Question

Fill in the blanks.

(i) The sum of the angles of a triangle is ...... .
(ii) The sum of any two sides of a triangle is always ...... than the third side.
(iii) In ∆ABC, if ∠A = 90°, then BC² = (......) + (......).
(iv) In ∆ABC, AB = AC and AD ⊥ BC, then BD = ...... .
(v) In the given figure, side BC of ∆ABC is produced to D and CE || BA. If ∠BAC = 50°
then ∠ACE = ...... .

Answer 1

(i) The sum of the angles of a triangle is 180°.
(ii) The sum of any two sides of a triangle is always greater than the third side.
(iii) In ∆ABC, if ∠A = 90°, then:
BC² = (AB²) + (BC²)

$$\begin{gathered} \begin{array}{l}\text{In}△\text{ABC, if}\angle \text{A}=90°,\mathrm{then}\mathrm{by}\mathrm{phythagoras}\mathrm{theorem}:\\ \end{array} \\ {\mathrm{BC}}^2={\text{AB}}^2+{\text{AC}}^2 \end{gathered}$$

(iv) In ∆ABC:
AB = AC
AD ⊥ BC
Then, BD = DC

$$\begin{gathered} \begin{array}{l}\text{BD}=\text{DC}\\ \end{array} \\ \text{This is because}\mathrm{in\; an\; isosceles\; triangle},\mathrm{the}\mathrm{perpendicular\; dropped\; from}\mathrm{the\; vertex\; joining\; the\; equal\; sides},bisects\; the\; base. \end{gathered}$$

(v) In the given figure, side BC of ∆ABC is produced to D and CE || BA.
If ∠BAC = 50°, then ∠ACE = 50°.

$\begin{array}{l}\text{AB II CE}\\ \text{∴}\angle \text{BAC}=\angle \text{ACE}=50{°}^{}(\text{alternate angles)}\end{array}$