Q1 of 99 Page 258

Look at the figures below and let us write by calculating the area.

(i) Given triangle is a right angle triangle whose area would be


Area of triangle


In a right angled triangle, according to the Pythagoras theorem


Hypothenuse2 = Base2 + height2 ………. eq 1


In the given triangle base = 5cm


Hypotenuse = 13cm


Height = ?


By eq 1 we have


132 = 52 + height2


169 - 25 = height2


height = √144 cm


height = 12cm


Area of the given triangle


= 30cm2


(ii) Given triangle has all the sides equal as 6 cm so it`s a equilateral triangle


Area of a equilateral triangle


Here side = 6cm


So area of triangle



= 9√3 cm2


(iii) Two sides of the given triangle are equal = 6cm and base = 8 cm


So it’s an isosceles triangle


Area of an isosceles triangle


( half of base = 4cm)


=


= 4× √20


= 8 √5 cm2


(iv) Given figure is a trapezium with one angle ABC = 90°



In right triangle ABC


CB = 5cm, is the base


AB = 12 cm, is the height of the triangle


AC is the hypothenuse which will be given by Pythagoros theorem


Hypothenuse2 = Base2 + height2 ………………….eq1


In the given figure


AC2 = CB2 + AB2


AC2 = 52 + 122


AC2 = 25 + 144


AC = √169


AC = 13cm


Area of right - angled triangle ABC




= 30cm2


Now in triangle ACD


AC = 13cm


DC = 10cm given


AD = 7 cm given


Area of triangle with given three sides


Where a, b, and c are the sides of the triangle


and s (semi perimeter of triangle )




s = 15 cm


Area of ADC triangle


=


=


= 30√2 cm2


Area of given figure = area of triangle ABC + area of triangle ACD


Area of given figure = 30cm2 + 30√2 cm2


= 60√2 cm2


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