Mathematics

Constructions of Quadrilaterals – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise-92

Solution 1:

Solution 2:

Solution 3:

Exercise-93

Solution 1:
Steps of construction:

1. Draw seg PQ of length 4 cm.
2. Using a protractor, draw a ray PX ⊥ PQ at point P and a ray QY ⊥ PQ at point Q.
3. Placing the point of the compass on point P and taking a radius of 2 cm, draw an arc of circle to cut AX at point S. Taking the same radius and placing the point of the compass on point Q, draw an arc of circle to cut AY at point R
4. Draw seg RS.

Thus a rectangle PQRS with PQ = 4 cm and RS = 2 cm is constructed.

Solution 2:
Steps of construction:

1. Draw seg LM of length 5.5 cm.
2. Using a protractor, draw a ray LX ⊥ LM at point L and a ray MY ⊥ LM at point M.
3. Placing the point of the compass on point L and taking a radius of 3.5 cm, draw an arc of circle to cut LX at point P. Taking the same radius and placing the point of the compass on point M, draw an arc of circle to cut MY at point N.
4. Draw seg NP.

Thus a rectangle LMNP with LM = 5.5 cm and MN = 3.5 cm is constructed.

Solution 3:
Steps of construction:

1. Draw seg WX of any length, say, 4.5 cm.
2. Using a protractor, draw a ray WA ⊥ WX at point W and a ray XB ⊥ WX at point X.
3. Placing the point of the compass on point W and taking a radius of 2.5 cm, draw an arc of circle to cut WA at point Z. Taking the same radius and placing the point of the compass on point X, draw an arc of circle to cut XB at point Y.
4. Draw seg YZ.

Thus a rectangle WXYZ with WX = 4.5 cm and WZ = 2.5 cm is constructed.

Identity – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise 71:

Solution 1:

Exercise 72:

Solution 1:

Solution 2:

Solution 3:

Exercise 73:

Solution 1:

Solution 2:

Solution 3:

Exercise 74:

Solution 1:

Exercise 75:

Solution 1(1):

Solution 1(2):

Solution 1(3):

Solution 1(4):

Solution 1(5):

Solution 1(6):

Solution 1(7):

Types of Quadrilaterals – Maharashtra Board Class 7 Solutions for Mathematics (English Medium)

MathematicsGeneral ScienceMaharashtra Board Solutions

Exercise 65:

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Triangles – Maharashtra Board Class 9 Solutions for Geometry

AlgebraGeometryScience and TechnologyHindi

Exercise – 2.1

Solution 1:

Solution 2:

Solution 3:

The ratio of the angles of the triangle is 3 : 3 : 6.
Let the measures of the angles be 3x°, 3x° and 6x°.
Then 3x° + 3x° + 6x° = 180°
∴12 x° = 180°
∴ x° = 15°
∴3 x° = 3 × 15° = 45° and 6x° = 6 × 15x° = 90x°
Two angles of the triangle are equal and one angle is a right angle.
The triangle is an isosceles right angled triangle.

Solution 4:

In the ∆ABC, m∠B = 90°.
So, AB is the altitude on side BC.
BC is the altitude on side AB.
BP is the altitude on side AC.
All the altitudes meet at the point B.

In a right angled triangle, the altitudes meet at the vertex forming a right angle.

Solution 5:

∆ABC is an obtuse angled triangle, ∠B being an obtuse angle.
The orthocenter O of the obtuse angled triangle lies outside the triangle.

Solution 6:

In an equilateral triangle, altitudes, medians and angle bisectors are one and the same.
So, in an equilateral triangle, the orthocenter, centroid and incentre lie at the same point.
In the above diagram, point O is the orthocenter, centroid and incentre of ∆ABC.

Solution 7:

∆ABC is an isosceles triangle in which AB = AC.
Side BC is its non-congruent side.
Seg AD is the median drawn on the side BC.
Median AD is the altitude and angle bisector of ∠A also.
[In an isosceles triangle, the centroid, orthocenter and incentre lie on the same line. In the given figure, all the three points lie on line AD.]

Solution 8:

The perpendicular bisectors of the sides of a triangle do not always pass through the opposite vertex except in an equilateral triangle. In the case of an isosceles triangle, the perpendicular bisector of the non-congruent side is a cevian and in a scalene triangle the perpendicular bisector of its sides are not cevians. Hence, in general, the perpendicular bisectors of the sides of a triangle are not cevians.
So, I agree with statement.

Exercise – 2.2

Solution 1(i):

The sum of the measures of the angles of a triangle is 180°.
∴m∠V + m∠A + m∠T = 180°
∴a° + a° + 40° = 180°
∴2a° + 40° = 180°
∴2a° = 180° – 40° = 140°
∴a° = 70°

Solution 1(ii):

The sum of the measures of the angles of a triangle is 180°.
∴m∠P + m∠U + m∠X = 180°
∴90° + 50° + a° = 180°
∴a° = 180° – 140° = 40°
∴a° = 40°

Solution 1(iii):

The sum of the measures of the angles of a triangle is 180°.
∴m∠T + m∠A + m∠P = 180°
∴a° + 110° + 25° = 180°
∴a° = 180° – 135°
∴a° = 45°

Solution 2:

The sum of the measures of the angles of a triangle is 180°.
∴m∠M + m∠T + m∠G = 180°
∴76° + 48° + m∠G = 180°
∴124° + m∠G = 180°
∴m∠G = 180° – 124°
∴m∠G = 56°

Solution 3:

Solution 4:

Solution 5:

m∠M + m∠N + m∠K = 180°
…(The sum of the measures of the angles of a triangle) ….(1)
m∠M + m∠N = 125° …(Given)..(2)
From (1) and (2),
125° + m∠K = 180°
∴m∠K = 55° …(3)
m∠M + m∠K = 113° ….(Given)…(4)
m∠M + 55° = 113° …[From (3)]
∴m∠M = 58°
From (1) and (4),
m∠N + 113° = 180°
∴m∠N = 67°
m∠M = 58°; m∠N = 67°; m∠K = 55°.

