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Elementary Mathematics

21). The mid-points of AB and AC of a \( \Large \triangle ABC \) are respectively X and Y. If BC + XY = 12 units, then the value of BC - XY is

A). 6

B). 8

C). 4

D). 12

View Answer

Correct Answer: 4

Since, X and Y are the mid-points of AB and AC, respectively.

Therefore, \( \Large XY = \frac{1}{2} BC \)

Now, BC + XY = 12

or \( \Large BC + \frac{1}{2}BC = 12 \) [\( \Large \because XY = \frac{1}{2}BC \)]

=> \( \Large \frac{3}{2}.BC = 12 => BC = 12 \times \frac{2}{3} = 8 \)

and \( \Large XY = \frac{1}{2} \times 8 = 4 \)

\( \Large \therefore BC - XY = 8 - 4 = 4 \)

22). In the figure given below, \( \Large \angle ABC \) = \( \Large \angle AED \) = \( \Large 90 ^{\circ} \)

Consider the following statements
I. ABC and ADE are similar triangles.
II. The four points B, C, E and D may lie on a circle.
Which of the above statements is/are correct?

A). Only I

B). Only II

C). Both I and ll

D). Neither I nor ll

View Answer

Correct Answer: Both I and ll

Given, \( \Large \angle AED = 90 ^{\circ} \)

=> \( \Large \angle DEC = 90 ^{\circ} \)

=> \( \Large \angle ACB = \angle ADE \)

Now, EDCB is a quadrilatral.

\( \Large \therefore EDCB, \)

\( \Large \angle E + \angle D + \angle B + \angle C = 360 ^{\circ} \)

\( \Large \angle E + \angle B = 180 ^{\circ} \)

=> \( \Large \angle C + \angle D = 180 ^{\circ} \)

\( \Large \therefore \angle C\ and \angle D\ are\ supplementary \)

\( \Large \therefore \) EDCB may lie on a circle, since, EDCB is a cyclic quadrilateral.

23). In a \( \Large \triangle ABC \), \( \Large \angle BCA = 60 ^{\circ} \) and \( \Large B^{2} = BC^{2} + CA^{2} + X \). What is the value of X?

A). \( \Large \left(BC\right) \left(CA\right) \)

B). \( \Large - \left(BC\right) \left(CA\right) \)

C). \( \Large \left(AB\right) \left(BC\right) \)

D). Zero

View Answer

Correct Answer: \( \Large - \left(BC\right) \left(CA\right) \)

By cosine law,

\( \Large \cos 60 ^{\circ} = \frac{AC^{2}+BC^{2}-AB^{2}}{2.AC.BC} = \frac{1}{2} \)

=> \( \Large AC^{2} + BC^{2} - AB^{2} = AC.BC \)

\( \Large \therefore By\ comparing,\ we\ get \)

\( \Large X = - \left(AC\right) \left(BC\right) \)

24). In a \( \Large \triangle ABC \), XY is drawn parallel to BC, cutting sides at X and Y, where AB = 4.8 cm, BC = 7.2 cm and BX = 2 cm. What is the length of XY?

A). 4 cm

B). 4.1 cm

C). 4.2 cm

D). 4.3 cm

View Answer

Correct Answer: 4.2 cm

Given that. AB = 4.8 cm. BC = 7.2 cm, BX = 2 cm

\( \Large \therefore AX = AB - BX \)

\( \Large 4.8 - 2 = 2.8 cm \)

From figure, \( \Large \triangle AXY = \triangle ABC \)

\( \Large \therefore \frac{XY}{BC} = \frac{AX}{AB} \)

=> \( \Large XY = \frac{AX}{AB}.BC \)

= \( \Large \frac{2.8}{4.8} \times 7.2\)

\( \Large \therefore XY = 4.2 cm \)

25). The angles \( \Large x ^{\circ} \), \( \Large a ^{\circ} \), \( \Large c ^{\circ} \) and \( \Large \left( \pi - b\right) ^{\circ} \) are indicated in the figure given below. Which one of the following is correct?

