Question

Match the two columns.
$\begin{array}{cc}Column-I& Column-II\\ \text{(A) The area of a triangle with base 4 cm and height 6 cm is:}& \left(p\right)\frac{5}{4}\sqrt{11}c{m}^2\\ \text{(B) If the perimeter of the isosceles triangle is 11cm and the}& \left(q\right)16\sqrt{3}c{m}^2\\ \text{base is 5 cm, then the area of the isosceles triangle is:}& \\ \text{(C) The area of the equilateral triangle, whose each side is}& \left(r\right)12c{m}^2\\ 8cmis:& \\ \text{(D) Area of an equilateral triangle with side a units:}& \left(s\right)\frac{\sqrt{3}}{4}{a}^{2}sq.unit\end{array}$ [4 Mark]

Answer 1

(A) Area of triangle $=\frac{1}{2}\times base\times height$
$\frac{1}{2}\times 4\times 6=12c{m}^2$ (1 mark)

(B) Perimeter of isoscele triangle = 11
$\Rightarrow x+x+5=11\Rightarrow 2x=6\Rightarrow x=3$
Area of isoscele triangle $=\frac{5}{4}\sqrt{36-25}c{m}^2$
$=\frac{5}{4}\sqrt{11}c{m}^2$
$\left(A\right)\to r;\left(B\right)\to p$ (1 mark)

(C) Area of equilateral triangle
$=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\left(8\right)\left(8\right)$
$=16\sqrt{3}c{m}^2$ (1 mark)

(D) Area of equilateral triangle with side a units is $\frac{\sqrt{3}}{4}{a}^2$ (1 mark)

$\left(A\right)\to r;\left(B\right)\to p;\left(C\right)\to q;\left(D\right)\to s$