required to make ten such caps.
SOL:- Let r be the base radius, h the height, and l the slant height of the cone.
r = 7 cm and h = 24 cm
Slant height (l) = √r² + h²
= √7² + 24²
= √49 + 576
= 25 cm … (1)
Curved surface area of one cap = Πrl = 22/7 X 7 X 25 cm2 [Using (1)] Curved surface area of 10 caps = 22/7 X 7 X 25 X 10 = 5500 cm2
Hence, the area of the sheet required for 10 caps is 5500 cm2.
26. In a cricket match, a batsman hits a boundary 6 times off the 30 balls that he plays. Find the probability that he does not hit a boundary off a delivery.
SOL:- The batsman played 30 balls, so the total number of trials is 30.
He hits a boundary 6 times, so he did not hit a boundary (30 – 6) = 24 times.
Let E be the event that the batsman did not hit a boundary.
Then the number of trials in which event E happened = 24
P(Batsman did not hit a boundary) = P(E) =
Number of trials in which event E happened/ Total number of trials = 24/ 30 = 4 /5
27. Three coins are tossed simultaneously 200 times, with the following frequencies of different outcomes:
Sol:- The three coins are tossed simultaneously 200 times.
Therefore, the total number of trials is 200.
Let E be the event of getting two heads
Then the number of trials in which event E happens = 72
P(Getting two heads) = P(E) = Number of trials in which event E happened/ Total number of trials
= 72 / 200
= 9 /25
SECTION D [ FIVE MARK QUESTIONS ]
28. Use this table to draw a graph.
find the values of a and b.
SOL:- Plot the points A (–5, –2), B (3, 2) and C (13, 7).
Draw a straight line passing through A, B and C.
The y coordinate that corresponds to the x coordinate – 1 is 0:a = 0
Through y = 5, draw a horizontal line that meets the graph at a point, say P. Through P, draw a vertical line that meets the X-axis. It will meet the X-axis at x = 9:b = 9
23. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
SOL:- Let ABCD be a quadrilateral such that P, Q, R and S are the mid-points of sides AB, BC, CD and DA, respectively. PR and SQ are joined. Join AC, PQ, QR, RS and SP.
In ΔABC, P is the mid-point of AB, and Q is the mid-point of BC.
PQ AC and [By mid-point Theorem] …. (1)
In Δ ADC, S is the mid-point of DA and R is the mid-point of DC. [Given] SRAC and [By mid-point Theorem] … (2)
From equations (1) and (2), we get
PQ SR and PQ = SR
Thus, in quadrilateral PQRS, one pair of opposite sides is parallel and equal. PQRS is a parallelogram.
We know that the diagonals of a parallelogram bisect each other so PR and SQ bisect each other.
Hence, the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.
30.In the figure given here, CDis parallel to AB and ΔCBA = 25°. Find x.
SOL:- Join AC and BE.
ΔACB = 90° [Angle inscribed in a semi-circle] …. (1)
In ΔACB, we have
∠ACB + ∠CBA + ∠CAB = 180° [Sum of angles of a triangle] ⇒ 90° + 25° + ∠CAB = 180° [Using (1) and ∠CBA = 25°, given] ∠CAB = 180° – 115 °
∠CAB = 65° … (2)
∠CEB = ∠CAB [Angles in the same segment] …. (3)
∠CEB = 65° [Using (2) and (3)] … (4)
CD is parallel AB (given) and CB is the transversa
∠DCB = ∠CBA [Alternate angles] …. (5)
∠CBA = 25° [Given] … (6)
∠DCB = 25° [Using (5) and (6)] …. (7)
∠DEB = ∠DCB [Angles in the same segment] … (8)
∠DEB = 25° [Using (7) and (8) …. (9)
Now x = ∠CED = ∠CEB – ∠DEB
x = 65° – 25° [Using (4) and (9)] x = 40°
Hence, x = 40°.
31. If each diagonal of a quadrilateral separates it into two triangles of equal areas, then show that the quadrilateral is a parallelogram.
Sol:- Let ABCD be the given quadrilateral. Diagonal AC separates it into two equal areas such that ar(ΔACD) = ar(CAB), while diagonal BD separates it into two equal areas such that ar(DAB) = ar(BCD).
Draw CP ⊥ AB produced, and DQ ⊥AB.
ar(quad ABCD) = ar(ΔCAB) + ar(ΔACD)] ar(quad ABCD) = 2ar (ΔCAB) [ ar(ΔACD) = ar(ΔCAB)] ….. (1)
Also, ar (quad ABCD) = ar(ΔDAB) + ar(ΔBCD)
ar(quad ABCD) = 2ar (ΔDAB) [ ar(ΔDAB) = ar(ΔBCD) ….. (2)
From equations (1) and (2), we get
ar(ΔCAB) = ar(ΔDAB)
= 1/ 2 AB X CP = ½ ABX DQ
CP = DQ
Altitude from C of ΔABC = Altit