Directions: Study the following questions carefully and choose the right answer.
1
The area of an isosceles ΔABC with AB = AC and altitude AD = 3 cm is 12 sq cm. What is its perimeter ?
» Explain it
A
 
Given, Area of triangle ABC = 12 cm2

∴   1  × h × b = 12
2

⇒   1  × 3 × b = 12
2

⇒  b = 8 cm

Here, BD =

CD =  b  =  8  = 4 cm
2 2

In right-angled ΔABD, by pythagoras theorem,

AB =  AD2 + BD2

a =  32 + 42  =  25  = 5 cm

Now, perimeter of an isosceles triangle = 2a + b = 2 × 5 + 8 = 18 cm

Hence, option A is correct.

2
What is the area between a square of side 10 cm and two inverted semi-circular, cross-sections each of radius 5 cm inscribed in the square ?
» Explain it
D

Given, each side of a square (a) = 10 cm

and, radius of each semi-circular (r) = 5 cm

Area between square and semi-circules = Area of square – 2 Area of semi-circle
= a2 – 2 ×  1  πr2
2
= (10)2 – 2 ×  1  ×  22  × (5)2
2 7

= 100 – 78.5 = 21.5 cm2

Hence, option D is correct.

3
The perimeter of a rectangle having area equal to 144 cm2 and sides in the ratio 4 : 9 is
» Explain it
A
Let, length of rectangle (l) = 4x    and    breadth of rectangle (b) = 9x

∴  Area of rectangle = l × b

⇒  144 = 4x × 9x = 36x2

⇒  x2 = 4

⇒  x = 2

Now, l = 4x = 4 × 2 = 8 cm    and    b = 9x = 9 × 2 = 18 cm

∴  Perimeter of rectangle = 2(l + b) = 2 (8 + 18) = 52 cm2

Hence, option A is correct.

4
One side of a parallelogram is 8.06 cm and its perpendicular distance from opposite side is 2.08 cm. What is the approximate area of the parallelogram ?
» Explain it
C
Area of parallelogram = Base × Height = 8.06 × 2.08 = 16.76 cm2

Hence, option C is correct.

5
In the figure given below, the area of rectangle ABCD is 100 sq cm, O is any point on AB and CD = 20 cm. Then, the area of ΔCOD is
 
» Explain it
C

Given that, CD = 20 cm

And, area of rectangle ABCD = 100 cm2

⇒  AD × CD = 100

⇒  AD × 20 = 100

⇒  AD = 5 cm

∵  AD = OP = 5 cm

∴  Area of ΔCOD =

1  × CD × OP =
2

1  × 20 × 5 = 50 cm2
2

Hence, option C is correct.