Question

the area of a triangle is 9 cm sq . find the base if the altitude exceeds the base by 7cm​

Answer 1

Answer:

The answer is base $=2$ cm

Step-by-step explanation:

Let the base of the triangle be $x$ cm

So the altitude $=x+7$, given in the question

Also the area $9$ sq. cm

$\Rightarrow \frac{1}{2}\times \text{base }\times \text{altitude }=9\\\Rightarrow \frac{1}{2}x(x+7)=9\\\Rightarrow x^2+7x=18\\\Rightarrow x^2+7x-18=0\\\Rightarrow (x-2)(x+9)=0\\\Rightarrow x=2,-9$

But we know x cannot be negative, thus the base  of the triangle is 2 cm

Answer 2

Answer:

Question :

The area of a triangle is 9 cm². Find the base if the altitude exceeds the base by 7cm.

$\begin{gathered}\end{gathered}$

Solution :

Here we have given that area of triangle is 9 cm² and the altitude exceeds base by 7cm. We need to find the base and altitude. So we'll use area of triangle formula :

$\bull \: {\sf{\underline{\underline{\red{Formula\: : - }}}}}$

${\longrightarrow{\small{\underline{\boxed{\sf{Area \: of \: triangle = \dfrac{1}{2} \times base \times altitude}}}}}}$

$\bull \: {\sf{\underline{\underline{\red{Assuming\: : - }}}}}$

Let the base of triangle be x cm. So, the altitude will be x+7.

$\bull \: {\sf{\underline{\underline{\red{According \: to \: the \: question\: : - }}}}}$

${\rightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times base \times altitude}}}$

${\rightarrow{\sf{9 = \dfrac{1}{2} \times x \times \bigg(x + 7 \bigg)}}}$

${\rightarrow{\sf{9 \times 2= {x}^{2} + 7x}}}$

${\rightarrow{\sf{18= {x}^{2} + 7x}}}$

${\rightarrow{\sf{{x}^{2} + 7x - 18 = 0}}}$

${\rightarrow{\sf{\bigg(x - 2 \bigg)\bigg(x + 9 \bigg) = 0}}}$

${\rightarrow{\sf{\underline{\underline{x = 2, - 9}}}}}$

Since, the altitude can't be negative.So, the value of x is 2.

$\bull \: {\sf{\underline{\underline{\red{Hence\: : - }}}}}$

• Altitude = 2 cm
• Base = 2+7 = 9 cm

Hence, the base and altitude of triangle is 2 cm and 9 cm.

$\begin{gathered}\end{gathered}$

Vefication :

Let's check our answer by substituting all values in the formula :

${\rightarrow{\sf{Area \: of \: triangle = \dfrac{1}{2} \times base \times altitude}}}$

${\rightarrow{\sf{9 \: {cm}^{2} = \dfrac{1}{2} \times 9\times 2}}}$

${\rightarrow{\sf{9 \: {cm}^{2} = \dfrac{1}{\cancel{2}}\times 9\times \cancel{2}}}}$

${\rightarrow{\sf{9 \: {cm}^{2} = 1 \times 9}}}$

${\rightarrow{\sf{9 \: {cm}^{2} = 9 \: {cm}^{2}}}}$

${\rightarrow{\sf{\underline{\underline{LHS = RHS}}}}}$

Hence Verified!

$\begin{gathered}\end{gathered}$

Learn More :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin{array}{l}\\ \large\dag\quad\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star \: \: \sf Circle = \pi r^2 \\ \\ \star \: \; \sf Square=(side)^2\\ \\ \star\; \; \sf Rectangle=Length\times Breadth \\\\ \star \: \: \sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \: \: \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \: \: \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star \: \: \sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star \: \: \sf Parallelogram =Breadth\times Height\\\\ \star \: \: \sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star \: \: \sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\rule{220pt}{3.5pt}$