31-07-2021 Answers

 1. (REF=25847) :

MN is a mid-segment of triangle ABC. If MN is 48 cm, what is the length of BC?



Answer: 96 cm

Solution: As MN is mid-segment, BC = 2*MN = 2*48 = 96 cm


2. (REF=25846) :

Find the largest angle in the triangle given below.



Answer: MAR

Solution: The angle opposite to the largest side is the largest angle in a triangle.

The largest side is MR (or RM) which is 62 m.

Hence the angle opposite to it MAR (or RAM) is the largest angle


3. (REF=25853) :

ABC is a right triangle. AM is perpendicular to BC. The size of angle ABC is equal to 55 degrees. Find the size of angle MAC.


Answer: 55 degrees

Solution: As ABC=55 degree, MCA = 180 - (90+55) = 35 degrees.

MAC = 180 - (AMC+MCA) = 180 - (90+35) = 55 degrees


4. (REF=25862) :

The vertices of the inscribed (inside) square bisect the sides of the second (outside) square. Find the ratio of the area of the outside square to the area of the inscribed square.


Answer: 2:1

Solution: Let the side of the outside square be a.

Area of the outside square = a^2.


The inside square area is sum of two triangles whose area is 1/2 * base * height, where base = a and height=a/2

Hence inside square area = 2 * 1/2 * a * a/2 = a^2/2


So ratio of area = a^2 : a^2/2 = 2:1


5. (REF=25855) :

Find the size of angle MBD in the figure below.


Answer: 72 degrees

Solution: Angle AMC = 180 - (56+78) = 46 degrees.

Hence DMB = AMC = 46 degrees.

Angle MBD = 180 - (62+46) = 72 degrees.


6. (REF=25848) :

In the triangle below ABC is a right angle. If area of triangle ABD is half of the area of triangle ABC, then choose the option which is true.


Answer: BD = DC

Solution: Area of right angled triangle ABC = 1/2 * AB * BC

Area of right angled triangle ABD = 1/2 * AB * BD

Given 1/2 * AB * BC = 2 * 1/2 * AB * BD,

So BC = 2*BD and hence BD = DC


7. (REF=25859) :

ABC is a right triangle with the size of angle ACB equal to 74 degrees. The lengths of the sides AM, MQ and QP are all equal. Find the measure of angle QPB.


Answer: 148 degrees

Solution: Angle QAM = 180 - (74+90) = 16 degrees.

As AMQ is an isosceles triangle, angle AQM = 16 degrees.

Hence angle AMQ = 180 - (16+16) = 148 degrees.


AMQ + QMP = QPM + QPB (As both add upto 180 degrees).

As QM = QP, in the isosceles triangle QMP=QPM, So AMQ=QPB=148 degrees.


8. (REF=25858) :

The rectangle below is made up of 12 congruent (same size) squares. Find the perimeter of the rectangle if the area of the rectangle is equal to 432 square cm.



Answer: 84 cm

Solution: Area of one square = a^2

12 * a^2 = 432,

a^2 = 36, a = 6 cm

Perimeter of the rectangle = 4a+3a+4a+3a = 14a = 14*6 = 84 cm


9. (REF=25861) :

Find the area of the shaded region.


Answer: 208 square cm

Solution: Area the bigger triangle = 30*15 = 450

Area of unshaded region = (15-4)*(30-8) = 11*22 = 242

Area of shaded region = 450-242 = 208 sq.cm


10. (REF=2947) :

What is the value of angle AED in the given image below? Angle EDC is 120 degrees and angle DCB is 100 degrees


Answer: 140 degree

Solution: Sum of angles in pentagon = (5-2) * 180  = 540 degrees.

Hence angle AED = 540 - 120 - 100 - 90 - 90 = 140 degrees.


11. (REF=25865) :

In the given figure BC is produced to D and  angle BAC = 40 degree and angle ABC = 70 degree. Find the value of  ACD:


Answer: 110 degrees

Solution: Angle ACB = 180 - (40+70) = 70 degrees.

Angle ACD = 180 - ACB = 180 - 70 = 110 degrees.


12. (REF=12035) :

The sides of a triangle measure 8, 9 and 10. Find the longest side of a similar triangle whose perimeter is 81.

Answer: 30

Solution: The sides of the similar triangle will be 8x,9x,10x.

Now 8x+9x+10x = 81 units, x=3 units.

Hence longest side = 10x = 30 units


13. (REF=3956) :

The length of the 3 sides of 4 triangles A,B,C,D are given below. Find the triangle that is impossible?


TriangleA - 10 cm, 210 cm and 200 cm               TriangleB - 23 cm, 45 cm and 20 cm 

TriangleC - 10 cm, 10 cm and 18 cm                   TriangleD - 20 cm, 60 cm and 55 cm 

Answer: TriangleB

Solution: In a triangle the sum of two sides should be larger than the third side. Hence TriangleB is impossible.


14. (REF=25856) :

The size of angle AOB is equal to 132 degrees and the size of angle COD is equal to 141 degrees. Find the size of angle DOB.


Answer: 93 degrees

Solution: Angle AOD+DOB+BOC = 180 degrees

Given AOD+DOB+BOC+DOB = 132+141 = 273 degrees.

Hence DOB = 273-180 = 93 degrees


15. (REF=25860) :

Find the area of the given shape.


Answer: 270 square cm

Solution: Area of big rectangle = 20*15 = 300 sq.cm

Area of cut right angled triangle = 1/2 * 5 * 12 = 30 sq.cm

So area of given shape = 300-30 = 270 sq.cm


16. (REF=25857) :

Find the size of angle x in the figure.


Answer: 24 degrees

Solution: The other angle in the smaller triangle = 180 - (41+94) = 45 degrees.

Hence x = 180 - (111+45) = 24 degrees


17. (REF=25851) :

ABCD is a parallelogram such that AB is parallel to DC and DA parallel to CB. The length of side AB is 20 cm. E is a point between A and B such that the length of AE is 3 cm. F is a point between points D and C. Find the length of DF such that the segment EF divide the parallelogram in two regions with equal areas.


Answer: 17 cm

Solution: Area of trapezium ADFE = Area of trapezium EFCB

Both trapeziums have same height H.

1/2 * H * (AE+DF) = 1/2 * H * (BE+FC),

AE+DF = BE+FC,

3+DF=17+FC,

DF-FC=14.

Also DF+FC=20 (As ABCD is a parallelogram opposite sides are equal).

Solving DF=17 cm


18. (REF=25852) :

Find the measure of angle A in the figure below


Answer: 87 degrees

Solution: The other angles of the triangle are 180-129=51 and 180-138=42

Hence angle A = 180 - (51+42) = 87 degrees.


19. (REF=25863) :

In the given trapezium, AEFB is a square. If AB = 10 cm, CE and FD are 3 cm, what is the area of the trapezium ACDB?


Answer: 130 sq.cm

Solution: Area of trapezium = 1/2 * height * (Sum of bases)

As AEFB is a square, height = AE = 10 cm.

CD = 3+10+3 = 16 cm

Required area = 1/2 * 10 * (16+10) = 130 sq.cm


20. (REF=25854) :

If the two triangles given are similar, find the value of X.


Answer: 132

Solution: Similar triangles have their side proportional.

So going through the options we find only for 132, 80/110 = 112/154 = 96/132 = 8/11.

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