1). In, two similar triangles \( \Large \triangle \) ABC and \( \Large \triangle \) DEF, DE = 3 cm, EF = 5 cm, DF = 4 cm and BC = 20 cm, then length of AB is equal to
A). 15 cm 
B). 12 cm 
C). 10 cm 
D). 6 cm 
Correct Answer: 12 cm
From similar triangles ABC and DEF
\( \Large \frac{AB}{BC} = \frac{DE}{EF} \)
\( \Large \therefore \frac{AB}{20} = \frac{3}{5} \)
=> \( \Large AB = \frac{3 \times 20}{5} \) = 12 cm.

2). If angles of one triangle are respectively equal to the angles of another triangle, then ratio of the corresponding sides is ratio of the corresponding
A). medians 
B). bisector angles (angle bisector segments) 
C). altitude 
D). all of these 
Correct Answer: all of these

3). The locus of the vertex A of an isosceles triangle, ABC which has BC as its fixed base is
A). a line parallel to BC 
B). a line perpendicular to BC 
C). a circle with BC as a diameter 
D). perpendicular bisector of BC 
Correct Answer: perpendicular bisector of BC

4). If AB and CD are two chords intersecting at a point P inside the circle such that AP = CP, then consider the following statements : Assertion (A) : AB = CD, Reason (R) : APC and DPB are similar triangles Of these statements
A). both A and R are true, and R is correct explanation of A. 
B). both A and R are true, but R is not a correct explanation of A. 
C). A is true, but R is false. 
D). A is false, but R is true 
Correct Answer: both A and R are true, and R is correct explanation of A.

5). If a triangle and a rectangle have equal areas and equal altitude, then base of the triangle is equal to
A). base of the rectangle. 
B). twice the base of the rectangle. 
C). thrice the base of the rectangle. 
D). four times the base of the rectangle 
Correct Answer: twice the base of the rectangle.
Let h be the common height of triangle and the rectangle.
If a and b respectively be the bases of triangle and rectangle, then by hypothesis
\( \Large \frac{1}{2}a \times h = b \times h \)
=> \( \Large \frac{1}{2}a = b \)
=> a = 2b 
6). Area of an isosceles rightangled triangle is 800 sq. metres. The greatest possible square has been cut out from it. The length of the diagonal of this square will be
A). \( \Large 10\ \sqrt{2} m\) 
B). \( \Large 10\ \sqrt{3} m\) 
C). 20 m 
D). \( \Large 20\ \sqrt{2} m\) 
Correct Answer: \( \Large 20\ \sqrt{2} m\) Let x be the base and side of isosceles \( \Large \triangle ABC \).
By hypothesis,
\( \Large \frac{1}{2}x \times x = 800 \)
=> \( \Large x^{2} = 1600 \)
=> x = 40 m
Therefore, Hypotenuse, BC = 2 x = 40 x/2 m
If s be the side of the maximum square, then
\( \Large s^{2}+\frac{1}{2} \left(xs\right)s+\frac{1}{2} \left(xs\right)2=800 \)
=> \( \Large s^{2}+xss^{2} = 800 \)
=> \( \Large xs = 800 \)
=> \( \Large s = \frac{800}{40} = 20 m \)
Therefore, Length of diagonal of the square = \( \Large 20 \sqrt{2} m \)

7). If perimeter of a triangle is 100 m and its sides are in the ratio 1 : 2 : 2, then area of the triangle (in \( \Large m^{2} \)) is
A). \( \Large 100 \sqrt{3} \) 
B). \( \Large 100 \sqrt{15} \) 
C). \( \Large 100 \sqrt{5} \) 
D). \( \Large 100 \sqrt{7} \) 
Correct Answer: \( \Large 100 \sqrt{15} \) Let the Sides. be x, 2x and 2x.
Then x + 2x + 2x = 100
=> x = 20
Hence sides are 20, 40 and 40.
Therefore, s = 50 [Because, 2s = x+b+c]
Area of the triangle
= \( \Large \sqrt{s \left(sa\right) \left(sb\right) \left(sc\right) } \)
= \( \Large \sqrt{50 \left(5020\right) \left(5040\right) \left(5040\right) } \)
= \( \Large \sqrt{50 \times 30 \times 10 \times 10} \)
= \( \Large 100 \sqrt{15} m^{2} \)

8). In the given figure, AD is the internal bisector and AE is the external bisector of \( \Large \angle \)BAC of any \( \Large \triangle \) ABC. Then which one of the following statements is not correct?
A). \( \Large AC^{2} = DC \times CE \) 
B). \( \Large AC \times CD = AC \times BD \) 
C). \( \Large AB : AC = BE : CE \) 
D). \( \Large \angle ADE  90 ^{\circ} \) 
Correct Answer: \( \Large AC^{2} = DC \times CE \)

9). In the given figure, \( \Large \angle \)ABC = \( \Large \angle \)ADB = \( 90^{\circ} \), which one of the following statements does not hold good?
A). \( \Large \triangle \) ABC, ADB and CDB are similar. 
B). \( \Large AD \times DC \) 
C). \( \Large \triangle ADB : \triangle CDB = AB : BC \) 
D). \( \Large AB^{2} = AD \times AC \) 
Correct Answer: \( \Large \triangle ADB : \triangle CDB = AB : BC \) (a) \( \Large \triangle ABC  \triangle ADB  \triangle CDB \)
From \( \Large \triangle ABC\ and\ \triangle ADB \)
\( \Large \angle ABC = \angle ADB = 90 ^{\circ} \)
AB = AB, common
From ASA, \( \Large \triangle ABC  \triangle ADB \)
Again, from \( \Large \triangle ABC\ and\ \triangle CDB \)
\( \Large \angle C = common \)
\( \Large BC = common \)
\( \Large \angle ABC = \angle CBD \)
From ASA, \( \Large \triangle ABC  \triangle ADB \)
From (i) and (ii)
\( \Large \triangle ABC = \triangle ADB  \triangle CDB \)
(b) From \( \Large \triangle ADB  \triangle BDC \)
\( \Large \frac{BD}{AD} = \frac{DC}{BD} \)
\( \Large \therefore BD^{2} = AD \times DC \)
And from \( \Large \triangle ADB  \triangle ABC \)
\( \Large \frac{AB}{AD} = \frac{AC}{AB} \)
(c) \( \Large \frac{\triangle ADB}{\triangle CDB} = \frac{\frac{1}{2}AD \times BD}{\frac{1}{2}CD \times BD} \)
= \( \Large \frac{AD}{CD} = \frac{AB^{2}}{BC^{2}} \ne \frac{AB}{BC} \)

10). Area of an equilateral triangle of side x is
A). \( \Large \frac{x^{2}}{\sqrt{2}} \) 
B). \( \Large \frac{\sqrt{3}}{4} x^{2} \) 
C). \( \Large \frac{\sqrt{3}}{2} x^{2} \) 
D). \( \Large \frac{\sqrt{3}}{2} x^{3} \) 
Correct Answer: \( \Large \frac{\sqrt{3}}{4} x^{2} \) \( \Large AD = \sqrt{AB^{2}BD^{2}} \) = \( \Large \sqrt{x^{2}\frac{x^{2}}{4}} \) = \( \Large \frac{\sqrt{3}}{2}x \) Therefore, Area of \( \Large \triangle ABC = \frac{1}{2} \times BC \times AD \) = \( \Large \frac{1}{2} \times x \times \frac{x\sqrt{3}}{2} \) = \( \Large \frac{\sqrt{3}}{4}x^{2} \) 