Question

Margaret and sam each drewca triangle with a base of length 1 cm.The hight of sams triangle is one-forth the height of margrets triangle. how many times greater is the area of margrets triangle then the area of sams triangle?




Answer
A.2
B.4
C.6
D.8
E.16
​

Answer 1

Answer:

Step-by-step explanation:

c

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Answer 2

Answer:

• Base of both triangle = 1 cm
• Height of Sam's △ = ¼ the Height of Margret's △

$\longrightarrow\sf Height \:of \: Sam's \:\Delta = \dfrac{1}{4} \times Height \:of \: Margret's \: \Delta \\\\\\\longrightarrow\sf \dfrac{Height \:of \: Sam's \:\Delta}{Height \:of \: Margret's \: \Delta} = \dfrac{1}{4}\\\\\\\longrightarrow\sf Height \:of \: Sam's \:\Delta :Height \:of \: Margret's \: \Delta = 1:4$

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Let's the Height of Sam's and Margret's △ be n and 4n respectively.

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$\setlength{\unitlength}{12mm}\begin{picture}\thicklines\put(8,1){\line(1,0){3}}\put(8,1){\line(1,1){1.5}}\put(11,1){\line(-1,1){1.5}}\put(9.5,1){\line(0,1){1.5}}\put(9.3,0.65){\sf1 cm}\put(9.25,1.6){\sf1}\put(9.25,3){\sf Sam}\end{picture}\setlength{\unitlength}{15mm}\begin{picture}\thicklines\put(6.5,0.8){\line(1,0){3}}\put(6.5,0.8){\line(1,1){1.5}}\put(9.5,0.8){\line(-1,1){1.5}}\put(8,0.8){\line(0,1){1.5}}\put(8,0.5){\sf1 cm}\put(7.8,1.4){\sf4}\put(7.8,2.5){\sf Margret}\end{picture}$

Here is a thing to understand, as Base of both triangle is equal, that's why Area will be all dependent on Height only. In simple words while solving it ½ and Base which are common will be cancelled.

$\underline{\bigstar\:\textsf{According to the given Question :}}$

$:\implies\sf Greater=\dfrac{Margret\:\Delta's\:Height}{Sam\:\Delta's\:Height}\\\\\\:\implies\sf Greater = \dfrac{4n}{n}\\\\\\:\implies\sf Greater = 4$

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$\therefore\:\underline{\textsf{Area of Margret's \Delta area is B) \textbf{4 times} more than Sam's \Delta }}.$