lled a:
7. In the figure given here, if O is the centre of the circle, find the value of x
A. 65 degree
B. 50 degree
C. 60 degree
D. 70 degree
8. ABCD is a trapezium in which AB || DC. Then ar (∆AOD)= ____.
C. ar (BDC)
D. ar (ABD)
9. If 5.5 cm is the edge of a cube then its surface area is:
A. 172 m2
B. 181.5 m2
C. 160 m2
D. 150.2 m2
10. The probability of getting a number less than 5 in a single throw of a die is:
SECTION B [ THREE MARK QUESTIONS ]
11. Write four solutions of the equation 3x + y = 4.
SOL:- Given: 3x + y − 4 = 0, y = 4 − 3x
Putting x = 0, we get y = 4 − 3(0) = 4− 0 = 4
Putting x = 1, we get y = 4 −3(1) = 4− 3 = 1
Putting x = 2, we get y = 4−3(2) = 4− 6 = −2
Putting x = 3, we get y = 4 −3(3) = 4−9 = −5
Solutions of the given equation are (0, 4), (1, 1), (2, −2) and (3, −5).
12. In ΔABC, D and E are the mid-points of AC and BC, respectively. If DE = 11.5 cm, then find the length of AB.
SOL:- Given, DE = 11.5cm
By the mid-point theorem,
DE is parallel to AB and half of it
= 2 × 11.5
AB = 23 cm
Hence, the length of AB is 23 cm.
13. ABCD is a parallelogram. P is a point on AD such that AP = 1/3 AD. If Q is a point on BC such that CQ = 1/3 BC , then show that AQCP is a parallelogram.
AP=1/3 AD —– (i)
CQ=1/3 BC —– (ii)
AD = BC ( Opposite sides of a parallelogram are equal)
Multiplying by 1/3 on both the sides
1/3 AD = 1/3BC
AP = CQ —– (iii) (from (i) and (ii))
AD is parallel to BC
so AP is parallel to CQ —– (iv)
From (iii) and (iv):
AQCP is a quadrilateral in which one pair of opposite sides AP and CQ are equal and parallel.
Hence, AQCP forms a parallelogram.
14. In ∆ABC, AD is a median. Prove that ar (ΔABD) = ar (ΔACD).
SOL:- Given: AD is a median of ΔABC.
To prove: ar(ΔABD) = ar(ΔACD)
ar(ΔABD) = ½ BD X AN – – – – (1)
ar(ΔACD) = ½ CD X AN – – – – – (2)
We have BD = DC – – – – -(3) [∵AD is the median on side BC] From equations (1), (2) and (3) we get
ar(ΔABD) = ar(ΔACD)
15. In the figure given here, PQR = 75 and PRQ = 35. Find QSR.
SOL:- Do it by yourself.
16. The length of the sides of a triangle are 9 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 parallel