Question

Match the following columns.

Column I	Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm.	(p) 6
(b) The area of an equilateral triangle is $4\sqrt{3}$ cm2. The perimeter of the triangle is ...... cm.	(q) 4
(c) If the height of an equilateral triangle is $3\sqrt{3}$ cm, then each side of the triangle measures ...... cm.	(r) 24
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm.	(s) 12

(a) ......,
(b) ......,
(c) ......,
(d) ......,

Answer 1

(a)
$$\begin{gathered} \mathrm{Let}: \\ a=26\mathrm{cm},b=28\mathrm{cm}\mathrm{and}c=30\mathrm{cm} \\ s=\frac{a+b+c}{2}=\frac{26+28+30}{2}=42\mathrm{cm} \\ \mathrm{By}\mathrm{Heron}\text{'}\mathrm{s}\mathrm{formula},\mathrm{we}\mathrm{have}: \\ \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC=\sqrt{s(s-a)(s-b)(s-c)} \\ =\sqrt{42(42-26)(42-28)(42-30)} \\ =\sqrt{42\times 12\times 16\times 14} \\ =\sqrt{14\times 3\times 2\times 3\times 2\times 4\times 4\times 14} \\ =14\times 4\times 2\times 3 \\ =336{\mathrm{cm}}^2 \end{gathered}$$

Hence, the height of the triangle corresponding to 28 cm is given by:
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=336{\mathrm{cm}}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=336 \\ \Rightarrow \mathrm{Height}=\frac{336\times 2}{28}=24\mathrm{cm} \end{gathered}$$

(b)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2 \\ \Rightarrow \frac{\sqrt{3}}{4}\times (\mathrm{Side}{)}^2=4\sqrt{3} \end{gathered}$$
$$\begin{gathered} \Rightarrow {\left(\mathrm{Side}\right)}^2=16 \\ \Rightarrow \mathrm{Side}=4\mathrm{cm} \end{gathered}$$

Perimeter = 3 × Side = 3 × 4 = 12 cm

(c)
$$\begin{gathered} \mathrm{Height}=\frac{\sqrt{3}}{2}\times \mathrm{Side} \\ \Rightarrow \mathrm{Side}=\frac{2\times \mathrm{Height}}{\sqrt{3}}=\frac{2\times 3\sqrt{3}}{\sqrt{3}}=6\mathrm{cm} \end{gathered}$$

(d)
$$\begin{gathered} \mathrm{Height}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{1}{2}\sqrt{4a^2-b^2} \\ =\frac{1}{2}\sqrt{4{\left(5\right)}^2-6^2}\left(a=5\mathrm{cm}\mathrm{and}b=6\mathrm{cm}\right) \\ =\frac{1}{2}\times \sqrt{100-36} \\ =\frac{1}{2}\times \sqrt{64} \\ =\frac{1}{2}\times 8 \\ =4\mathrm{cm} \end{gathered}$$

Column I	Column II
(a) The lengths of three sides of a triangle are 26 cm, 28 cm and 30 cm. The height corresponding to base 28 cm is ....... cm.	(r) 24
(b) The area of an equilateral triangle is $4\sqrt{3}$ cm2. The perimeter of the triangle is ...... cm.	(s) 12
(c) If the height of an equilateral triangle is $3\sqrt{3}$ cm, then each side of the triangle measures ...... cm.	(p) 6
(d) Let the base of an isosceles triangle be 6 cm and each of the equal sides be 5 cm. Then, its height is ...... cm.	(q) 4