Q. 14.2( 6 Votes )

# Look at the figur

Answer :

(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

Hypothenuse^{2} = Base^{2} + height^{2} ………. eq 1

In the given triangle base = 5cm

Hypotenuse = 13cm

Height = ?

By eq 1 we have

⇒ 13^{2} = 5^{2} + height^{2}

⇒ 169 - 25 = height^{2}

⇒ height = √144 cm

⇒ height = 12cm

Area of the given triangle

= 30cm^{2}

(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 cm^{2}

(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 cm^{2}

(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

Hypothenuse^{2} = Base^{2} + height^{2} ………………….eq1

In the given figure

AC^{2} = CB^{2} + AB^{2}

⇒ AC^{2} = 5^{2} + 12^{2}

⇒ AC^{2 =} 25 + 144

⇒ AC = √169

⇒ AC = 13cm

Area of right - angled triangle ABC

= 30cm^{2}

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 cm^{2}

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

Area of given figure = 30cm^{2} + 30√2 cm^{2}

= 60√2 cm^{2}

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