Question

Find the area of the triangle in square centimeters, whose base and altitude are as under :
(i) 1.5 m and 0.8 m
(ii) 32 cm and 105 mm​

Answer 1

Question :-

Find the area of the triangle in square centimeters, whose base and altitude are as under :

(i) 1.5 m and 0.8 m

(ii) 32 cm and 105 mm

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Solution (i) :-

➲ Given Information :-

• Base of the triangle ➢ 1.5 m
• :⇒ Base of the triangle ( in cm. ) ➢ 150 cm
• Altitude of the triangle ➢ 0.8 m
• :⇒ Altitude of the triangle ( in cm. ) ➢ 80 cm

➲ To Find :-

• Area of the triangle ( in cm. )

➲ Concept :-

• Area And Perimeter of Plane Figures

➲ Formula Used :-

• $\boxed{ \boxed{ \bf{ \red{area \: of \: triangle} \: = \dfrac{ \blue{base }\times \green{ altitude}}{ \pink2} }}}$

➲ Explanation :-

• Simply substitute the given values in the formula mentioned above. After some minor calculations, the resultant will be our answer i.e., area of the triangle. Now, let's proceed towards our calculation.

➲ Calculation :-

Formula Used is...

$\rm{:\implies \boxed{ \boxed{ \bf{ \red{area \: of \: triangle} \: = \dfrac{ \blue{base }\times \green{ altitude}}{ \pink2} }}}}$

Substituting the values given in the formula mentioned above, We get,

$\rm{:\implies \red{area \: of \: triangle} \: = \dfrac{ \blue{150\: cm }\times \green{ 80 \: cm}}{ \pink2} }$

Cancelling & Calculating further, We get,

$\rm{ :\implies\red{area \: of \: triangle} \: = \dfrac{ \blue{12,000\: cm²}}{ \pink2} }= \rm{\purple{6,000\: cm²}}$

∴ Area of triangle is 6,000 cm²

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Solution (ii) :-

➲ Given Information :-

• Base of the triangle ➢ 32 cm
• Altitude of the triangle ➢ 105 mm
• :⇒ Altitude of the triangle ( in cm. ) ➢ 10.5 cm

➲ To Find :-

• Area of the triangle ( in cm. )

➲ Concept :-

• Area And Perimeter of Plane Figures

➲ Formula Used :-

• $\boxed{ \boxed{ \bf{ \red{area \: of \: triangle} \: = \dfrac{ \blue{base }\times \green{ altitude}}{ \pink2} }}}$

➲ Explanation :-

• Simply substitute the given values in the formula mentioned above. After some minor calculations, the resultant will be our answer i.e., area of the triangle. Now, let's proceed towards our calculation.

➲ Calculation :-

Formula Used is...

$\rm{:\implies \boxed{ \boxed{ \bf{ \red{area \: of \: triangle} \: = \dfrac{ \blue{base }\times \green{ altitude}}{ \pink2} }}}}$

Substituting the values given in the formula mentioned above, We get,

$\rm{:\implies \red{area \: of \: triangle} \: = \dfrac{ \blue{32\: cm }\times \green{ 10.5 \: cm}}{ \pink2} }$

Cancelling & Calculating further, We get,

$\rm{ :\implies\red{area \: of \: triangle} \: = \dfrac{ \blue{336\: cm²}}{ \pink2} }= \rm{\purple{168\: cm²}}$

∴ Area of triangle is 168 cm²

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Final Answers :-

1. Area of the triangle is 6,000 cm²
2. Area of the triangle is 168 cm²

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Note :-

Please scroll from right to left to see the whole solution.

Answer 2

Question :-

Find the area of the triangle in square centimeters, whose base and altitude are as under :

(i) 1.5 m and 0.8 m

Given :-

• Base = 1.5 m = 1.5 x 100 = 150 cm
• Altitude = 0.8 m = 0.8 x 100 = 80 cm

To Find :-

• Area of the triangle

$\sf ✯ \: \red{Formula \: used : - } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies \pink {area = \frac{altitude \times base}{2} }$

Putting the values, we get :-

$\leadsto \sf area = \frac{150 \times 80}{2} \\ \leadsto \sf \frac{150 \times \cancel{ 80}}{ \cancel2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf\leadsto \: 150 \times 80 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \leadsto \sf12000 \: {cm}^{2} \: \: \: \: \: \: \: \: \: \: \: \:$

$\rm\therefore \: \orange{the \: area \: of \: triangle \: = 12000 \: {cm}^{2} }$

(ii) 32 cm and 105 mm

Given :-

• Base = 32 cm
• Altitude = 105 = 105 ÷ 10 = 10.5 cm

To Find :-

• Area of the triangle

$\sf ✯ \: \red{Formula \: used : - } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf : \implies \pink {area = \frac{altitude \times base}{2} }$

Putting the values, we get :-

$\sf \leadsto {area = \frac{32 \times 10.5}{2} } \\ \sf \leadsto \: \frac{32 \times 105}{2 \times 10} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto \: \frac{ \cancel{32 }\times \cancel{105}}{ \cancel2 \times \cancel{10}} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto\frac{16 \times 21}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto \: \frac{ \cancel{16} \times 21}{ \cancel2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto \: 8 \times 21 = 168 {cm}^{2}$

$\rm\therefore \orange{area \: of \: the \: triangle \: = {168 \: cm}^{2} }$

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