ogram.
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 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, respecti