-angled at A
Solution:<\/strong>
(d)<\/strong>\u00a0In \u2206ABC, AD \u22a5 BC
BD = 8 cm, DC = 2 cm, AD = 4 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-54.png\")
In right \u2206ACD,
AC\u00b2 = AD\u00b2 + CD\u00b2 (Pythagoras Theorem)
= (4)\u00b2 + (2)\u00b2 = 16 + 4 = 20
and in right \u2206ABD,
AB\u00b2 = AD\u00b2 + DB\u00b2
= (4)\u00b2 + (8)2 = 16 + 64 = 80
and BC\u00b2 = (BD + DC)\u00b2 = (8 + 2 )\u00b2 = (10)\u00b2 = 100
AB\u00b2 + AC\u00b2 = 80 + 20 = 100 = BC\u00b2
\u2206ABC is a right triangle whose \u2220A = 90\u00b0<\/p>\nQuestion 44.<\/strong>
In a \u2206ABC, point D is on side AB and point G is on side AC, such that BCED is a trapezium. If DE : BC = 3:5, then Area (\u2206ADE) : Area (BCED) =
(a) 3 : 4
(b) 9 : 16
(c) 3 : 5
(d) 9 : 25
Solution:<\/strong>
(b)<\/strong>
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-55.png\")
In \u2206ABC, D and E are points on the side AB and AC respectively, such that BCED is a trapezium DE: BC = 3: 5
In \u2206ABC, DE || BC
\u2206ADE ~ \u2206ABC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-56.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-57.png\")
<\/p>\nQuestion 45.<\/strong>
If ABC is an isosceles triangle and D is a point on BC such that AD \u22a5 BC, then
(a) AB\u00b2 \u2013 AD\u00b2 = BD.DC
(b) AB\u00b2 \u2013 AD\u00b2 = BD\u00b2 \u2013 DC\u00b2
(c) AB\u00b2 + AD\u00b2 = BD.DC
(d) AB\u00b2 + AD\u00b2 = BD\u00b2 \u2013 DC\u00b2
Solution:<\/strong>
(a)<\/strong>\u00a0If \u2206ABC, AB = AC
D is a point on BC such that
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-58.png\")
AD \u22a5 BC
AD bisects BC at D
In right \u2206ABD,
AB\u00b2 = AD\u00b2 + BD\u00b2
AB\u00b2 \u2013 AD\u00b2 = BD\u00b2 = BD x BD = BD x DC (BD = DC)<\/p>\nQuestion 46.<\/strong>
\u2206ABC is a right triangle right-angled at A and AD \u22a5 BC. Then ,\u00a0BDDC\u00a0=
(a)\u00a0(ABAC)2
(b)\u00a0ABAC
(c)\u00a0(ABAD)2
(d)\u00a0ABAD
Solution:<\/strong>
(a)<\/strong>\u00a0In right angled \u2206ABC, \u2220A = 90\u00b0
AD \u22a5 BC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-59.png\")
<\/p>\nQuestion 47.<\/strong>
If E is a point on side CA of an equilateral triangle ABC such that BE \u22a5 CA, then AB\u00b2 + BC\u00b2 + CA\u00b2 =
(a) 2 BE\u00b2
(b) 3 BE\u00b2
(c) 4 BE\u00b2
(d) 6 BE\u00b2
Solution:<\/strong>
(c)<\/strong>\u00a0\u2206ABC is an equilateral triangle
BE \u22a5 AC
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-60.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-61.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-62.png\")
<\/p>\nQuestion 48.<\/strong>
In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then
(a) AQ\u00b2 + CP\u00b2 = 2 (AC\u00b2 + PQ\u00b2)
(b) 2 (AQ\u00b2 + CP\u00b2) = AC\u00b2 + PQ\u00b2
(c) AQ\u00b2 + CP\u00b2 = AC\u00b2 + PQ\u00b2
(d) AQ + CP =\u00a012\u00a0(AC + PQ)
Solution:<\/strong>
(c)<\/strong>\u00a0In right \u2206ABC, \u2220B = 90\u00b0
P and Q are points on AB and BC respectively
AQ, CP, and PQ are joined
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-63.png\")
