Draw perpendiculars of equal lengths upward at one end of the line and downwards at the other end. (iii) Draw a line segment of length 5.5 cm. ∴ The third angle of ∆APM and ∆BMQ and opposite sides of equal angles are equal. ∠A = ∠B (alternate angles formed by cutting the parallel lines AP and QB by PQ and AB.) Using a set square, locate the midpoint of this line. (ii) What is special about the position of M on AB. (i) Are the sides of ∆AMP equal to the sides of ∆BMQ? Why? The point of inter section of PQ and AB is marked as M. In the figure, AP and BQ equal and parallel are lines drawn at the ends of the line AB. ∠M = ∠Z(Sides opposite to equal angles are also equal) The sides AC and PR opposite to the 70° angle are also equal. Also BC and PQ, opposite to the 50° angle are also equal. So the third angles of the triangles ∠C and ∠R are also equal. In each pair of triangles below, find matching pairs of sides and write their names.ĪC = PR (The two angles and the side in between them in ∆ ABC are equal to the two angles and the side in between them in ∆PQR. (i) The angles of ∆AMD, ∆MBC and ∆DCM are 40°, 60° and 80° respectively. ∴ ∠AMD = ∠CDM (Alternate interior angles) Given AB = 12 cm and M is the mid-point of AB. (ii) What is special about the quadrilateral AMCD and MBCD? (i) Compute the angle of ∆AMD, ∆MBC and ∆DCM? In the figure below, the lines AB and CD are parallel and M is the mid point of AB. So the third side and other angles are equal. ∴ The two sides in ∆AMC and ∆BMC and the angle made by them are equal. In the figure below, M is the midpoint of the line AB. So the opposite angles in quadrilateral ACBD are also equal. The angles opposite to equal sides of triangles ∆ABC and ∆ABD are equal. The opposite sides of quadrilateral ∆CBD are equal. The angles between the sides AC, AB and BD, AB are equal. Is ABCD in the figure, a parallelo¬gram? Why? ∠ABC = ∠BDE (The angles opposite to the equal sides of equal triangles are equal) But they are corresponding angles. So the thirif side of triangle are also equal.) When we consider the triangles ∆CAB, ∆EBD (Corresponding angles)īC = DE (The two sides of ∆ CAB and the angle made by them are equal to the two sides of ∆ EBD and the angle made by them. In the figure below, AC and BE are parallel lines. ∠N = ∠X (If two sides of a triangle are equal, the angles opposite to these sides are also equal) (ii) MN = XY (If two sides of triangle and the angle made by then are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal) ∴ ∠A = ∠Q (The opposite angles of equal sides of two triangles are also equal) (i) BC = PR (If two sides of a triangle and the angle made by them are equal to two sides of another triangle and the angle made by them, then the third sides of the triangle are also equal.) In each pair of triangles below find the pairs of matching angles and write them down. ∴ ∠BAC = ∠DAC = 30° (Angles opposite to equal sides are equal in a triangle ) ∴ ∠ACD = ∠ACB = 50° (Angles opposite to equal sides of a triangle are equal) The sides of the triangles ABC and ADC are equal.
In the quadrilateral ABCD shown below, AB = AD, BC = CDĬompute all the angles of the quadrilateral? You can go online and type the temperature you want to convert into Google using this method: “ 350 degrees F to C” or vice versa. So the angles of ∆ABC are equal to the sides of ∆ABC. The side of ∆ABC are equal to the sides of ∆ABD. The side AB is common to both the triangles in the figure.
∠A = 60, ∠B = 70° then ∠C = 180 – (60° + 70°)Īre the angles of ∆ABC and ∆ABDequal in the figure above? Why? (The angle opposite to equal sides are equal)Ĭompute the remaining angles of both the triangles. Solution: C = 80° (Use the property that the sum of three angles of a triangle is 180°) In the triangles below AB = QR, BC = RP, CA = PQĬompute ∠C of ∆ABC and all angle of ∆PQR. ∠N = ∠X (The angles opposite to the side of length 8cm) ∠M = ∠Z (The angles opposite to the side of length 4 cm) (ii) ∠L = ∠Y (The angles opposite to the sides of length 10cm) ∠C = ∠Q (The angles opposite to the sides of length 6 cm) ∠B = ∠P (The angles opposite to the sides of length 4 cm) (i) ∠A = ∠R (The angles opposite to 5cm sides) In each pair of triangles below, find all pair of matching angles and write them down.
Kerala State Syllabus 8th Standard Maths Solutions Chapter 1 Equal Triangles Equal Triangles Text Book Questions and Answers
Lesson 23 base angles of isosceles triangles download#
You can Download Equal Triangles Questions and Answers, Activity, Notes, Kerala Syllabus 8th Standard Maths Solutions Chapter 1 help you to revise complete Syllabus and score more marks in your examinations.
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