1.

Base of a triangle is 9 and height is 5. Base of another triangle is 10 and height is 6. Find the ratio of the areas of these trianlge. *​

Answer»

GIVEN :-

  • Base of 1st triangle = 9 units.
  • Height of 1st triangle = 5 units
  • Base of 2nd triangle = 10 units.
  • Height of 2nd triangle = 6 units

TO FIND :-

  • RATIO of area of both triangles.

SOLUTION :-

☯ Area For First Triangle,

\star \: \sf \: base \:(b) = 9 \: unit \\ \star \sf \: height \: (h) \: = 5 \: units \\ \\ \sf \to \: area \: ( 1st \triangle) = \dfrac{1}{2} \times b \times h \\ \\ \to \sf \: \dfrac{1}{2} \times 9 \times 5 \\ \\ \to \boxed{ \red{ \bf \: area \: (1st \triangle) = \dfrac{45}{2}..........(1) }}

☯ Area of second triangle,

\star \: \sf \: base \:(b) = 10\: unit \\ \star \sf \: height \: (h) \: = 6 \: units \\ \\ \sf \to \: area \: ( 2nd \triangle) = \dfrac{1}{2} \times b \times h \\ \\ \to \sf \: \dfrac{1}{ \cancel2} \times 10 \times \cancel{6 }\\ \\ \to \boxed{ \red{ \bf \: area \: (2nd \triangle) = 30 \: ..........(1) }}

☯ ACCORDING to the Question :-

\huge { \boxed{{\purple{ \implies \: (1) \div (2)}}}} \\ \\ \implies \: \sf \left(\dfrac{ \dfrac{45}{2} }{30} \right) \\ \\ \implies \sf \: \dfrac{45}{2} \times \dfrac{1}{30} \\ \\ \implies \sf \: \dfrac{ \cancel{45}}{ \cancel{60}} \\ \\ \implies \sf \: \dfrac{<klux>3</klux>}{4} \\ \\ \implies \boxed{ \red{ \bf \:ratio \: of \: both \: \triangle = 3 \ratio 4 }}

Hence the ratio of both triangle is 3 : 4.


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