30/07/2021 Answers

 1. (REF=10499) :

The length of the longest chord that can be inscribed in a circle of radius 12cm is

Answer: 24 cms

Solution: The longest chord the diameter and hence reqd length = 2r = 24 cms.

2. (REF=10492) :

If each interior angle of a regular polygon is 150 degrees, find the number of sides in the polygon.

Answer: 12

Solution: Interior angle = (n-2)180/n = 150,

Solving n=12

3. (REF=9214) :

In a decagon, how many triangles can be formed using the vertices of the decagon such that only one side of the triangle is same as one side of the decagon?

Answer: 60

Solution: The shortcut formula is N (N-4).

Here as it is a decagon N=10.

Hence num of triangles = 10 * 6 = 60

4. (REF=10497) :

In the below figure, x:y:z = 3:4:5, then find w.

Answer: 60 degrees

Solution: x+y+z = 180, 3x+4x+5x = 180, x=15,

So x=45, y = 60, z=75.

w is alternatively opposite to y and hence is 60 degrees

5. (REF=10494) :

If a parallelogram can be circumscribed by a circle, then it must be a

Answer: Rectangle

Solution: (Rectangle is a super set of square and hence rectangle is considered as the circle can also circumscribe rectangles which are not squares)

6. (REF=10495) :

The sum of all the angles in a convex polygon is 540 degrees. How many side does the polygon have?

Answer: 5

Solution: Sum of all angles = (2n-4)*90 = 540,

Solving n=5

7. (REF=10496) :

A convex polygon has 9 sides. What is the sum of all the angles of the polygon?

Answer: 1260 degrees

Solution: Sum of all angles = (2n-4)*90.

As n=9, sum of angles = 14*90 = 1260

8. (REF=9793) :

A ladder 100 feet long is leaning against a wall and it's lower end is 60 feet from the bottom of the wall. What is the length of the side of largest cubical box that can be placed between wall and ladder without disturbing the position of the ladder?

Answer: 34.29 ft 

Solution: Using Pythagoras theorem, we know the larger triangle formed by wall, ladder and the ground has sides as 100 ft, 80 and 60 ft.

There are two methods of solving this. Method 1) Using similar triangles   2) Using equal area.

Method 1) Let side of the cube be a.

(80-a)/a = 80/60, Solving  34.285 feet

Method 2)

Area of bigger triangle = 1/2 * base * height = 2400 sq.ft

Area of the smaller triangles = 1/2 * x * (60-x) + 1/2 * x * (80-x)

= 70x - x^2

Area of the square = x^2

Now 2400 = 70x - x^2 + x^2

Solving x=2400/70 = 34.285 feet

9. (REF=2949) :

What is the area of the circle in the diagram provided below, if AB is 14 metres?

Answer: 154 sq.m

Solution: The diameter of the circle is equal to the side of the square which is 14 metres.

Hence the radius of the circle is 7 metres.

Area = 22/7 * 7* 7 = 22*7 = 154 sq.m

10. (REF=10500) :

What is the length of the inradius of a triangle whose sides are 18, 24 and 30 cms?

Answer: 6 cms

Solution: s = (a+b+c)/2 = 72/2 = 36

Inradius = SQRT[(s-a)(s-b)(s-c)/s]

= SQRT(18*12*6/36) = 6 cms

11. (REF=9329) :

Find the length of AB in the diagram if the length of the sides of the squares is 4cms.

Answer: 11.31 cms

Solution: Diagonal of 1 square = sqrt(16+16).

Hence AB = 2* sqrt(32) = 8*sqrt(2) cms = 11.31 cms approx

12. (REF=2951) :

What is the length of the side of the given square, if AK is equal to BK and the area of triangle KBC is 16 sq.metres ?

Answer: 8m

Solution: As AK=BK overall area of square is 4* area of triangle KBC = 64 sq.metres

As area of the square = 64, side is 8 metres

13. (REF=2945) :

What is the value for side AB in the triangle below ?

Answer: 8

Solution: By Pythagoras theorem,

AB2 = 102 - 62

=> AB = 8

14. (REF=9423) :

A rope makes 70 rounds of the circumference of a cylinder whose radius of the base is 14 cm. how many times can it go round a cylinder with radius 20 cm?

Answer: 49

Solution: Length of rope = 70 * 2 * 22/7 * 14

Number of times it can go round the other cylinder = 70 * 2 * 22/7 * 14 divided by 2 * 22/7 * 20

= 49 times

15. (REF=10491) :

If each interior angle of a regular polygon is 4 times the exterior angle, find the number of sides in the polygon.

Answer: 10

Solution: Let the num of sides be n.

Exterior angle = 360/n

Interior angle = (n-2)180/n

Given (n-2)180/n = 4*360/n,

Solving n=10

16. (REF=7823) :

How many diagonals are in a hexagon?

Answer: 9

Solution: The formula to find the number of diagonals in a polygon with N sides is N*(N-3)/2.

Hence num of diagonals = 6 * (6-3) * 1/2 = 9.

17. (REF=2941) :

What is the side of the square if BD is 5√ 2 metres

Answer: 5 metres

Solution: By pythagoras theorem,

BD2 = DC2 + CB2

As DC = CB,

25*2 = 2DC2

=> DC = side of the square = 5 metres

18. (REF=9331) :

What is the area of triangle ABC?

Answer: 8 sq.units

Solution: Area of the triangle = 1/2 * base * height = 1/2 *  4  *  4 = 8 sq.units

19. (REF=10498) :

Lines pm and cq are parallel. If angles zmp = 30 degrees, cqm = 110 degrees, what is the value of qmy?

Answer: 80 degrees

Solution: As pm and cq are parallel, angle pmq=180-110=70 deg.

Hence qmy=180-30-70=80

20. (REF=10493) :

A regular polygon has 18 sides. Find the exterior angle of the polygon.

Answer: 20 degrees

Solution: Exterior angle = 360/n = 360/18 = 20 degrees