Question

find the altitude of a triangle whose base is 40cm and area is 240cm????​

Answer 1

Given :-

• A triangle whose base is 40cm and area is 240cm²

To find :-

• Altitude of triangles

Solution :-

• Area of triangle = 240 cm²

• Base of triangle = 40 cm

As we know that

→ Area of triangle = ½ × b × h

Where " b " is based and " h " is height or altitude of triangle.

• According to question

→ Area of triangle = 240

→ ½ × b × h = 240

→ ½ × 40 × h = 240

→ 20 × h = 240

→ h = 240/20

→ h = 12 cm

Hence,

• Altitude of triangle is 12cm

Verification :-

→ Area of triangle = 240 cm²

→ ½ × b × h = 240

→ ½ × 40 × 12 = 240

→ 20 × 12 = 240

→ 240 = 240

• Hence verified

Answer 2

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\sf Find\: the\: altitude\: of\: a\: triangle\: whose\\\sf base\: is\: 40\:cm\: and\: area\: is\: 240\:cm????$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf Area\:of \:triangle(A) =240\:{cm}^{2}$
• $\sf \bf Base\:of \:triangle (b) =40\:cm$

$\sf \bf {\boxed {\mathbb {TO\:FIND}}}$

• $\sf \bf Altitude \:of\:the\:triangle(h)$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

${\color {gold} \underline {\tt Let \:the \:altitude \:of \:the \:triangle \:be\:x}}$

${\boxed {\boxed {\color {blue} {\sf \bf A=\dfrac{1}{2}\times b\times h}}}}$

• $\sf A=area\:of\:triangle$
• $\sf b=base\:of\:triangle$
• $\sf h=altitude \:of \:triangle$

${\underbrace {\overbrace {\color {orange} {\bf Substitute \:the \:values}}}}$

$\sf \bf \implies 240=\dfrac{1}{2}\times 40 \times x$

$\sf \bf \implies 240 \times 2=40x$

$\sf \bf \implies 480 =40x$

$\sf \bf \implies x=\dfrac{480}{40}$

$\sf \bf \implies x=\dfrac{48\cancel {0}}{4\cancel {0}}$

$\sf \bf \implies x=\dfrac{48}{4}$

$\implies {\boxed {\color {green} {\sf \bf x=12\:cm}}}$

${\color {purple} \underline {\sf \therefore The\:altitude \:of \:the \:triangle \:is \:12\:cm}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

__________________________________________

$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\bf Area\:of \:triangle =\dfrac{1}{2}\times b \times h$

$\bf Area\:of \:rectangle =l\times b$

$\bf Area\:of \:square ={a}^{2}$

$\bf Area\:of \:circle =\pi {r}^{2}$