A). x = a + c - b

B). x = b - a - c

C). x = a + b + c

D). x = a - b + c

View Answer

Correct Answer: x = a + b + c

\( \Large \because \angle PCT + \angle PCB = \pi \) [linear pair]

\( \Large \angle PCB = \pi - \left( \pi - b ^{\circ} \right) = b ^{\circ} \) ...(i)

In \( \Large \triangle BPC, \)

\( \Large \angle PCB + \angle BPC + \angle PBC = \pi \)

=> \( \Large \angle PBC = \pi - \angle PCB + \angle PBC = \pi \)

=\( \Large \pi - b ^{\circ} - a ^{\circ} \) ...(ii)

\( \Large \because \angle ABE + \angle EBC = \pi \)

\( \Large \because [ \angle PBC = \angle EBC] \) [linear pair]

=> \( \Large \angle ABE = \pi \angle PBC = \pi - \left( \pi -b ^{\circ} -a ^{\circ} \right) \)

= \( \Large a ^{\circ} + b ^{\circ} \) ...(iii)

Now, in \( \Large \triangle ABE, \)

Sum of two interior angles = Exterior angle

\( \Large \angle EAB + \angle ABE = \angle BES \)

=> \( \Large c ^{\circ} + b ^{\circ} + a ^{\circ} = x ^{\circ} \)

\( \Large \therefore x ^{\circ} = a ^{\circ} + b ^{\circ} + c ^{\circ} \)

26). In the figure given below, \( \Large YZ \parallel MN \), \( \Large XY \parallel \ LM \ and \ XZ \parallel LN\) Then, MY is

A). the median of ALMN

B). the angular bisector of ALMN

C). perpendicular to LN

D). perpendicular bisector ol LN

View Answer

Correct Answer: the median of ALMN

Since,

\( \Large ZY \parallel MN \ and \ ZX \parallel YN \)

\( \Large \therefore XNYZ \ is\ a\ parallelogram. \)

=> ZX = YZ ...(i) ...(i)

Also,

\( \Large ZY \parallel YN and XY \parallel ZL \)

\( \Large \therefore XYLZ\ is\ a\ parallelogram. \)

\( \Large \therefore XZ = YL \) ...(ii)

From Eqs. (i) and (ii), we get

YN = LY

Therefore, MY is a mjedian of \( \Large \triangle LMN \)

27). In the figure given below, AB is parallel to CD \( \Large AB \parallel CD \), \( \Large \angle ABC \) = \( \Large 65 ^{\circ} \), \( \Large \angle CDE \) = \( \Large 15 ^{\circ} \) and AB = AE. What is the value of \( \Large \angle AEF \)?

A). \( \Large 30 ^{\circ} \)

B). \( \Large 35 ^{\circ} \)

C). \( \Large 40 ^{\circ} \)

D). \( \Large 45 ^{\circ} \)

View Answer

Correct Answer: \( \Large 35 ^{\circ} \)

Given that,

\( \Large \angle ABC = 65 ^{\circ} and \angle CDE = 15 ^{\circ} \)

Here, \( \Large \angle ABC + \angle TCB = 180 ^{\circ} \) [\( \Large \because AB \parallel CD \)]

\( \Large \therefore \angle TCB = 180 ^{\circ} - 65 ^{\circ} = 115 ^{\circ} \)

\( \Large \because \angle TCB + \angle DCB = 180 ^{\circ} \) [linear\ pair]

\( \Large \therefore \angle DCB = 65 ^{\circ} \)

Now, in \( \Large \triangle CDE, \)

\( \Large \angle CED = 180 ^{\circ} - \left( \angle ECD + \angle EDC\right) \)

[\( \Large \because \angle ECD = \angle BCD \)]

= \( \Large 180 ^{\circ} - \left(65 ^{\circ} +15 ^{\circ} \right) = 100 ^{\circ} \)

\( \Large \because \angle DEC + \angle FEC = 180 ^{\circ} \) [linear pair]