Solution 6:

Solution 7(i):

m∠NME + m∠EMR = 180°
…(Angles in a linear pair)
∴m∠NME + 140° = 180°
∴m∠NME = 180° – 140°
∴m∠NME = 40° …(1)
∠TEN = ∠ENM + ∠NME
….(Remote interior angle theorem)
∴100° = x + 40°
…[Given and from(1)]
∴x = 100° – 40°
∴x = 60°.

Solution 7(ii):

∠RQS = ∠PQX
…(Vertically opposite angles)
m∠PQX = 100° …(Given)
∴m∠RQS = 100°
∠QST = ∠QRS + ∠RQS
…(Remote interior angle theorem)
∴ x = 50° + 100°
∴ x = 150°

Solution 7(iii):

In ∆NYX, m∠NYX = 90° …(Given)
∴m∠N + m∠X = 90°
…(Acute angles of a right angled triangle)
∴m∠N + 45° = 90°
…(Given : m∠X = 45°)
∴m∠N = 45°
Consider the DNMZ.
Now, m∠N + m∠NMZ + m∠Z = 180°
…(The sum of the measures of the angles of a triangle)
∴45° + 110° + x = 180°
…[From (1) and given]
∴155° + x = 180°
∴x = 180° – 155°
∴x = 25°

Solution 7(iv):

AB‖DE …(Given)
AD is the transversal.
∴∠BAD = ∠ADE
…(Alternate angles)
∠BAD = 70°
∴∠ADE = 70° i.e.∠EDC = 70° …(1)
Now consider the DCDE:
∠CED + ∠ECD + ∠EDC = 180°
…(The sum of the measures of the angles of a triangle)
∴35° + x + 70° = 180°
…[Given and from (1)]
∴x + 105° = 180°
∴x = 180° – 105°
∴x = 75°

Solution 8:

m∠TPR + m∠RPQ = 180°
…(Angles in a linear pair) …(1)
Similarly, m∠PRQ + m∠SRQ = 180° …(2)
and m∠RQP + m∠PQU = 180° …(3)
From (1), (2) and (3),
m∠TPR + m∠RPQ + m∠PRQ + m∠SRQ + m∠RQP + m∠PQU = 540°
∴(m∠TPR + m∠SRQ + m∠PQU)
+ (m∠RPQ + m∠PRQ + m∠RQP) = 540° …(4)
But, m∠RPQ + m∠PRQ + m∠RQP = 180°
…(The sum of the measures of the angles of a triangle) …(5)
From (4) and (5),
(m∠TPR + m∠SRQ + m∠PQU) + 180° = 540°
∴m∠TPR + m∠SRQ + m∠PQU = 540° – 180°
= 360°
∴∠TPR + ∠SRQ + ∠PQU = 4 right angles.

Solution 9:

Solution 10:

Exercise – 2.3

Solution 1:

Solution 2:

Solution 3:

Solution 4:

Solution 5:

Solution 6:

Solution 7:

Solution 8:

Solution 9:

Solution 10:

Solution 11:

Solution 12:

Solution 13:

seg SK ‖ seg MP ‖ seg TN.
and AT is the transversal. ….(Given)
∴∠ASK begin mathsize 12px style approximately equal to end style ∠AMP begin mathsize 12px style approximately equal to end style ∠ATN …(Corresponding angles) ….(1)
∠A is common to the triangles, ∆ASK,
∆AMP and ∆ATN.
∴∆ASK~∆MAP~∆TAN ….(AA test for similarity)

Solution 14:

Solution 15:

Seg RV ⊥ seg PS
∴ m∠RVP = m∠SVR = 90°
In ∆SRP and ∆SVR,
m∠SRP = m∠SVR = 90° …[Given and from(1)]
∠PSR = ∠VSR …(Common angle)
∴∆SRP ~ ∆SVR …(AA test for similarity)
In ∆SRP and ∆RVP,
m∠SRP = m∠RVP = 90° …[Given and from(1)]
∠SPR = ∠RPV …(Common angle)
∴∆SRP ~ ∆RVP …(AA test for similarity)
In ∆SPR, m∠S + m∠P = 90°
In ∆RVP, m∠P + m∠VRP = 90° …(2) …(Acute angles of a right angled triangle)
From (2), ∠RSV = ∠VRP …(3)
In ∆SVR and ∆RVP
m∠SVR = m∠RVP = 90°
and ∠RSV = ∠VRP …[From (3)]
∴∆SVR ~ ∆RVP …(AA test for similarity)
The pairs of the given triangles are similar.

Solution 16:

Solution 17: