Question

solve 19 question........​

Answer 1

Answer:

108cm²

Step-by-step explanation:

hypotenuse =15cm

base =9cm

perpendicular =

h²=p²+b²

15²=p²+9²

225=p²+81

p²=144

p=12

the area of triangle =1/2×base×height

=1/2×18×12

=108cm²

Answer 2

⟼ Let ABC be the given isoceles traingle

⟼ AB= AC= 15cm

⟼ AD ⊥ BC and its bisector BC.

BD=DC= 9cm

⟼ By Phthagores theoram:-

(AB)²= (BD)²+(AD)²

15²= 9²+(AD)²

(AD)²= 225-81

(AD)²= 144

AD= 12

⟼ Area of traingle:- 1÷2×b×h

1÷2×18×12

108 cm²

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So the area of Isoceles traingle is:- 108cm².

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Some Important Formulae:-

⟼ Area of rectangle= Length×Breadth

⟼ Length of a rectangle= Area÷Breadth

⟼ Breadth of a rectangle= Area÷Length

⟼ Area of a square= s×s

⟼ Side of a square= √Area

⟼ Area of a parallelogram= Base×Height

⟼ Area of a traingle= 1÷2×base×height

⟼ Area of an equilateral triangle= √3a²÷4

⟼ Area of a trapezium= 1÷2× height× sum of parallel sides

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