=> \( \Large \angle FEC = 180 ^{\circ} -100 ^{\circ} = 80 ^{\circ} \)

Given that, AB = AE

i.e., \( \Large \triangle ABE\ an \ isosceles\ triangle. \)

\( \Large \therefore \angle ABE = \angle AEB = 65 ^{\circ} \)

\( \Large \because \angle AEB + \angle AEF + \angle FEC = 180 ^{\circ} \) [straighrt line]

=> \( \Large 65 ^{\circ} +x ^{\circ} +80 ^{\circ} = 180 ^{\circ} \)

\( \Large \therefore x ^{\circ} = 180 ^{\circ} - 145 ^{\circ} = 35 ^{\circ} \)

28). External angle of a regular polygon is \( \Large 72^{\circ} \). Find the sum of all the internal angles of it.

A). \( \Large 360 ^{\circ} \)

B). \( \Large 480 ^{\circ} \)

C). \( \Large 352 ^{\circ} \)

D). \( \Large 540 ^{\circ} \)

View Answer

Correct Answer: \( \Large 540 ^{\circ} \)

External angle of any polygon

\( \Large 72 ^{\circ} = \frac{360 ^{\circ} }{n} \)

=> \( \Large n = 5 \)

\( \Large \therefore \) Given, polygon is regular pentagon

\( \Large \therefore \) Every interior angle of it = \( 180 ^{\circ} \) - External angle

= \( \Large 180 ^{\circ} - 72 ^{\circ} = 108 ^{\circ} \)

\( \Large \therefore \) Sum of total interior angles of it \( = 5 \times 108 ^{\circ} = 540 ^{\circ} \)

29). The ratio of the numbers of sides of two regular polygons is 1 : 2. If each interior angle of the first polygon is \( \Large 120 ^{\circ} \), then the measure of each interior angle of the second polygon is

A). \( \Large 140 ^{\circ} \)

B). \( \Large 135 ^{\circ} \)

C). \( \Large 150 ^{\circ} \)

D). \( \Large 160 ^{\circ} \)

View Answer

Correct Answer: \( \Large 150 ^{\circ} \)

Given, interior angle of the first polygon = \( \Large 120 ^{\circ} \)

Let number of sides in first polygon be \( \Large n_{1} \) .

Then, \( \Large \frac{n_{1} - 2}{n_{1}} \times 180 ^{\circ} = 120 ^{\circ} \)

=> \( \Large 3n_{1} - 6 = 2n_{1} \)

=> \( \Large n_{1} = 6 \)

\( \Large \therefore \) Sides of the second polygon \( = 6n = 6 \times 2 = 12\)

\( \Large \therefore \) Interior angle of the second polygon\( \Large = \frac{12-2}{12} \times 180 ^{\circ} = 150 ^{\circ} \)

30). In the quadrilateral ABCD shown below, \( \Large \angle \)DAB = \( \Large \angle \)DCX = \( \Large 120 ^{\circ} \). If \( \Large \angle \)ABC = \( \Large 105 ^{\circ} \), then what is the value of \( \Large \angle \)ADC?

A). \( \Large 45 ^{\circ} \)

B). \( \Large 60 ^{\circ} \)

C). \( \Large 75 ^{\circ} \)

D). \( \Large 95 ^{\circ} \)

View Answer

Correct Answer: \( \Large 75 ^{\circ} \)

Given, \( \Large \angle ABC = 105 ^{\circ} \)

\( \Large \angle DAB = 120 ^{\circ} \)

\( \Large \angle DCX = 120 ^{\circ} \)

=> \( \Large \angle DCB = 180 ^{\circ} - 120 ^{\circ} = 60 ^{\circ} \)

Angles of a quadrilatral is equal to \( \Large 360 ^{\circ} \)

\( \Large \therefore \angle ADC = 360 ^{\circ} - \left(120 ^{\circ} + 105 ^{\circ} + 60 ^{\circ} \right) \)

= \( \Large 360 ^{\circ} -285 ^{\circ} = 75 ^{\circ} \)

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