1. In a building, there are 24 cylindrical pillars. For each pillar, radius is 28 m and height is 4 m. Find the total cost of painting the curved surface area of the pillars at the rate of ₹ 8 per m2.
Solution:
4. Write 3.61492 x 106 in usual form.
Solution: 3.61492 x 106
= 3.61492 x 1000000
= 3614920
3. Express 0.00000000837 in standard form.
0.00000000837
= 0.00000000837 x 109 / 109
= 8.37 ×10-9
2. Simplify [25 x t-4]/[5-3 x 10 x t-8]
Solution:
We can write the given expression as;
[52x t-4]/[5-3 x 5 x 2 x t-8]
= [52 x t-4+8]/[5-3+1 x2]
= [52+2 x t4]/[2]
= [54 x t4]/[2]
= [625/2] t4
1.If a new-born bear weighs 4 kg, calculate how many kilograms a five-year-old bear weigh if its weight increases by the power of 2 in 5 years?
Weight of new-born bear = 4 kg
Rate of weight increase in 5 years = power to 2
Thus, the weight of the 5-year old bear = 42 = 16 kg
5. In Fig. 6.40, ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠OZY and ∠YOZ.
As we know, the sum of the interior angles of the triangle is 180°.
So, ?X +?XYZ + ?XZY = 180°
substituting the values as given in the question we get,
62° + 54° + ?XZY = 180°
Or, ?XZY = 64°
Now, As we know, ZO is the bisector so,
?OZY = ½ ?XZY
? ?OZY = 32°
Similarly, YO is a bisector and so,
?OYZ = ½ ?XYZ
Or, ?OYZ = 27° (As ?XYZ = 54°)
Now, as the sum of the interior angles of the triangle,
?OZY +?OYZ + ?O = 180°
Substituting their respective values we get,
?O = 180° – 32° – 27°
Or, ?O = 121°
4. In the Figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint: Draw a line parallel to ST through point R.]
Solution:
First, construct a line XY parallel to PQ.
As we know, the angles on the same side of the transversal are equal to 180°.
So, ?PQR + ?QRX = 180°
Or,?QRX = 180° – 110°
? ?QRX = 70°
Similarly,
?RST + ?SRY = 180°
Or, ?SRY = 180° – 130°
? ?SRY = 50°
Now, for the linear pairs on the line XY-
?QRX + ?QRS + ?SRY = 180°
Substituting their respective values we get,
?QRS = 180° – 70° – 50°
Or, ?QRS = 60°
3. In the Figure, if AB || CD, EF ⊥ CD and ∠GED = 126°, find ∠AGE, ∠GEF and ∠FGE.
Solution:
Since AB || CD GE is a transversal.
It is given that ?GED = 126°
So, ?GED = ?AGE = 126° (alternate interior angles)
Also,
?GED = ?GEF + ?FED
As
EF ? CD, ?FED = 90°
? ?GED = ?GEF + 90°
Or, ?GEF = 126° – 90° = 36°
Again, ?FGE + ?GED = 180° (Transversal)
Substituting the value of ?GED = 126° we get,
?FGE = 54°
So,
?AGE = 126°
?GEF = 36° and
?FGE = 54°
2. In the Figure, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that ∠ROS = 1/2(∠QOS – ∠POS).
In the question, it is given that (OR ? PQ) and ?POQ = 180°
So, ?POS + ?ROS + ?ROQ = 180° (Linear pair of angles)
Now, ?POS + ?ROS = 180° – 90° (Since ?POR = ?ROQ = 90°)
? ?POS + ?ROS = 90°
Now, ?QOS = ?ROQ + ?ROS
It is given that ?ROQ = 90°,
? ?QOS = 90° + ?ROS
Or, ?QOS – ?ROS = 90°
As ?POS + ?ROS = 90° and ?QOS – ?ROS = 90°, we get
?POS + ?ROS = ?QOS – ?ROS
=>2 ?ROS + ?POS = ?QOS
Or, ?ROS = ½ (?QOS – ?POS) (Hence proved).
In the Figure, lines XY and MN intersect at O. If ∠POY = 90° and a : b = 2 : 3, find c.
Solution:
As we know, the sum of the linear pair is always equal to 180°
So,
?POY + a + b = 180°
Substituting the value of ?POY = 90° (as given in the question) we get,
a + b = 90°
Now, it is given that a : b = 2 : 3 so,
Let a be 2x and b be 3x.
? 2x + 3x = 90°
Solving this we get
5x = 90°
So, x = 18°
? a = 2 × 18° = 36°
Similarly, b can be calculated and the value will be
b = 3 × 18° = 54°
From the diagram, b + c also forms a straight angle so,
b + c = 180°
=> c + 54° = 180°
? c = 126°
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