In right \u2206ABC,
AC\u00b2 = AB\u00b2 + BC\u00b2 \u2026.(i)
(Pythagoras Theorem)
Similarly in right \u2206PBQ,
PQ\u00b2 = PB\u00b2 + BQ\u00b2 \u2026\u2026\u2026(ii)
In right \u2206ABQ
AQ\u00b2 = AB\u00b2 + BQ\u00b2 \u2026.(iii)
and in right \u2206CPB,
CP\u00b2 = PB\u00b2 + BC\u00b2 \u2026.(iv)
Adding (iii) and (iv)
AQ\u00b2 + CP\u00b2 = AB\u00b2 + BQ\u00b2 + PB\u00b2 + BC\u00b2
= AB\u00b2 + BC\u00b2 + BQ\u00b2 + PB\u00b2
= AC\u00b2 + PQ2 {From (i) and (ii)}<\/p>\nQuestion 49.<\/strong>
If \u2206ABC ~ \u2206DEF such that DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm, then perimeter of \u2206ABC is
(a) 18 cm
(b) 20 cm
(c) 12 cm
(d) 15 cm
Solution:<\/strong>
(d)<\/strong>\u00a0\u2206ABC ~ \u2206DEF
DE = 3 cm, EF = 2 cm, DF = 2.5 cm, BC = 4 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-64.png\")
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-65.png\")
<\/p>\nQuestion 50.<\/strong>
If \u2206ABC ~ \u2206DEF such that AB = 9.1 cm and DE = 6.5 cm. If the perimeter of \u2206DEF is 25 cm, then the perimeter of \u2206ABC is
(a) 36 cm
(b) 30 cm
(c) 34 cm
(d) 35 cm
Solution:<\/strong>
(d)<\/strong>\u00a0\u2206ABC ~ \u2206DEF
AB = 9.1 cm and DE = 6.5 cm
Perimeter of \u2206DEF = 25 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-66.png\")
<\/p>\nQuestion 51.<\/strong>
In an isosceles triangle ABC, if AB = AC = 25 cm and BC = 14 cm, then the measure of altitude from A on BC is
(a) 20 cm
(b) 22 cm
(c) 18 cm
(d) 24 cm
Solution:<\/strong>
(d)<\/strong>\u00a0\u2206ABC is an isosceles triangle in which AB = AC = 25 cm, BC = 14 cm
![\"RD](\"https:\/\/www.learninsta.com\/wp-content\/uploads\/2018\/06\/RD-Sharma-Class-10-Solutions-Chapter-7-Triangles-MCQS-67.png\")
From A, draw AD \u22a5 BC
D is mid-point of BC
BD =\u00a012\u00a0BC =\u00a012\u00a0x 14 = 7 cm
Now in right \u2206ABD
AD\u00b2 = AB\u00b2 \u2013 BD\u00b2
= (25)\u00b2 \u2013 (7)\u00b2 = 625 \u2013 49 = 576 = (24)\u00b2
AD = 24 cm<\/p>\nWe have provided complete details of RD Sharma Class 10 Solutions Chapter 4 MCQs. If you have any queries related to CBSE<\/strong><\/a>\u00a0Class 10, feel free to ask us in the comment section below.<\/p>\n<\/span>FAQs on RD Sharma Class 10 Solutions Chapter 4 MCQs<\/span><\/h2>\n\n\n\n
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<\/span>Is it important to study all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 MCQs?<\/span><\/h3>\n\n\n
Yes, learning all of the concepts included in RD Sharma Class 10 Solutions Chapter 4 MCQs is required in order to achieve high scores on the Class 10 board exams. These RD Sharma Class 10 Solutions Chapter 4 MCQs were created by subject matter specialists who compiled model questions that covered all of the textbook’s exercise questions.<\/p>\n\n<\/div>\n<\/div>\n
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<\/span>Where can I download RD Sharma Class 10 Solutions Chapter 4 MCQs free PDF?
![\"RD](\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-4-MCQs.jpg\")
<\/p>\n