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Ratio of Areas of Two Triangles with Equal Heights is 2 : 3. If Base of the Smaller Triangle is 6 Cm Then What is the Corresponding Base of the Bigger Triangle ? - Geometry Mathematics 2

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प्रश्न

Ratio of areas of two triangles with equal heights is 2 : 3. If base of the smaller triangle is 6 cm then what is the corresponding base of the bigger triangle ?

उत्तर

\[\frac{\text{ Area of smaller triangle }}{\text{ Area of bigger triangle }} = \frac{2}{3}\]
\[ \Rightarrow \frac{\frac{1}{2} \times \text{Height of smaller triangle } \times \text{ Base of smaller triangle }}{\frac{1}{2} \times \text{ Height of bigger triangle } \times \text{ Base of bigger triangle }} = \frac{2}{3}\]
\[ \Rightarrow \frac{6}{\text{ Base of bigger triangle }} = \frac{2}{3}\] 

\[\Rightarrow \text{ Base of bigger triangle } = \frac{3}{2} \times 6\]
\[ = 9\]

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अध्याय 1: Similarity - Problem Set 1 [पृष्ठ २७]

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बालभारती Geometry (Mathematics 2) [English] 10 Standard SSC Maharashtra State Board
अध्याय 1 Similarity
Problem Set 1 | Q 3 | पृष्ठ २७

संबंधित प्रश्न

The ratio of the areas of two triangles with the common base is 14 : 9. Height of the larger triangle is 7 cm, then find the corresponding height of the smaller triangle.


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  2. `("A"(∆"PBC"))/("A"(∆"ABC"))`
  3. `("A"(∆"ABC"))/("A"(∆"ADC"))`
  4. `("A"(∆"ADC"))/("A"(∆"PQC"))`

 In trapezium PQRS, side PQ || side SR, AR = 5AP, AS = 5AQ then prove that, SR = 5PQ 

 

 


In the figure, PM = 10 cm, A(∆PQS) = 100 sq.cm, A(∆QRS) = 110 sq. cm, then find NR.


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`(A(∆ABD))/(A(∆ABC))`


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\[\frac{A\left( ∆ ADC \right)}{A\left( ∆ ABC \right)}\] 


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`(A(ΔPQB))/(A(ΔPBC)) = (1/2 xx square xx square)/(1/2 xx square xx square)`

= `square/square`


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