Triangles, Quadrilateral & Polygons:

Chapter-11


Triangles:
Classification of Triangles:
1. Equilateral Triangle: A triangle with 3 equal sides
2. Isosceles triangle:  A triangle with at least 2 equal sides
3. Scalene triangle: A triangle with no equal sides

* The sum of interior angles of a triangle is 180⁰ 
* An exterior angle of a triangle is equal to the sum of its interior opposite angles.



Q. 1. a) In the figure, ABD is a straight line.
Calculate the value of a.




Solution: 
a) a⁰  = 56⁰  + 50⁰ (ext. ∠ of △)
= 106⁰
a = 106

b) In the figure, ACE, BCD and DEF are straight lines. Calculate the value of b and of c.




            b) ACB = 94⁰ (vert. opp. ∠s) 
b⁰  +  38⁰  + 94⁰ = 180⁰ (∠sum of △ ABC)
                   b⁰ = 180⁰  -  38⁰  - 94⁰
= 48⁰
      c⁰  = 25⁰  + 94⁰
                              = 119 ( ext. ∠ of △CDE)

Practice:
a) In the figure, ABD is a straight line. Find the value of a.





Solution: 
a) a⁰  = 53⁰  + 48⁰ (ext. ∠ of △)
= 101⁰
a = 101


b) In the figure, ABC, ADF and BDE are straight lines. Find the value of b and of c.




             b) EDF = 93⁰ (vert. opp. ∠s) 
b⁰  +  33⁰  + 93⁰ = 180⁰ (∠sum of △ ABC)
                   b⁰ = 180⁰  -  33⁰  - 93⁰
= 54⁰
      c⁰  = 41⁰  + 93⁰
                              = 134 ( ext. ∠ of △CDE)



4. Angles in a Parallelogram:

Q. The figure shows a parallelogram ABCD where BAD = 64⁰. E lies on AB such that ADE = 49⁰.


Calculate
i) ABC
ii) CDE.






Solution:
i) ABC + 64⁰  = 180⁰ (int. ∠s, AD ∥ BC)
ABC = 180⁰ - 64
= 116

ii) ADC = 116⁰ (opp.  ∠s, of  ∥ gram)
CDE + 49⁰ = 116
CDE = 116⁰ - 49
= 67

Practice:
1. The figure shows a parallelogram ABCD where ADC = 108⁰. E lies on AB such that BCE = 38⁰.

i) Given that ABC = 9x⁰, find the value of x.
ii) Find DCE.








2. The figure shows a parallelogram ABCD. Find the value of x and of y.







Angles in a Rhombus:
The figure shows a rhombus ABCD. The diagonal BD is produced to E such that AD = DE.


If    ABE = 68⁰, calculate
i) BCD
ii) DAE.

i) CBD = 68⁰ (diagonals bisect interior angles of a rhombus)
BCD + 68⁰ + 68⁰  = 180⁰ (int. ∠s, AB ∥ DC)
BCD = 180⁰ - 68⁰ - 68
= 44

ii) ADB  = 68⁰ (base ∠Ñ• of isos. ΔABD)
DAE + AED = 68⁰ ext. ∠ of Î”)
DAE = 68/2
=34

* Base angles of an isosceles triangle are equal.

Practice:
1. The figure shows a rhombus ABCD where ACD = 32. AB is produced to E such that AC = CE.

Find
i) ABC
ii) BCE




2. The figure shows a rhombus ABCD where the diagonals AC and BD intersect at E. Find the value of x.







Construction of Quadrilateral:

Quadrilateral is a closed plane figure that has 4 sides, 4 verticales, 4 interior angles.


Any point on the perpendicular bisector of a line segment is equidistant from the two end points of the line segment.

Q.1. Construct a quadrilateral PQRS such that PQ = 6 cm, QR = 7.5 cm, RS = 8.2 cm and the diagonal PR = 9.2 cm. Measure and write down the size of QRS. 

Solution:
Construction Steps:
a) Using a ruler, draw PR = 9.2 cm. 
b) Since S is 5.8 cm away from P, with P as centre and 5.8 cm as radius, draw arc 1.
c) Since S is 8.2 cm away from R, with R as centre and 8.2 cm as radius, draw arc 2 to cut arc 1 at S. 
d) Join PS and RS.
e) Since Q is 6 cm away from P, with P as centre and 6 cm as radius, draw arc 3.
f) Since Q is 7.5 cm away from R, with R as centre and 7.5 cm as radius, draw arc 4 to cut arc 3 at Q.
g) Join PQ and QR.
QRS = 79


                                   Polygons:
* A polygon is a closed plane figure with three or more straight line segments as its sides.
* The sum of interior angles of an n-sided polygon is (n-2) x 180



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