 cm, 12 cm and 15 then what is its area.
SOL:- Area of the triangle
A = √s(s-a) (s-b) (s-c)
S = (a+b+c) / 2
= (9+12+15) / 2
= 36 / 2 = 18 cm
A = √s(s-a) (s-b) (s-c)
= √18 (18-9) (18-12) (18- 15)
= √18 x 9 x 6 x 3
= 9 x 3 x 2
= 54 sq.cm
Hence, the area of the triangle is 54 sq. cm.
17. Find the total surface area of a hemispherical bowl which is 5 cm thick and has inner radius 35 cm.
SOL:- Let R be outer radius and r be the inner radius of the bowl.
Thickness of ring = 5 cm
R = ( r+5) cm
= 35 + 5
= 40 cm
Total surface area = CSA of outer hemisphere + CSA of inner hemisphere + Area of ring
= 2 ΠR² + 2Πr² + Π( R² – r²)
= Π [2R² + 2r² + ( R ² – r²)] = 22/7 (3R² + r²)
= 22/7 (4800 + 1225)
= 22/7 (6025)
= 18935.71 sq. cm
18. A coin is tossed 500 times with the following frequencies:Head: 245, Tail: 255. Compute the probability of each event.
SOL:- Since the coin is tossed 500 times, the total number of trials is 500.
Let the event of getting a head be E and the event of getting a tail be F.
Then the number of trials of getting E is 245.
And the number of trials of getting F is 255.
P (Getting a head)
P(E) = Number of trials which event E happens / Total number of trails= 245/500
= 49 /100
P (Getting a tail)
P (F) = Number of trials in which event F happens / Total number of trials= 255 / 500
SECTION C [ FOUR MARK QUESTIONS ]
19. Draw the graph of the equation 2x + 3y = 13, and determine whether x = 4, y = 2 is a solution not.
SOL:- The given equation is 2x + 3y = 13.
y = (13 – 2x) / 3
If x = 2,
Y = ( 13 – 4) / 3 = 3
If x = 5
Y = (13 – 10) / 3 = 1
Thus, we have the following table for 2x + 3y = 13:
Plot the points A (2, 3) and B (5, 1) on a graph paper. Draw a line that passes through points A and B.
Line AB represents the graph for 2x + 3y = 13. We can see that the point (4, 2) does not lie on the graph of the equation 2x + 3y = 13.
Hence, x = 4, y = 2 is not a solution of the equation 2x + 3y = 13.
20.Two opposite angles of a parallelogram measure (60 – x)° and (3x – 4)°. Find the measure of each angle of the parallelogram.
SOL:- Let ABCD be a parallelogram, where ∠A = (3x – 4)° and ∠C = (60 – x)°.
Then ∠A = ∠C
= 3x – 4 = 60 – x
= 4x = 64
x = 16
∠A = (3 × 16 – 4)°
= (48 – 4)°
∠A = 44°
∠C = (60 – 16)° = 44°
Now, ∠A + ∠B = 180°
44° + ∠B = 180°
∠B = 180° – 44° = 136°
Also, ∠D = ∠B = 136°
Thus, the angles of the parallelogram are 44°, 136°, 44°, 136°.
22. PQRS and ABRS are two parallelograms, and X is any point on side BR. Show that:
(i) ar(PQRS) = ar(ABRS)
SOL:= i) Parallelograms PQRS and ABRS are on the same base RS and between the same parallels RS and BP.
ar (PQRS) = ar (ABRS).
Hence, ar(PQRS) = ar(ABRS).
(ii) ar(AXS) =
SOL:= ΔAXS and parallelogram ABRS are on the same base AS and between the same parallels AS and BR.
Ar (ΔAXS) = ½ ar (ABRS)
Ar (ΔAXS) – ½ ar ( PQRS)
Hence, ar(AXS) = ½ ar ( PQRS) .
23. Construct a triangle ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.
SOL:- Steps of construction:
1. Draw a line segment BC of length 7 cm.
2. At B, construct ΔXBC = 75°.
3. With B as the centre and radius equal to 13 cm, draw an arc cutting XB at D.
4. Joint DC.
5. Draw PQ, the perpendicular bisector of DC, and let it intersect DB at A.
6. Join AC.
Then, ΔABC is the required triangle.
Since A lies on the perpendicular bisector of DC, it is equidistant from points D and C.
AD = AC … (1)
Now, BD = 13 cm [By construction] BA + AD = 13 cm
BA + AC = 13 cm [Using (1)] Hence, ΔABC is the required triangle
24. A cylinder is 12 cm high, and the circumference of its base is 44 cm. Find its curved surface area and total surface area.
SOL:- Let r be the base radius and h be the height of the cylinder.
Then h = 12 cm.
Given: Circumference of base = 44 cm
⇒ 2 π r = 44 cm
= 2 X 22/7 X r = 44 cm
r = 44 X 7/ 2 X 22
r = 7 cm … (1)
Curved surface area of a cylinder = 2Πrh
= 2 X 22/7 X 7 X 12
= 44 × 12
= 528 cm2
Total surface area a cylinder = 2Πrh + 2Πr2
= 2Πr(r + h)
= 44(7 + 12) [r = 7 cm, from (1)] = 44(19)
= 836 cm2
25. A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet 