Solution:
Construction:
Step I: Draw AC = 5 cm.
Step II: Draw the right bisector of AC at O.
Step III: Draw an angle of 50° at O and product both sides.
Step IV: Draw two arcs with centre O and of the same radius 2.5 cm to cut at B and D.
Step V: Join AB, BC, CD and DA.
Thus, ABCD is the required rectangle.
Q2. Construct a quadrilateral ABCD in which BC = 4 cm, ∠B = 60°, ∠C = 135°, AB = 5 cm and ∠A = 90°.
Solution.
Construction:
Step I: Draw AB = 5 cm.
Step II: Draw the angle of 60° at B and cut BC = 4 cm.
Step III: Draw an angle of 135° at C and angle of 90° at A which meet each other at D.
Thus, ABCD is the required quadrilateral.
Q3. Is it possible to construct a quadrilateral ABCD in which AB = 5 cm, BC = 7.5 cm, ZA = 80°, B = 140° and C=145°? If not, give reason.
Solution:
No, it is not possible to construct a quadrilateral ABCD with the given measurements.
Angle (A + B + C)= 80° +140° +145° =365° is greater than 360°.
The sum of all the four angles is 360°. quadrilateral cannot be constructed.
Question 4.
In the parallelogram given alongside if m∠Q = 110°, find all the other angles.
Solution:
Given m∠Q = 110°
Then m∠S = 110° (Opposite angles are equal)
Since ∠P and ∠Q are supplementary.
Then m∠P + m∠Q = 180°
⇒ m∠P + 110° = 180°
⇒ m∠P = 180° – 110° = 70°
⇒ m∠P = m∠R = 70° (Opposite angles)
Hence m∠P = 70, m∠R = 70°
and m∠S = 110°
Q5. Write true and false against each of the given statements.
(a) Diagonals of a rhombus are equal.
(b) Diagonals of rectangles are equal.
(c) Kite is a parallelogram.
(d) Sum of the interior angles of a triangle is 180°.
(e) A trapezium is a parallelogram.
(f) Sum of all the exterior angles of a polygon is 360°.
(g) Diagonals of a rectangle are perpendicular to each other.
(h) Triangle is possible with angles 60°, 80° and 100°.
(i) In a parallelogram, the opposite sides are equal.
Solution:
(a) False
(b) True
(c) False
(d) True
(e) False
(f) True
(g) False
(h) False
(i) True
Q6. If AM and CN are perpendiculars on the disgonal BD of a parallelogram ABCD,Is △AMD≅△CNB?Give reason.
Solution
In triangles AMD and CNB
AD=BC(opposite sides of a parallelogram)
∠AMB=∠CNB=90∘
∠ADM=∠NBC since AD∥BC and BD is the transversal.
So, △AMD≅△CNB by AAS
Q7.Verify whether a polyhedron can have 10 faces, 20 edges and 15 vertices.
Solution:
We have
Number of faces F = 10
Number of edges E = 20
and number of vertices V = 15
Euler’s formula:
V + F – E = 2
⇒ 15 + 10 – 20 = 2
⇒ 5 ≠ 2
Hence, it is not possible to have a polyhedron satisfying the above data
Q8. Draw the nets of the following polyhedrons.
(i) Cuboid
(ii) Triangular prism with a base equilateral triangle.
(iii) Square pyramid.
Solution:
i.
ii.
Q9. Name the solids that have:
(i) 4 faces
(ii) 8 triangular faces
(iii) 6 faces
(iv) 1 curved surface
(v) 5 faces and 5 vertices
(vi) 6 rectangular faces and 2 hexagonal faces
Solution:
(i) Tetrahedron
(ii) Regular octahedron
(iii) Cube and cuboid
(iv) Cylinder
(v) Square and a rectangular pyramid
(vi) Hexagonal prism
Q10. The scale of a map is given as 1 : 50,000. Two villages are 5 cm apart on the map. Find the actual distance between them.
Solution
Let the map distance be x cm and actual distance be y.
1 : 50,000 =x:y
Solution:
Let the required number of workers be x.
The number of workers, faster will they do the work.
So, the two quantities are inversely proportional
x1y1 = x2y2
⇒ 42 × 15 = 30 × x
⇒ x = 21
Hence, the required number of men = 21
Q12. The cost of 5 metres of cloth is ₹ 210. Tabulate the cost of 2, 4, 10 and 13 metres of cloth of the same type.
Solution:
Let the length of the cloth be x m and its cost be ₹ y. We have the following table.
Q12.Mohit deposited a sum of ₹ 12000 in a Bank at a certain rate of interest for 2 years and earns an interest of ₹ 900. How much interest would be earned for a deposit of ₹ 15000 for the same period and at the same rate of interest?
Solution:
Let the required amount of interest be ₹ x.
No comments:
Post a Comment