Question

3. Base of a triangle is 12 &
height is 7 base of another
triangle is 14 & height is 8. Find
the ratios of areas of these
triangles
O A) 3/4
O B) 4/3
O C) 2/3
O D) 3/2​

Answer 1

Answer:

A)

Step-by-step explanation:

Your answer is 3:4 or 3/4

Answer 2

Given :

• Base of first triangle = 12 cm
• Height of first triangle = 7 cm
• Base of second triangle = 14 cm
• Height of second triangle = 8 cm

To Find :

• The ratio of areas of these triangles.

Solution :-

As we know that, formula of area of triangle is given by :-

$\star \: \boxed{ \sf{Area_{(\triangle)} = \dfrac{1}{2} \times base \times height }}$

Calculating the area of first triangle,

$\dashrightarrow \: \: \rm \: Area_{(first \: \triangle)} = \dfrac{1}{2} \times 12 \times 7 \\ \\ \\ \dashrightarrow \: \: \rm \: Area_{(first \: \triangle)} = \dfrac{1}{ \not2} \times \not12 \times 7 \\ \\ \\ \dashrightarrow \: \: \rm \: Area_{(first \: \triangle)} = 6 \times 7 \\ \\ \\ \dashrightarrow \: \: \rm \: \underline{ \boxed{ \rm{Area_{(first \: \triangle)} = 42 \: cm^2}}} \blue{ \bigstar}$

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Calculating the area of second triangle,

$\dashrightarrow \: \: \rm \: Area_{(second \: \triangle)} = \dfrac{1}{2} \times 14 \times 8 \\ \\ \\ \dashrightarrow \: \: \rm \: Area_{(second \: \triangle)} = \dfrac{1}{ \not2} \times \not14 \times 7 \\ \\ \\ \dashrightarrow \: \: \rm \: Area_{(second \: \triangle)} = 7 \times 7 \\ \\ \\ \dashrightarrow \: \: \rm \: \underline{ \boxed{ \rm{Area_{(second \: \triangle)} = 49\: cm^2}}} \green{ \bigstar}$

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$\bigstar \: {\underline{ \sf{ \pmb{According \: to \: the \: Question : }}}}$

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$\dashrightarrow \: \: \sf \dfrac{Area \: of \: first \: \triangle}{Area \: of \: second \: \triangle}$

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$\dashrightarrow \: \: \sf \cancel \dfrac{42}{49}$

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$\dashrightarrow \: \: \underline{ \boxed{\bf \dfrac{6}{7}}} \: \:$

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$\underline{ \sf{ \therefore \: The \: ratio \: of \: areas \: of \: these \: triangles \: = \bf{6 : 7}}}$